High School Biology — Molecular Genetics — Solutions

Companion to the Practice Set · Mark-by-mark walkthroughs · AP-Feeder / ON / BC / AB styles练习题配套答案 · 逐分讲解 · AP 衔接 / 安 / 卑 / 阿省考风格

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 DNA structureDNA 结构 · HS-LS1-1 [3 marks][3 分]

A double-stranded DNA molecule contains 30% adenine (A) bases. What percentage of the bases are cytosine (C)?一个双链 DNA 分子含 30% 腺嘌呤（A）碱基。胞嘧啶（C）碱基占多少百分比？

Answer:答案： (B) 20%

(a) Apply Chargaff's rules应用查哥夫定律 A1+A1+A1

By Chargaff's rules, A pairs with T, so %T = %A = 30%. The remaining bases must be G and C, which also pair with each other: %G + %C = 100% - 30% - 30% = 40%. Since %G = %C, each is 20%.根据查哥夫定律，A 与 T 配对，故 %T = %A = 30%。剩余碱基为 G 和 C，相互配对：%G + %C = 100% - 30% - 30% = 40%。由于 %G = %C，故各为 20%。

Why the distractors fail.干扰项分析。

(A) 30%: confuses C with T, incorrectly applying A = T = C.混淆了 C 与 T，错误地认为 A = T = C。

(D) 40%: gives the combined G+C percentage, not the C percentage alone.给出的是 G+C 的合计百分比，而非单独 C 的百分比。

Chargaff's rules: A = T and G = C in any double-stranded DNA.查哥夫定律：在任意双链 DNA 中，A = T 且 G = C。 This follows directly from Watson-Crick base pairing: adenine always hydrogen-bonds to thymine (two H-bonds), and guanine always bonds to cytosine (three H-bonds). Because every base on one strand is paired with a complementary base on the other, the molar ratios must be equal. The four percentages always sum to 100%, so knowing one purine immediately fixes both purines and both pyrimidines. A useful shortcut: if you know %A, then %T = %A, and %G = %C = (100 - 2 %A) / 2.这直接源于沃森-克里克碱基配对：腺嘌呤始终与胸腺嘧啶氢键结合（两个氢键），鸟嘌呤始终与胞嘧啶氢键结合（三个氢键）。由于一条链上的每个碱基都与另一条链上的互补碱基配对，摩尔比必然相等。四种碱基百分比之和始终为 100%，因此已知一种嘌呤的百分比即可推出其余三种。实用简算：若已知 %A，则 %T = %A，%G = %C = (100 - 2 %A) / 2。

Q2EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §3 Transcription转录 · HS-LS1-1 [3 marks][3 分]

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？

Answer:答案： (A) 5'-AUG CCU GGA-3'

(a) Read template 3' to 5' and write complementary RNA 5' to 3'从模板链 3' 至 5' 方向读取，写出互补 RNA（5' 至 3'） A1+A1+A1

RNA polymerase reads the DNA template in the 3' to 5' direction and builds mRNA in the 5' to 3' direction. The RNA complement uses U instead of T. Working codon by codon:RNA 聚合酶沿 3' 至 5' 方向读取 DNA 模板，并沿 5' 至 3' 方向合成 mRNA。RNA 互补使用 U 替代 T。逐密码子推导：

Template TAC (3'→5') → mRNA AUG (5'→3')模板 TAC（3'→5'）→ mRNA AUG（5'→3'）
Template GGA (3'→5') → mRNA CCU (5'→3')模板 GGA（3'→5'）→ mRNA CCU（5'→3'）
Template CCT (3'→5') → mRNA GGA (5'→3')模板 CCT（3'→5'）→ mRNA GGA（5'→3'）

Result: 5'-AUG CCU GGA-3'.结果：5'-AUG CCU GGA-3'。

Why the distractors fail.干扰项分析。

(B) 5'-TAC GGA CCT-3': copies the template directly instead of writing the complementary sequence; also uses T instead of U.直接抄写了模板链而非互补序列，且使用了 T 而非 U。

(C) 5'-ATG CCT GGA-3': correct complementary sequence but uses DNA bases (T not U), so this is the non-template (coding) strand, not the mRNA.互补序列正确，但使用了 DNA 碱基（T 而非 U），这是非模板链（编码链），而非 mRNA。

(D) 5'-AUG GGA CCU-3': swaps the second and third codons; arises from reading the template in the wrong direction for the last two codons.第二和第三密码子顺序互换；来自对后两段模板读取方向错误。

The mRNA sequence mirrors the non-template (coding) strand but with U replacing T.mRNA 序列与非模板（编码）链相同，但用 U 替换 T。 A reliable two-step method: (1) identify the non-template strand (same sequence as the template but complementary on the other strand); (2) replace every T with U. This shortcut works because the mRNA is complementary to the template strand, which makes it identical to the non-template strand apart from the sugar-phosphate backbone and the U/T substitution. Always state the 5' to 3' direction of the mRNA product, as direction is part of a complete answer.可靠的两步法：(1) 确认非模板链（与模板链互补）；(2) 将每个 T 替换为 U。此捷径之所以有效，是因为 mRNA 与模板链互补，从而与非模板链相同（除磷酸糖骨架和 U/T 替换外）。务必标注 mRNA 产物的 5' 至 3' 方向，方向是完整答案的一部分。

Q3MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §1 DNA structure / ChargaffDNA 结构 / 查哥夫定律 · C3.2k [6 marks][6 分]

A researcher analyses a double-stranded DNA sample and finds that 18% of the bases are guanine (G). State Chargaff's rules and determine percentages of A, T, C; then state one structural feature the rules help explain.研究人员分析双链 DNA 样品，发现 18% 的碱基是鸟嘌呤（G）。陈述查哥夫定律并推导 A、T、C 各自的百分比；再陈述该定律有助于解释的一个 DNA 双螺旋结构特征。

Answer:答案： (a) G = C = 18%, A = T = 32%G = C = 18%，A = T = 32% · (b) complementary antiparallel strand structure互补反向平行双链结构

(a) State Chargaff's rules and calculate each base percentage陈述查哥夫定律并计算各碱基百分比 A1+A1+A1+A1

