Cambridge IGCSE IGCSE & A-Level Sheet preview

Gene Expression and Regulation

AP Biology · Topic 6

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Handout Vocabulary
6.1

DNA and RNA Structure
Syllabus
Big IdeaLearning ObjectiveEssential Knowledge

Big Idea 3 — Information Storage and Transmission
Living systems store, retrieve, transmit, and respond to information essential to life processes.

6.1.A
Describe the structures involved in passing hereditary information from one generation to the next.

  • 6.1.A.1 Genetic information is stored in and passed to subsequent generations through DNA molecules and, in some cases, RNA molecules.
    • 6.1.A.1.i Prokaryotic organisms typically have circular chromosomes.
    • 6.1.A.1.ii Eukaryotic organisms typically have multiple linear chromosomes that are comprised of DNA. These chromosomes are condensed using histones and associated proteins.
  • 6.1.A.2 Prokaryotes and eukaryotes can contain plasmids, which are extra-chromosomal circular molecules of DNA.

6.1.B
Describe the characteristics of DNA that allow it to be used as hereditary material.

  • 6.1.B.1 Nucleic acids exhibit specific nucleotide base pairing that is conserved through evolution.
    • 6.1.B.1.i Purines (guanine and adenine) have a double ring structure.
    • 6.1.B.1.ii Pyrimidines (cytosine, thymine, and uracil) have a single ring structure.
    • 6.1.B.1.iii Purines pair with pyrimidines: adenine with thymine (or uracil in RNA) and guanine with cytosine.

Source: College Board AP Course and Exam Description

DNA carries genetic information as a double helix 双螺旋 of two strands. Its nucleotides 核苷酸 pair by rule – A with T, G with C (complementary base pairing 互补配对) – so one strand specifies the other. The strands run antiparallel 反平行. RNA is single-stranded, uses uracil (U) instead of thymine, and has ribose sugar.

DNA: two antiparallel strands held by complementary base pairs DNA: two antiparallel strands held by complementary base pairs

Vocabulary Train
English Chinese Pinyin
double helix 双螺旋 shuāng luó xuán
nucleotides 核苷酸 hé gān suān
complementary base pairing 互补配对 hù bǔ pèi duì
antiparallel 反平行 fǎn píng xíng
Exercise sheet
6.2

DNA Replication
Syllabus
Big IdeaLearning ObjectiveEssential Knowledge

Big Idea 3 — Information Storage and Transmission
Living systems store, retrieve, transmit, and respond to information essential to life processes.

6.2.A
Describe the mechanisms by which genetic information is copied for transmission between generations.

  • 6.2.A.1 DNA replication ensures continuity of hereditary information.
    • 6.2.A.1.i DNA is synthesized in the 5' to 3' direction.
    • 6.2.A.1.ii Replication is a semiconservative process, meaning one strand of DNA serves as the template for a new strand of complementary DNA.
    • 6.2.A.1.iii Helicase unwinds the DNA strands.
    • 6.2.A.1.iv Topoisomerase relaxes supercoiling in front of the replication fork.
    • 6.2.A.1.v DNA polymerase requires RNA primers to initiate DNA synthesis.
    • 6.2.A.1.vi DNA polymerase synthesizes new strands of DNA continuously on the leading strand and discontinuously on the lagging strand.
    • 6.2.A.1.vii Ligase joins the fragments on the lagging strand.
    • Exclusion statement: The names of the steps and particular enzymes involved, excluding DNA polymerase, ligase, RNA polymerase, helicase, and topoisomerase, are beyond the scope of the AP Exam.

Source: College Board AP Course and Exam Description

Before a cell divides, DNA is copied semiconservatively 半保留复制: the helix unwinds and each old strand templates a new one, so each daughter helix has one old and one new strand. DNA polymerase 聚合酶 adds nucleotides following base-pairing rules, building the new strand and proofreading as it goes.