Chargaff's rules state that in any double-stranded DNA: (1) %A = %T, and (2) %G = %C. Given %G = 18%, therefore %C = 18%. The remaining percentage is 100% - 18% - 18% = 64%, split equally between A and T: %A = %T = 32%.查哥夫定律指出，在任意双链 DNA 中：(1) %A = %T，(2) %G = %C。已知 %G = 18%，故 %C = 18%。剩余百分比 = 100% - 18% - 18% = 64%，A 和 T 各占一半：%A = %T = 32%。

%C = 18% A1
%A = %T = 32% A1+A1
Rules stated (A pairs with T; G pairs with C)定律陈述（A 与 T 配对；G 与 C 配对） A1

(b) Structural feature explained by Chargaff's rules查哥夫定律解释的结构特征 A1+A1

Chargaff's rules explain the antiparallel complementary strand structure of the double helix: the two strands run in opposite directions (one 5' to 3', the other 3' to 5') and each base on one strand is always paired with a specific complementary base on the other strand via hydrogen bonds (A-T, G-C). This strict pairing is what makes the two strands complementary.查哥夫定律解释了双螺旋的反向平行互补双链结构：两条链方向相反（一条 5' 至 3'，另一条 3' 至 5'），且一条链上的每个碱基通过氢键与另一条链上的特定互补碱基配对（A-T，G-C）。正是这种严格配对使两条链互补。

Chargaff's ratios were the numerical clue that solved the DNA structure puzzle.查哥夫比例是破解 DNA 结构谜题的数字线索。 Erwin Chargaff measured base compositions of DNA from many species and found the A:T and G:C ratios were always 1:1, but the A:G ratio varied between species. Watson and Crick used this as direct evidence for specific base pairing. On an Alberta diploma exam, part (b) expects you to name the feature precisely: "complementary antiparallel double-stranded structure" earns full credit; vague answers like "it is a double helix" do not.查哥夫测量了多个物种 DNA 的碱基组成，发现 A:T 和 G:C 之比始终为 1:1，但 A:G 之比因物种而异。沃森和克里克将此作为特异碱基配对的直接证据。在阿尔伯塔毕业考中，(b) 小问要求精确命名该特征："互补反向平行双链结构"可得满分；"是双螺旋"这类模糊答案不能得分。

Q4MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §2 DNA replicationDNA 复制 · A&P 12 [10 marks][10 分]

The Meselson-Stahl experiment. Bacteria grown in $^{15}$N then switched to $^{14}$N. (a) Density after one generation. (b) Density pattern after two generations. (c) Model supported and ruling out the conservative model.梅塞尔森-斯塔尔实验。细菌在 $^{15}$N 中培养后转移至 $^{14}$N。(a) 一代后 DNA 密度。(b) 两代后密度模式。(c) 支持的复制模型及排除保留性模型的理由。

Answer:答案： (a) all hybrid density全为杂合密度 · (b) half hybrid, half light一半杂合，一半轻链 · (c) semiconservative replication半保留复制

(a) After one generation in $^{14}$N在 $^{14}$N 培养基中繁殖一代后 A1+A1+A1

Each original double-stranded $^{15}$N/$^{15}$N DNA molecule unwinds. Each old $^{15}$N strand serves as a template for a new $^{14}$N strand. The result is two hybrid ($^{15}$N/$^{14}$N) molecules. All DNA molecules have intermediate (hybrid) density, forming a single band at a position between heavy and light in a cesium chloride gradient. This is the expected result of semiconservative replication: each daughter molecule retains one parental strand.每条原始双链 $^{15}$N/$^{15}$N DNA 解开，每条旧 $^{15}$N 链作为模板合成新的 $^{14}$N 链。结果产生两条杂合（$^{15}$N/$^{14}$N）分子。所有 DNA 分子均为中等（杂合）密度，在氯化铯梯度中在重链和轻链之间形成单一条带。这是半保留复制的预期结果：每条子分子保留一条亲代链。

(b) After two generations in $^{14}$N在 $^{14}$N 培养基中繁殖两代后 A1+A1+A1

Each hybrid molecule again unwinds. The $^{15}$N strand templates a new $^{14}$N strand (producing one hybrid molecule), and the $^{14}$N strand from the first generation templates another new $^{14}$N strand (producing one light molecule). Out of four total molecules: two are hybrid ($^{15}$N/$^{14}$N) and two are light ($^{14}$N/$^{14}$N). The cesium chloride gradient shows two bands: one at intermediate density and one at light density, in equal amounts (1:1 ratio).每条杂合分子再次解开。$^{15}$N 链作为模板合成新的 $^{14}$N 链（产生一条杂合分子），第一代的 $^{14}$N 链作为模板合成另一条新的 $^{14}$N 链（产生一条轻链分子）。四条分子中：两条为杂合（$^{15}$N/$^{14}$N），两条为轻链（$^{14}$N/$^{14}$N）。氯化铯梯度中出现两条条带：中等密度和轻密度各一条，比例相等（1:1）。

(c) Model supported and ruling out conservative replication支持的模型及排除保留性复制的理由 A1+A1+A1+A1

The results support the semiconservative model, in which each daughter molecule inherits one original (parental) strand and one newly synthesised strand. The conservative model predicts that after one generation, the original heavy ($^{15}$N/$^{15}$N) double helix is preserved intact, and a new light ($^{14}$N/$^{14}$N) double helix forms. This would produce two bands (heavy and light) after one generation. Meselson and Stahl observed only a single intermediate band after one generation, directly ruling out the conservative model.结果支持半保留复制模型：每条子分子保留一条亲代链（旧链）和一条新合成链。保留性模型预测一代后原始重链（$^{15}$N/$^{15}$N）双螺旋完整保留，并形成一条新的轻链（$^{14}$N/$^{14}$N）双螺旋，即出现两条条带（重和轻）。梅塞尔森和斯塔尔在一代后只观察到一条中等密度的条带，直接排除了保留性模型。

The Meselson-Stahl experiment is the gold standard proof of semiconservative replication.梅塞尔森-斯塔尔实验是半保留复制的金标准证明。 The three proposed models in 1958 were: semiconservative (each daughter gets one old + one new strand), conservative (parent is restored, daughter is all new), and dispersive (old and new DNA intermixed throughout both strands). The single intermediate band at generation 1 ruled out conservative; the persistence of a hybrid band at generation 2 (rather than dispersive strands of uniform intermediate density in all molecules) helped rule out dispersive replication. Examiners award marks for naming the model correctly, giving a mechanistic reason (one old, one new strand per molecule), and citing the specific evidence that refutes the alternative.1958 年提出的三种模型为：半保留（每条子分子获得一条旧链和一条新链）、保留性（亲代链复原，子代链全新）和弥散性（新旧 DNA 混合分布于两条链中）。第一代只出现一条中等密度条带排除了保留性；第二代仍出现杂合条带（而非所有分子均为中等密度的弥散性条带）有助于排除弥散性复制。考官按以下要点给分：正确命名模型、给出机制性原因（每分子一条旧链一条新链），以及引用排除另一模型的具体证据。