Semi-conservative DNA replication at a replication fork Semi-conservative DNA replication at a replication fork

Vocabulary Train
English Chinese Pinyin
semiconservatively 半保留复制 bàn bǎo liú fù zhì
DNA polymerase 聚合酶 jù hé méi
Exercise sheet
6.3

Transcription and RNA Processing
Syllabus
Big IdeaLearning ObjectiveEssential Knowledge

Big Idea 3 — Information Storage and Transmission
Living systems store, retrieve, transmit, and respond to information essential to life processes.

6.3.A
Describe the mechanisms by which genetic information flows from DNA to RNA to protein.

  • 6.3.A.1 The sequence of the RNA bases, together with the structure of the RNA molecule, determines RNA function.
    • 6.3.A.1.i Messenger RNA (mRNA) molecules carry information from DNA in the nucleus to the ribosome in the cytoplasm.
    • 6.3.A.1.ii Distinct transfer RNA (tRNA) molecules bind specific amino acids and have anticodon sequences that base pair with the codons of mRNA. tRNA is recruited to the ribosome during translation to generate the primary peptide sequence based on the mRNA sequence.
    • 6.3.A.1.iii Ribosomal RNA (rRNA) molecules are functional building blocks of ribosomes.
  • 6.3.A.2 RNA polymerases use a single template strand of DNA to direct the inclusion of bases in the newly formed RNA molecule. This process is known as transcription.
  • 6.3.A.3 The enzyme RNA polymerase synthesizes mRNA molecules in the 5' to 3' direction by reading the template DNA strand in the 3' to 5' direction.
  • 6.3.A.4 In eukaryotic cells the mRNA transcript undergoes a series of enzyme-mediated modifications.
    • 6.3.A.4.i The addition of a poly-A tail makes mRNA more stable.
    • 6.3.A.4.ii The addition of a GTP cap helps with ribosomal recognition.
    • 6.3.A.4.iii The excision of introns, along with the splicing and retention of exons, generates different versions of the resulting mature mRNA molecule. This process is known as alternative splicing.

Source: College Board AP Course and Exam Description

Transcription 转录 copies a gene's DNA into messenger RNA 信使RNA (mRNA). RNA polymerase reads the template strand and builds a complementary RNA. In eukaryotes the mRNA is then processed: a cap and tail are added, and introns 内含子 (non-coding parts) are spliced out, leaving the exons 外显子.

Introns are removed and exons joined to make mature mRNA Introns are removed and exons joined to make mature mRNA

Explore

Transcribe DNA into messenger RNA

Transcription copies a DNA template into mRNA, pairing A→U, T→A, C→G, G→C. Step through to build the RNA strand base by base.

Vocabulary Train
English Chinese Pinyin
Transcription 转录 zhuǎn lù
messenger RNA 信使 xìn shǐ
introns 内含子 nèi hán zi
exons 外显子 wài xiǎn zi
Exercise sheet
6.4

Translation
Syllabus
Big IdeaLearning ObjectiveEssential Knowledge

Big Idea 3 — Information Storage and Transmission
Living systems store, retrieve, transmit, and respond to information essential to life processes.

6.4.A
Explain how the phenotype of an organism is determined by its genotype.

  • 6.4.A.1 Translation of the mRNA to generate a polypeptide occurs on ribosomes that are present in the cytoplasm of both prokaryotic and eukaryotic cells, as well as the cytoplasmic surface of the rough ER of eukaryotic cells.
  • 6.4.A.2 In prokaryotic organisms, translation of the mRNA molecule occurs while it is being transcribed.
  • 6.4.A.3 Translation involves many sequential steps, including initiation, elongation, and termination. The salient features of translation include:
    • 6.4.A.3.i Translation is initiated when the rRNA in the ribosome interacts with the mRNA at the start codon (AUG, coding for the amino acid methionine).
    • 6.4.A.3.ii The sequence of nucleotides on the mRNA is read in triplets, called codons.
    • 6.4.A.3.iii Each codon encodes a specific amino acid, which can be deduced by using a genetic code chart. Many amino acids are encoded by more than one codon.
    • 6.4.A.3.iv Nearly all living organisms use the same genetic code, which is evidence for the common ancestry of all living organisms.
    • 6.4.A.3.v tRNA brings the correct amino acid to the place specified by the codon on the mRNA.
    • 6.4.A.3.vi The amino acid is transferred to the growing polypeptide chain.
    • 6.4.A.3.vii The process continues along the mRNA until a stop codon is reached.
    • 6.4.A.3.viii Translation terminates with the release of the newly synthesized protein.
    • Exclusion statement: The details and names of the enzymes and factors involved in each of these steps are beyond the scope of the AP Exam.
    • Exclusion statement: Memorization of the genetic code, with the exception of the start codon AUG, is beyond the scope of the AP Exam.
  • 6.4.A.4 Genetic information in retroviruses is a special case and has an alternate flow of information: from RNA to DNA, made possible by reverse transcriptase, an enzyme that copies the viral RNA genome into DNA. This DNA integrates into the host genome and is transcribed and translated for the assembly of new viral progeny.

Source: College Board AP Course and Exam Description

Translation 翻译 builds a protein from the mRNA at the ribosome 核糖体. The mRNA is read in three-base codons 密码子, each specifying one amino acid (the genetic code). Transfer RNA 转运RNA brings the matching amino acid, and the ribosome links them into a polypeptide until a stop codon ends it. This is the "central dogma": DNA → RNA → protein.

mRNA is read by a ribosome to build a protein from amino acids mRNA is read by a ribosome to build a protein from amino acids

Vocabulary Train
English Chinese Pinyin
Translation 翻译 fān yì
ribosome 核糖体 hé táng tǐ
codons 密码子 mì mǎ zi
Transfer RNA 转运 zhuǎn yùn
Exercise sheet
6.5

Regulation of Gene Expression
Syllabus
Big IdeaLearning ObjectiveEssential Knowledge

Big Idea 3 — Information Storage and Transmission
Living systems store, retrieve, transmit, and respond to information essential to life processes.

6.5.A
Describe the types of interactions that regulate gene expression.

  • 6.5.A.1 Regulatory sequences are stretches of DNA that interact with regulatory proteins to control transcription. Some genes are constitutively expressed, and others are inducible.
  • 6.5.A.2 Epigenetic changes can affect gene expression through reversible modifications of DNA or histones.
  • 6.5.A.3 The phenotype of a cell or an organism is determined by the combination of genes that are expressed and the levels at which they are expressed.
    • 6.5.A.3.i Observable cell differentiation results from the expression of genes for tissue-specific proteins.
    • 6.5.A.3.ii Induction of transcription factors during development results in sequential gene expression.
    • 6.5.A.3.iii The function and amount of gene products determine the phenotype of organisms.

6.5.B
Explain how the location of regulatory sequences relates to their function.

  • 6.5.B.1 Both prokaryotes and eukaryotes have groups of genes that are coordinately regulated.
    • 6.5.B.1.i Prokaryotes regulate operons in an inducible or repressible system.
    • 6.5.B.1.ii In eukaryotes, groups of genes may be influenced by the same transcription factors to coordinately regulate expression.

Source: College Board AP Course and Exam Description

Cells control which genes are expressed and when. In prokaryotes, operons 操纵子 switch groups of genes on or off. In eukaryotes, regulation happens at many levels – which genes are transcribed (transcription factors, promoters, enhancers), RNA processing, and after translation. This lets a cell respond to its environment without changing its DNA.