Section B · Worked SolutionsB 部分 · 详细解答

Q5EASY易 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §3 Transcription转录 · HS-LS1-1 [6 marks][6 分]

A gene on the non-template strand reads 5'-ATG GCC TAA-3'. (a) DNA template strand. (b) mRNA produced. (c) Location of transcription and enzyme.DNA 非模板链上的基因读序为 5'-ATG GCC TAA-3'。(a) DNA 模板链。(b) 转录产生的 mRNA。(c) 转录位置与酶。

Answer:答案： (a) 3'-TAC CGG ATT-5' · (b) 5'-AUG GCC UAA-3' · (c) nucleus; RNA polymerase细胞核；RNA 聚合酶

(a) Write the DNA template strand (3' to 5')写出 DNA 模板链（3' 至 5'） A1+A1

The template strand is antiparallel and complementary to the non-template strand. Write the complement of each base and reverse the direction:模板链与非模板链反向互补。写出每个碱基的互补碱基并反转方向：

Non-template: 5'-ATG GCC TAA-3'非模板链：5'-ATG GCC TAA-3'
Template: 3'-TAC CGG ATT-5'模板链： 3'-TAC CGG ATT-5'

(Award A1 for correct sequence, A1 for correct 3'-5' direction labelling.)（序列正确得 A1，正确标注 3'-5' 方向得 A1。）

(b) Write the mRNA sequence (5' to 3')写出 mRNA 序列（5' 至 3'） A1+A1

RNA polymerase reads the template 3' to 5' and synthesises RNA 5' to 3', replacing T with U:RNA 聚合酶从 3' 至 5' 读取模板，从 5' 至 3' 合成 RNA，以 U 替换 T：

Template: 3'-TAC CGG ATT-5' (read 3'→5')模板：3'-TAC CGG ATT-5'（从 3' 至 5' 读取）
mRNA: 5'-AUG GCC UAA-3'mRNA：5'-AUG GCC UAA-3'

Note: AUG is the start codon (Met) and UAA is a stop codon. (A1 for correct sequence with U substitution; A1 for 5'→3' direction.)注：AUG 为起始密码子（Met），UAA 为终止密码子。（序列正确含 U 替换得 A1；标注 5'→3' 方向得 A1。）

(c) Location and enzyme位置与酶 A1+A1

In eukaryotic cells, transcription takes place in the nucleus. The enzyme that carries out transcription is RNA polymerase (specifically RNA polymerase II for protein-coding genes).在真核细胞中，转录发生在细胞核中。执行转录的酶是 RNA 聚合酶（蛋白质编码基因由 RNA 聚合酶 II 负责）。

The non-template strand is a shortcut to the mRNA sequence.非模板链是推导 mRNA 序列的捷径。 Because the mRNA is complementary to the template strand, its sequence is identical to the non-template (coding) strand except that every T becomes U. Here the coding strand already reads 5'-ATG GCC TAA-3', so the mRNA is simply 5'-AUG GCC UAA-3'. This shortcut is only valid when you are given the non-template strand; if given only the template strand, you must complement and substitute U for T. Mentioning RNA polymerase specifically (not just "polymerase") and the nucleus (not the cytoplasm, where translation occurs) are common mark-losing mistakes.由于 mRNA 与模板链互补，其序列与非模板（编码）链相同，只需将 T 替换为 U。本题编码链已为 5'-ATG GCC TAA-3'，故 mRNA 直接为 5'-AUG GCC UAA-3'。此捷径仅在给出非模板链时有效；若只给出模板链，必须先互补再将 T 替换为 U。常见失分点：酶应写"RNA 聚合酶"而非仅写"聚合酶"；位置应写"细胞核"而非翻译所在地"细胞质"。

Q6MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §4 Translation翻译 · SBI4U D3.3 [8 marks][8 分]

mRNA: 5'-AUG-UGC-GAA-CCG-UAG-3'. UGC = Cys, GAA = Glu, CCG = Pro. (a) Start and stop codons, roles. (b) Polypeptide sequence. (c) Anticodon for Glu; codon-anticodon pairing.mRNA：5'-AUG-UGC-GAA-CCG-UAG-3'。UGC = Cys，GAA = Glu，CCG = Pro。(a) 起始和终止密码子及作用。(b) 多肽链序列。(c) Glu 的反密码子；密码子-反密码子配对。

Answer:答案： (a) AUG (start), UAG (stop)AUG（起始），UAG（终止） · (b) Met-Cys-Glu-Pro · (c) 3'-CUU-5'

(a) Start and stop codons and their roles起始和终止密码子及其作用 A1+A1+A1

AUG is the start codon. It signals the ribosome to begin translation and codes for the amino acid methionine (Met), which is the first residue of every polypeptide. UAG is a stop codon (one of three: UAA, UAG, UGA). It does not code for an amino acid; instead, it signals the ribosome to terminate translation and release the completed polypeptide chain. (A1 for each codon identified with its role.)AUG 是起始密码子，指示核糖体开始翻译，同时编码氨基酸甲硫氨酸（Met），是每条多肽的第一个残基。UAG 是终止密码子（三种之一：UAA、UAG、UGA），不编码氨基酸，而是指示核糖体终止翻译并释放完整的多肽链。（每个密码子连同其作用各得 A1。）

(b) Polypeptide sequence (N-terminus to C-terminus)多肽链序列（从 N 端到 C 端） A1+A1

Reading the codons in order (UAG is stop, so not translated):按顺序读取密码子（UAG 为终止，不翻译）：

AUG = Met
UGC = Cys
GAA = Glu
CCG = Pro
UAG = stop (no amino acid added)

Polypeptide: Met-Cys-Glu-Pro (4 amino acids; N-terminus Met, C-terminus Pro). (A1 for correct sequence; A1 for Met first/stop not included.)多肽：Met-Cys-Glu-Pro（4 个氨基酸；N 端为 Met，C 端为 Pro）。（序列正确得 A1；Met 为首且不含终止密码子对应氨基酸得 A1。）

(c) Anticodon for Glu codon GAA; codon-anticodon pairingGlu 密码子 GAA 的反密码子；密码子-反密码子配对 A1+A1+A1

The codon is 5'-GAA-3'. The tRNA anticodon is antiparallel and complementary: 3'-CUU-5' (written 5' to 3': 5'-UUC-3'). During translation, the tRNA anticodon base-pairs with the mRNA codon at the ribosomal A site: G pairs with C, A pairs with U, A pairs with U. The anticodon is on the loop of the tRNA molecule and aligns antiparallel to the mRNA codon (anticodon 3'→5' against codon 5'→3'). (A1 for correct anticodon; A1 for antiparallel; A1 for G:C and A:U pairing explanation.)密码子为 5'-GAA-3'。tRNA 反密码子反向互补：3'-CUU-5'（从 5' 至 3' 写为：5'-UUC-3'）。翻译时，tRNA 反密码子在核糖体 A 位与 mRNA 密码子碱基配对：G 与 C 配对，A 与 U 配对，A 与 U 配对。反密码子位于 tRNA 分子的环上，与 mRNA 密码子反向平行排列（反密码子 3'→5' 对应密码子 5'→3'）。（反密码子正确得 A1；反向平行得 A1；G:C 和 A:U 配对解释得 A1。）

The anticodon is always written antiparallel to the codon, and uses U (not T) for all pairings.反密码子始终与密码子反向排列，且所有配对均使用 U（而非 T）。 A common error is to write the anticodon as 3'-CTT-5' (using DNA bases). tRNA is an RNA molecule, so its anticodon uses U throughout: 3'-CUU-5'. Another trap is to forget the antiparallel convention: if the codon runs 5' GAA 3', the anticodon runs 3' CUU 5', meaning the two sequences are read in opposite directions when they base-pair. The ribosome positions the A-site codon and A-site tRNA anticodon in exact antiparallel alignment, and hydrogen bonds form between complementary bases (G:C three bonds; A:U two bonds), which is what holds the tRNA-mRNA complex together long enough for peptide bond formation.常见错误是将反密码子写为 3'-CTT-5'（使用 DNA 碱基）。tRNA 是 RNA 分子，其反密码子全部使用 U：3'-CUU-5'。另一陷阱是忘记反向平行约定：若密码子为 5'-GAA-3'，则反密码子为 3'-CUU-5'，即两条序列在配对时方向相反。核糖体将 A 位密码子与 A 位 tRNA 反密码子严格反向平行对齐，互补碱基间形成氢键（G:C 三个；A:U 两个），使 tRNA-mRNA 复合体在肽键形成期间保持稳定。

Q7MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §5 Mutations突变 · C3.6k [8 marks][8 分]

Original mRNA: 5'-AUG-GCU-AAG-UAA-3' (Met-Ala-Lys-stop). (a) AAG to AAA substitution. (b) GCU to GAU substitution. (c) Cytosine insertion between codons 1 and 2.原始 mRNA：5'-AUG-GCU-AAG-UAA-3'（Met-Ala-Lys-终止）。(a) AAG 变为 AAA 的替换。(b) GCU 变为 GAU 的替换。(c) 第一和第二密码子之间插入一个胞嘧啶碱基。

Answer:答案： (a) silent mutation; no change沉默突变；蛋白质不变 · (b) missense mutation; Ala replaced by Asp错义突变；Ala 被 Asp 替换 · (c) frameshift insertion; likely non-functional protein移码插入突变；蛋白质可能失去功能

(a) AAG to AAA substitution (third codon)第三密码子 AAG 变为 AAA 的替换 A1+A1+A1

This is a point substitution (specifically a transition, G to A). Both AAG and AAA code for lysine (Lys). Therefore this is a silent (synonymous) mutation: the amino acid sequence of the protein is unchanged (still Met-Ala-Lys), and protein function is very likely unaffected. This illustrates the degeneracy (redundancy) of the genetic code.这是一个点替换突变（具体为转换，G 变为 A）。AAG 和 AAA 均编码赖氨酸（Lys）。因此这是一个沉默（同义）突变：蛋白质的氨基酸序列不变（仍为 Met-Ala-Lys），蛋白质功能很可能不受影响。这体现了遗传密码的简并性（冗余性）。

(b) GCU to GAU substitution (second codon)第二密码子 GCU 变为 GAU 的替换 A1+A1+A1

This is a point substitution (C to A, a transversion). GCU codes for alanine (Ala); GAU codes for aspartate (Asp). Because the amino acid changes, this is a missense mutation. The protein produced is Met-Asp-Lys instead of Met-Ala-Lys. Alanine is nonpolar; aspartate is negatively charged at physiological pH. This change in chemical properties may alter protein folding and could reduce or eliminate protein function depending on the location in the protein.这是一个点替换突变（C 变为 A，颠换）。GCU 编码丙氨酸（Ala）；GAU 编码天冬氨酸（Asp）。由于氨基酸改变，这是一个错义突变。产生的蛋白质为 Met-Asp-Lys，而非 Met-Ala-Lys。丙氨酸为非极性；天冬氨酸在生理 pH 下带负电荷。这种化学性质的变化可能改变蛋白质折叠，并可能降低或消除蛋白质功能，具体影响取决于突变在蛋白质中的位置。

(c) Cytosine insertion between codons 1 and 2第一和第二密码子之间插入一个胞嘧啶碱基 A1+A1

Inserting a single base shifts the reading frame of every codon downstream of the insertion point. This is a frameshift mutation (insertion type). The resulting mRNA reads: 5'-AUG-CGC-UAA-G...-3'. The new second codon CGC codes for Arg, and the next codon becomes UAA, a stop codon, terminating translation prematurely. The protein produced is Met-Arg (only two amino acids) instead of three, so the protein is almost certainly non-functional. Frameshift mutations are typically more severe than substitutions because they scramble all downstream codons.插入一个碱基使插入点下游所有密码子的阅读框发生移位。这是一个移码突变（插入型）。产生的 mRNA 读序变为：5'-AUG-CGC-UAA-G...-3'。新的第二密码子 CGC 编码 Arg，下一个密码子变为 UAA（终止密码子），翻译提前终止。产生的蛋白质为 Met-Arg（仅两个氨基酸），原蛋白质为三个氨基酸，几乎肯定失去功能。移码突变通常比替换突变更严重，因为其扰乱了所有下游密码子。