The lac operon switches genes on only when lactose is present The lac operon switches genes on only when lactose is present

Vocabulary Train
English Chinese Pinyin
operons 操纵子 cāo zòng zi
Exercise sheet
6.6

Gene Expression and Cell Specialization
Syllabus
Big IdeaLearning ObjectiveEssential Knowledge

Big Idea 3 — Information Storage and Transmission
Living systems store, retrieve, transmit, and respond to information essential to life processes.

6.6.A
Explain how the binding of transcription factors to promoter regions affects gene expression and the phenotype of the organism.

  • 6.6.A.1 RNA polymerase and transcription factors bind to promoter or enhancer DNA sequences to initiate transcription. These sequences can be upstream or downstream of the transcription start site.
  • 6.6.A.2 Negative regulatory molecules inhibit gene expression by binding to DNA and blocking transcription.

6.6.B
Explain the connection between the regulation of gene expression and phenotypic differences in cells and organisms.

  • 6.6.B.1 Gene regulation results in differential gene expression and influences cell products and functions.
  • 6.6.B.2 Certain small RNA molecules have roles in regulating gene expression.

Source: College Board AP Course and Exam Description

Every cell in a body has the same DNA, yet cells differ because they express different genes – differential gene expression 差异表达. This is how one fertilized egg produces many specialized cell types (muscle, nerve, skin); signals during development turn specific genes on and off.

A stem cell differentiates into specialised cell types A stem cell differentiates into specialised cell types

Vocabulary Train
English Chinese Pinyin
differential gene expression 差异表达 chā yì biǎo dá
Exercise sheet
6.7

Mutations
Syllabus
Big IdeaLearning ObjectiveEssential Knowledge

Big Idea 3 — Information Storage and Transmission
Living systems store, retrieve, transmit, and respond to information essential to life processes.

6.7.A
Describe the various types of mutation.

  • 6.7.A.1 Alterations in a DNA sequence are mutations that can cause changes in the type or amount of the protein produced and the consequent phenotype. DNA mutations can be beneficial, detrimental, or neutral based on the effect or the lack of effect they have on the resulting nucleic acid or protein and the phenotypes that are conferred by the protein.
    • 6.7.A.1.i Point mutations occur when one nucleotide has been substituted for a different nucleotide.
    • 6.7.A.1.ii Frameshift mutations occur when one or more nucleotides are inserted or deleted, causing the reading frame to be shifted.
    • 6.7.A.1.iii Nonsense mutations occur when there is a point mutation that causes a premature stop.
    • 6.7.A.1.iv Silent mutations occur when the change in the nucleotide sequence has no effect on the amino acid sequence.
    • Illustrative examples for 6.7.A.1:
      • Mutations in the CFTR gene disrupt ion transport and result in cystic fibrosis.
      • Mutations in the MC1R gene give adaptive melanism in pocket mice.
    • Exclusion statement: Knowledge of specific mutations and their effects is beyond the scope of the AP Exam.

6.7.B
Explain how changes in genotype may result in changes in phenotype.

  • 6.7.B.1 Errors in DNA replication or DNA repair mechanisms as well as external factors, including radiation and reactive chemicals, can cause random mutations in the DNA.
    • 6.7.B.1.i Whether a mutation is beneficial, detrimental, or neutral depends on the environmental context.
    • 6.7.B.1.ii Mutations are a source of genetic variation.
  • 6.7.B.2 Errors in mitosis or meiosis can result in changes in phenotype.
    • 6.7.B.2.i Changes in chromosome number resulting from nondisjunction often result in new phenotypes caused by triploidy (aneuploidy).
    • 6.7.B.2.ii Changes in chromosome number often result in disorders with developmental limitations.
    • 6.7.B.2.iii Alterations in chromosome structure lead to genetic disorders.
    • Exclusion statement: Knowledge of specific disorders related to changes in chromosome number is beyond the scope of the AP Exam.

6.7.C
Explain how alterations in DNA sequences contribute to variation that can be subject to natural selection.