Three mutation categories by effect on protein: silent, missense, nonsense, and frameshift.按对蛋白质的影响分类：沉默突变、错义突变、无义突变和移码突变。 Silent (synonymous) mutations change a codon to a synonym coding for the same amino acid; the protein is unchanged. Missense mutations change one amino acid to a different one; effect ranges from neutral to lethal. Nonsense mutations change a sense codon to a stop codon, truncating the protein. Frameshift mutations (insertions or deletions not in multiples of three) shift the reading frame, usually producing a completely different and non-functional protein from the mutation point onwards. On the Alberta diploma exam, you are expected to name the mutation type, state its effect on the amino acid sequence, and justify the severity of the effect.沉默（同义）突变将密码子改变为编码同一氨基酸的同义密码子，蛋白质不变。错义突变将一种氨基酸替换为另一种，影响从中性到致命不等。无义突变将有义密码子改变为终止密码子，使蛋白质截短。移码突变（非三的倍数的插入或缺失）使阅读框移位，通常从突变点起产生完全不同且无功能的蛋白质。在阿尔伯塔毕业考中，需命名突变类型、陈述其对氨基酸序列的影响，并论证影响的严重程度。

Q8HARD难Honors荣誉级 🇨🇦 ON安 ON Provincial-style安大略省考风格 §6 Gene regulation (lac operon)基因调控（lac 操纵子） · SBI4U D3.3 [6 marks][6 分]

The lac operon in E. coli. (a) What happens to gene expression when lactose IS present; include the role of the inducer molecule. (b) Why the operon model illustrates differential gene expression; one advantage for the bacterium.大肠杆菌的 lac 操纵子。(a) 乳糖存在时基因表达发生的变化；包括诱导分子的作用。(b) 操纵子模型为何体现差异基因表达；对细菌的一个优点。

Answer:答案： (a) allolactose binds repressor, releases operator, transcription occurs异乳糖结合阻遏蛋白，解除操作子，转录开始 · (b) genes expressed only when substrate present; saves energy仅在底物存在时表达；节约能量

(a) Gene expression in the presence of lactose乳糖存在时的基因表达 A1+A1+A1

When lactose enters the cell, some of it is converted to allolactose, which acts as the inducer molecule. Allolactose binds to the lac repressor protein, changing its shape (allosteric change) so that the repressor can no longer bind to the operator region of the lac operon. With the operator unblocked, RNA polymerase can bind to the promoter and transcribe the structural genes (lacZ, lacY, lacA) into a single mRNA. This mRNA is translated to produce the enzymes needed for lactose metabolism (including beta-galactosidase). Gene expression is therefore switched ON in the presence of lactose. (A1 for allolactose as inducer; A1 for repressor shape change or release from operator; A1 for transcription of structural genes.)当乳糖进入细胞后，部分乳糖转化为异乳糖，后者作为诱导分子。异乳糖与 lac 阻遏蛋白结合，使阻遏蛋白构象改变（变构效应），导致阻遏蛋白无法再结合到 lac 操纵子的操作子区域。操作子解除封锁后，RNA 聚合酶得以结合启动子并转录结构基因（lacZ、lacY、lacA）为一条 mRNA。该 mRNA 翻译产生乳糖代谢所需的酶（包括 beta-半乳糖苷酶）。因此，乳糖存在时基因表达被开启。（异乳糖作为诱导分子得 A1；阻遏蛋白构象改变或脱离操作子得 A1；结构基因转录得 A1。）

(b) Differential gene expression and one advantage差异基因表达及一个优点 A1+A1+A1

The lac operon is an example of differential gene expression because the lactose metabolism genes are only expressed under certain conditions (when lactose is present) and are silenced under others (when lactose is absent). The same genome produces different gene products depending on the environment. One advantage for E. coli: producing the lactose-digesting enzymes only when lactose is available saves energy and metabolic resources. The bacterium does not waste cellular resources synthesising enzymes for a substrate that is not present.lac 操纵子是差异基因表达的例子，因为乳糖代谢基因仅在特定条件下（乳糖存在时）表达，在其他条件下（乳糖缺乏时）被沉默。同一基因组根据环境产生不同的基因产物。大肠杆菌的一个优点：仅在乳糖可用时才产生乳糖消化酶，节约了能量和代谢资源。细菌不会浪费细胞资源合成底物不存在时用不到的酶。

The lac operon is the canonical example of negative (inducible) gene regulation in prokaryotes.lac 操纵子是原核生物负调控（可诱导）基因调控的经典案例。 The default state is OFF (repressor bound to operator blocks transcription). Lactose flips the switch to ON by generating allolactose, which inactivates the repressor. This is called negative regulation because the repressor normally blocks (negatively controls) transcription, and the inducer removes the block. Contrast with positive regulation, where an activator protein is required to turn genes on. For full marks on an Ontario SBI4U question, you must name the inducer (allolactose, not lactose itself), describe the conformational change in the repressor, and use the term "differential gene expression" with a mechanism-based explanation rather than just restating the outcome.默认状态为关闭（阻遏蛋白结合操作子，封锁转录）。乳糖通过产生异乳糖使阻遏蛋白失活，从而将开关拨到开启位置。这称为负调控，因为阻遏蛋白通常阻断（负向控制）转录，而诱导分子解除封锁。与之对比的是正调控，其中需要激活蛋白将基因开启。在安大略 SBI4U 考题中，要得满分需命名诱导分子（异乳糖，非乳糖本身），描述阻遏蛋白的构象变化，并用基于机制的解释使用"差异基因表达"这一术语，而非仅重述结果。

Q9HARD难 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §7 Biotechnology (PCR + gel electrophoresis)生物技术（PCR + 凝胶电泳） · A&P 12 [4 marks][4 分]

PCR amplification and gel electrophoresis. (a) Three steps of one PCR cycle. (b) Which gel band represents the largest fragment and why.PCR 扩增与凝胶电泳。(a) 一个 PCR 循环的三个步骤。(b) 哪条凝胶条带代表最大片段及原因。

Answer:答案： (a) denaturation, annealing, extension变性、退火、延伸 · (b) band closest to wells = largest fragment最靠近上样孔的条带 = 最大片段

(a) Three steps of one PCR cycle一个 PCR 循环的三个步骤 A1+A1+A1

Denaturation变性 (~95 degrees C): The double-stranded DNA is heated to break the hydrogen bonds between complementary bases, separating the two strands into single-stranded templates.（约 95 摄氏度）：加热双链 DNA 以断开互补碱基间的氢键，将两条链分离为单链模板。
Annealing退火 (~50-65 degrees C): The temperature is lowered to allow short, synthetic DNA primers to bind (anneal) to their complementary sequences on each template strand, flanking the target region.（约 50-65 摄氏度）：降温使短的合成 DNA 引物与每条模板链上的互补序列结合（退火），位于目标区域两侧。
Extension延伸 (~72 degrees C): Taq polymerase extends the primers in the 5' to 3' direction, using the template strand to synthesise a new complementary DNA strand, copying the target region.（约 72 摄氏度）：Taq 聚合酶从 5' 至 3' 方向延伸引物，以模板链为基础合成新的互补 DNA 链，复制目标区域。

(b) Band representing the largest fragment代表最大片段的条带 A1

The band closest to the wells (the top of the gel) represents the largest fragment. In gel electrophoresis, an electric current pulls negatively charged DNA through the porous agarose gel toward the positive electrode (at the bottom). Smaller fragments move faster and travel further from the wells; larger fragments move more slowly through the pores and remain closer to the wells.最靠近上样孔（凝胶顶部）的条带代表最大片段。在凝胶电泳中，电流将带负电的 DNA 通过多孔琼脂糖凝胶向正极（底部）迁移。较小的片段移动更快，离上样孔更远；较大的片段因难以穿过孔隙而移动较慢，留在靠近上样孔的位置。

In gel electrophoresis, size and migration distance are inversely related.在凝胶电泳中，片段大小与迁移距离成反比。 This inverse relationship between size and distance is why a DNA ladder (molecular weight marker) is always run alongside samples: by comparing a band's position to the ladder, you can estimate the fragment's size in base pairs. The charge-to-mass ratio of all DNA fragments is approximately the same (each phosphate group carries one negative charge, and mass is proportional to length), so separation is purely by size through the molecular sieve of the gel. PCR amplifies the target region between the two primers, so all copies should be the same length, producing a single sharp band rather than a smear.大小与距离的反比关系正是为什么总是将 DNA 梯（分子量标记）与样品同时电泳的原因：通过将条带位置与梯相比较，可估算片段大小（以碱基对计）。所有 DNA 片段的电荷质量比大致相同（每个磷酸基团携带一个负电荷，质量与长度成正比），因此分离完全基于片段在凝胶分子筛中的大小。PCR 扩增两引物之间的目标区域，故所有拷贝长度相同，产生单一清晰条带而非弥散。

Section C · Worked SolutionsC 部分 · 详细解答

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §2 DNA replication (applied)DNA 复制（应用） · C3.2k [8 marks][8 分]

Key enzymes of DNA replication. (a) Enzyme that unwinds the double helix. (b) Why an RNA primer is needed before DNA polymerase III. (c) Fate of RNA primers after synthesis.DNA 复制中的关键酶。(a) 在复制叉处解开双螺旋的酶。(b) DNA 聚合酶 III 前需要 RNA 引物的原因。(c) DNA 合成完成后 RNA 引物的命运。

Answer:答案： (a) helicase解旋酶 · (b) DNA pol III cannot initiate; primase lays primerDNA 聚合酶 III 无法起始；引物酶铺设引物 · (c) removed by DNA pol I; gap filled; ligase seals由 DNA 聚合酶 I 切除；填补缺口；DNA 连接酶封闭

(a) Enzyme that unwinds the helix解开双螺旋的酶 A1+A1

Helicase unwinds the double helix at the replication fork. It does so by breaking (hydrolysing) the hydrogen bonds between the complementary base pairs, separating the two DNA strands and creating a replication fork with two single-stranded templates. Single-strand binding proteins (SSBPs) stabilise the unwound strands to prevent re-annealing.解旋酶在复制叉处解开双螺旋。它通过断裂（水解）互补碱基对之间的氢键，分离两条 DNA 链，形成具有两条单链模板的复制叉。单链结合蛋白（SSBPs）稳定解开的链，防止其重新退火。

(b) Why RNA primer is required为什么需要 RNA 引物 A1+A1+A1

DNA polymerase III can only add nucleotides to an existing 3'-OH group; it cannot start a new strand from scratch (de novo synthesis). Therefore, a short RNA primer (about 10 nucleotides) must first be laid down to provide the free 3'-OH end from which DNA polymerase III can extend. The enzyme that synthesises this RNA primer is primase (a type of RNA polymerase). Without the primer, the replication fork would stall because no 3'-OH end would be available.DNA 聚合酶 III 只能将核苷酸添加到已有的 3'-OH 基团上，无法从头开始合成新链。因此，必须先铺设一段短的 RNA 引物（约 10 个核苷酸），提供 DNA 聚合酶 III 可以延伸的游离 3'-OH 末端。合成此 RNA 引物的酶是引物酶（一种 RNA 聚合酶）。没有引物，复制叉将会停滞，因为没有可用的 3'-OH 末端。

(c) Fate of RNA primersRNA 引物的命运 A1+A1+A1

After DNA synthesis, the RNA primers are removed by DNA polymerase I, which has 5' to 3' exonuclease activity: it degrades the RNA primer while simultaneously filling the gap with DNA nucleotides (using the adjacent DNA as template). The small gap left at the 3' end of each removed primer cannot be filled by DNA polymerase I. DNA ligase then seals (joins) the nick between adjacent Okazaki fragments or between the primer-replaced segment and the main chain, forming a continuous strand with phosphodiester bonds.DNA 合成完成后，RNA 引物由 DNA 聚合酶 I 切除，后者具有 5' 至 3' 外切核酸酶活性：它在降解 RNA 引物的同时用 DNA 核苷酸填补缺口（以相邻 DNA 为模板）。DNA 聚合酶 I 无法填补每个被切除引物的 3' 末端留下的小缺口。DNA 连接酶随后封闭（连接）相邻冈崎片段之间或引物替换片段与主链之间的缺口，以磷酸二酯键形成连续链。

DNA polymerase III extends; DNA polymerase I replaces primers; DNA ligase seals nicks.DNA 聚合酶 III 延伸；DNA 聚合酶 I 替换引物；DNA 连接酶封闭缺口。 The three-enzyme relay for primer removal is a classic exam topic at the SBI4U level. The key distinction: DNA pol III is the main replicative polymerase (high speed, high processivity); DNA pol I is the "repair" polymerase that removes primer RNA and fills the gap; DNA ligase seals the final nick. On the lagging strand, this process repeats for every Okazaki fragment (roughly every 1,000-2,000 bp in prokaryotes). The need for primase explains why the lagging strand is synthesised discontinuously: each new Okazaki fragment needs its own primer, which then must be removed and replaced.引物切除的三酶接力是 SBI4U 层面的经典考点。关键区别：DNA 聚合酶 III 是主要的复制聚合酶（高速、高持续合成能力）；DNA 聚合酶 I 是"修复"聚合酶，切除引物 RNA 并填补缺口；DNA 连接酶封闭最终缺口。在滞后链上，此过程对每个冈崎片段重复（原核生物中每约 1,000-2,000 bp 一次）。对引物酶的需求解释了为何滞后链以不连续方式合成：每个新的冈崎片段都需要自己的引物，引物随后必须被切除和替换。

Q11MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 Translation (applied)翻译（应用） · HS-LS1-1 [8 marks][8 分]

Mutation at codon 12: GGA (Gly) to UGA (stop). Original protein = 18 amino acids. (a) Classify mutation type and effect on protein. (b) Mutant protein length and consequence. (c) Heritability if mutation is in a somatic cell.第 12 位密码子突变：GGA（Gly）变为 UGA（终止）。原始蛋白质 = 18 个氨基酸。(a) 突变类型分类及对蛋白质的影响。(b) 突变蛋白质长度及后果。(c) 突变发生在体细胞中是否可遗传。

Answer:答案： (a) substitution; nonsense mutation替换；无义突变 · (b) 11 amino acids; likely non-functional11 个氨基酸；可能无功能 · (c) no; somatic mutations are not transmitted to offspring否；体细胞突变不传给后代

(a) Classify the mutation突变分类 A1+A1

This is a point substitution: a single base change converts one codon to another. Specifically it is a nonsense mutation because the codon changes from a sense codon (GGA, which codes for glycine) to a stop codon (UGA). The effect on protein function is premature termination: translation stops before the full-length protein is produced, resulting in a truncated (shortened) protein that is very likely non-functional.这是一个点替换突变：单个碱基变化将一个密码子转变为另一个密码子。具体而言，这是一个无义突变，因为密码子从有义密码子（GGA，编码甘氨酸）变为终止密码子（UGA）。对蛋白质功能的影响是翻译提前终止：翻译在产生全长蛋白质之前停止，产生截短蛋白质，几乎肯定失去功能。

(b) Mutant protein length and consequence突变蛋白质长度及后果 A1+A1+A1

The original protein is 18 amino acids long. The mutation introduces a stop codon at position 12, so translation terminates before incorporating the 12th amino acid (UGA signals stop, so amino acids 1 through 11 are added, and codon 12 is the stop). The mutant protein contains 11 amino acids. Because it is missing 7 of the original 18 amino acids (codons 12-18), the protein's three-dimensional structure is altered. Depending on the functional domains, the protein is likely non-functional or severely impaired.原始蛋白质共 18 个氨基酸。突变在第 12 位引入终止密码子，翻译在添加第 12 个氨基酸之前终止（UGA 为终止信号，因此添加了第 1 至 11 个氨基酸，第 12 位密码子为终止）。突变蛋白质含 11 个氨基酸。由于缺失了原 18 个氨基酸中的 7 个（第 12-18 位密码子），蛋白质的三维结构改变。取决于功能域的位置，该蛋白质可能完全失去功能或严重受损。

(c) Heritability of a somatic cell mutation体细胞突变的可遗传性 A1+A1+A1

No, this mutation will not be inherited by the organism's offspring. Somatic (body) cells are non-reproductive cells; they do not contribute genetic material to gametes (sperm or egg cells). Only mutations that occur in germline cells (the cells that give rise to gametes) can be passed to the next generation. A somatic mutation is inherited only by the daughter cells produced by mitosis within the same organism (clonal expansion), and dies with the organism. It cannot enter the germline and is therefore not heritable in the evolutionary sense.不会，此突变不会遗传给后代。体细胞是非生殖细胞，不向配子（精子或卵细胞）提供遗传物质。只有发生在生殖细胞（产生配子的细胞）中的突变才能传递给下一代。体细胞突变只能由同一生物体内通过有丝分裂产生的子代细胞继承（克隆扩增），并随个体死亡而消失。它无法进入生殖系，因此在进化意义上不可遗传。

Nonsense mutations are among the most disruptive point mutations because they shorten the protein, often removing critical domains.无义突变是最具破坏性的点突变之一，因为它缩短蛋白质，通常去除关键功能域。 The three stop codons are UAA, UAG, and UGA. Any base substitution that creates one of these from a sense codon is a nonsense mutation. The earlier in the coding sequence the stop codon appears, the more truncated the protein and the more severe the effect. The somatic vs. germline distinction is fundamental to genetics: cancer is caused by somatic mutations accumulating in body cells, but cancer itself is not heritable through reproduction (with rare exceptions such as inherited cancer predisposition alleles, which are germline).三种终止密码子为 UAA、UAG 和 UGA。任何将有义密码子替换为上述密码子的碱基替换均为无义突变。终止密码子出现在编码序列越靠前的位置，蛋白质截短越严重，影响越大。体细胞突变与生殖细胞突变的区别是遗传学的基本概念：癌症由体细胞中积累的突变引起，但癌症本身无法通过生殖遗传（罕见例外是遗传性癌症易感等位基因，这类等位基因位于生殖细胞系中）。

Q12HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §5 + §7 Mutations + Biotechnology (PCR applied)突变 + 生物技术（PCR 应用） · HS-LS1-1 / HS-LS3-1 [10 marks][10 分]

Forensic PCR from one dsDNA molecule. (a) Molecule count after 5 cycles using $2^n$. (b) Why Taq polymerase instead of human DNA polymerase; key property. (c) Germline mutation: does band position change on gel? (d) Germline vs somatic mutation heritability; which type is more significant for forensics.从单个双链 DNA 分子进行法医 PCR。(a) 用 $2^n$ 计算 5 个循环后的分子数量。(b) 为什么使用 Taq 聚合酶而非人类 DNA 聚合酶；关键特性。(c) 生殖细胞突变：凝胶上条带位置是否改变？(d) 生殖细胞突变与体细胞突变的可遗传性；哪种突变对法医学更重要。

Answer:答案： (a) $2^5 = 32$ molecules个分子 · (b) heat stable; not denatured at 95 degrees C耐高温；在 95 摄氏度不变性 · (c) no change in position位置不变 · (d) germline; present in every cell of the body生殖细胞突变；存在于体内每个细胞中

(a) Number of dsDNA molecules after 5 cycles5 个循环后双链 DNA 分子的数量 A1+A1

Starting from 1 double-stranded DNA molecule, each PCR cycle doubles the number of molecules: after $n$ cycles, there are $2^n$ molecules.从 1 个双链 DNA 分子开始，每个 PCR 循环使分子数量翻倍：经过 $n$ 个循环后，有 $2^n$ 个分子。 $$ \text{molecules after 5 cycles} = 2^5 = 32 $$ (A1 for correct formula application; A1 for answer 32.)（正确应用公式得 A1；答案 32 得 A1。）

(b) Why Taq polymerase; essential property为什么使用 Taq 聚合酶；关键特性 A1+A1+A1

Taq polymerase is a thermostable (heat-stable) DNA polymerase isolated from the bacterium Thermus aquaticus, which lives in hot springs at temperatures approaching 95 degrees C. Human DNA polymerase is irreversibly denatured (unfolded and inactivated) at the high temperatures used in the denaturation step (approximately 94-96 degrees C). Taq polymerase retains its enzymatic activity at these temperatures and therefore does not need to be replaced after each denaturation step, making PCR practical and automatable. The essential property for the denaturation step is thermostability: the ability to withstand repeated heating to approximately 95 degrees C without losing activity.Taq 聚合酶是从嗜热菌（Thermus aquaticus）中分离的耐热 DNA 聚合酶，该细菌生活在接近 95 摄氏度的温泉中。人类 DNA 聚合酶在变性步骤（约 94-96 摄氏度）中会不可逆变性（展开并失活）。Taq 聚合酶在这些温度下保持酶活性，因此无需在每次变性步骤后更换，使 PCR 具有实际可操作性和可自动化性。对变性步骤至关重要的特性是热稳定性：能够耐受反复加热至约 95 摄氏度而不失活。

(c) Gel band position after amplifying a mutant target扩增突变目标后凝胶条带位置 A1+A1

The band position on the gel would NOT change. A single base substitution (point mutation) changes one nucleotide out of potentially hundreds or thousands in the target region. This negligibly changes the molecular weight of the amplified fragment. Since gel electrophoresis separates DNA fragments by size (number of base pairs), a one-base difference is too small to produce a detectable shift in band position. The fragment would migrate to essentially the same position as the wild-type. (To detect a single nucleotide change, techniques such as Sanger sequencing or allele-specific PCR would be required.)凝胶上的条带位置不会改变。单个碱基替换（点突变）在目标区域数百乃至数千个核苷酸中只改变一个核苷酸。这对扩增片段的分子量影响可以忽略不计。由于凝胶电泳按大小（碱基对数）分离 DNA 片段，一个碱基的差异太小，无法产生可检测的条带位置偏移。该片段将迁移到与野生型基本相同的位置。（要检测单核苷酸变化，需要使用桑格测序或等位基因特异性 PCR 等技术。）

(d) Germline vs somatic mutation heritability; forensic significance生殖细胞突变与体细胞突变的可遗传性；法医学重要性 A1+A1+A1

A germline mutation occurs in the cells that produce gametes and is therefore present in every cell of any offspring that inherits it. It is heritable across generations. A somatic mutation occurs in a non-reproductive body cell and is inherited only by daughter cells arising from that cell by mitosis; it cannot be passed to the organism's offspring and dies with the individual. For forensic identification, germline mutations are more significant. Because a germline mutation is present in every nucleated cell of the individual's body (blood cells, skin cells, hair follicles, semen), a DNA sample taken from any tissue at a crime scene will carry the same genetic profile. Somatic mutations would be present in only a subset of cells and would not be reliably detected across tissue samples.生殖细胞突变发生在产生配子的细胞中，因此存在于继承该突变的任何后代的每个细胞中，可跨代遗传。体细胞突变发生在非生殖性体细胞中，只能由该细胞通过有丝分裂产生的子代细胞继承，不能传给后代，随个体死亡而消失。对于法医学身份识别，生殖细胞突变更为重要。由于生殖细胞突变存在于个体体内每个有核细胞中（血细胞、皮肤细胞、毛囊、精液），从犯罪现场任何组织采集的 DNA 样本都将携带相同的遗传特征。体细胞突变只存在于部分细胞中，在不同组织样本中无法可靠检测。

PCR's power in forensics comes from exponential amplification of trace DNA and universal sample compatibility of germline profiles.PCR 在法医学中的强大之处在于对痕量 DNA 的指数扩增，以及生殖细胞遗传特征在所有组织样本中的普遍兼容性。 After 30 cycles of PCR, $2^{30} \approx 10^9$ copies are generated from a single starting molecule, producing enough DNA for gel analysis, sequencing, and profiling from a single hair or a few skin cells. The $2^n$ formula assumes perfect doubling efficiency, which real PCR approaches but never perfectly achieves due to primer competition and reagent depletion. For part (d), the key forensic insight is that germline mutations define a unique (or near-unique) genetic profile that is identical across all body tissues and is inherited by biological relatives, both of which are essential for suspect identification and familial DNA searching.经过 30 个 PCR 循环后，从单个起始分子产生约 $2^{30} \approx 10^9$ 个拷贝，足够从单根毛发或少量皮肤细胞进行凝胶分析、测序和遗传特征分析。$2^n$ 公式假设完美的倍增效率，真实 PCR 接近但从未完全达到，这是由于引物竞争和试剂耗尽。对于 (d) 小问，关键的法医学洞见是：生殖细胞突变定义了在所有体组织中相同且由生物学亲属共享的独特（或近乎独特）遗传特征，这两点对于嫌疑人识别和家族 DNA 搜索都至关重要。