  • 6.7.C.1 Changes in genotype may affect phenotypes that are subject to natural selection. Genetic changes that enhance survival and reproduction can be selected for by environmental conditions.
    • 6.7.C.1.i The horizontal acquisitions of genetic information in prokaryotes via transformation (uptake of DNA), transduction (viral transmission of genetic information), conjugation (cell-to-cell transfer of DNA), and transposition (movement of DNA segments within and between DNA molecules) increase genetic variation.
    • 6.7.C.1.ii Related viruses can recombine genetic information if they infect the same host cell.
    • 6.7.C.1.iii Reproductive processes that increase genetic variation are evolutionarily conserved and are shared by various organisms.
    • Illustrative examples for 6.7.C.1: Sickle cell anemia

Source: College Board AP Course and Exam Description

A mutation 突变 is a change in the DNA sequence. Point mutations change one base (silent, missense, or nonsense); insertions/deletions can cause a frameshift 移码 that garbles everything downstream. Mutations in gametes are heritable; they may be harmful, neutral, or beneficial – and beneficial ones supply the variation for natural selection.

Substitution, deletion, and insertion mutations Substitution, deletion, and insertion mutations

Vocabulary Train
English Chinese Pinyin
mutation 突变 tū biàn
frameshift 移码 yí mǎ
Exercise sheet
6.8

Biotechnology
Syllabus
Big IdeaLearning ObjectiveEssential Knowledge

Big Idea 3 — Information Storage and Transmission
Living systems store, retrieve, transmit, and respond to information essential to life processes.

6.8.A
Explain the use of genetic engineering techniques in analyzing or manipulating DNA.

  • 6.8.A.1 Genetic engineering techniques can be used to analyze and manipulate DNA and RNA.
    • 6.8.A.1.i Gel electrophoresis is a process that separates DNA fragments by size and charge.
    • 6.8.A.1.ii During polymerase chain reaction (PCR), DNA fragments are amplified by denaturing DNA, annealing primers to the original strand, and extending the new DNA molecule.
    • 6.8.A.1.iii Bacterial transformation introduces foreign DNA into bacterial cells.
    • 6.8.A.1.iv DNA sequencing technology determines the order of nucleotides in a DNA molecule. Typically, these techniques result in a DNA fingerprint that allows for the comparison of DNA sequences from various samples.
    • Illustrative examples for 6.8.A.1:
      • Amplified DNA fragments can be used to identify organisms and perform phylogenetic analysis.
      • Analysis of DNA can be used for forensic identification.
      • Genetically modified organisms include transgenic animals.
      • Gene cloning allows propagation of DNA fragments.
    • Exclusion statement: Knowledge of the details of each of these genetic engineering techniques is beyond the scope of the AP Exam.

Source: College Board AP Course and Exam Description

Biotechnology 生物技术 tools manipulate genetic material: PCR 聚合酶链反应 copies DNA, gel electrophoresis 凝胶电泳 separates DNA fragments by size, restriction enzymes and cloning move genes between organisms, and CRISPR edits sequences. These techniques enable genetic testing, engineered organisms, and medical treatments.

Gel electrophoresis separates DNA fragments by length Gel electrophoresis separates DNA fragments by length

The polymerase chain reaction doubles the DNA each cycle The polymerase chain reaction doubles the DNA each cycle

Worked example. A template DNA strand 3'-TAC GGA TTC-5' is transcribed into mRNA 5'-AUG CCU AAG-3'. A ribosome reads three codons: AUG = Met (start), CCU = Pro, AAG = Lys — coding for Met–Pro–Lys. Changing the third base of a codon often gives the same amino acid, because the genetic code is degenerate, which softens the effect of many mutations.

Vocabulary Train
English Chinese Pinyin
Biotechnology 生物技术 shēng wù jì shù
PCR 聚合酶链反应 jù hé méi liàn fǎn yìng
gel electrophoresis 凝胶电泳 níng jiāo diàn yǒng
Exercise sheet
6.8

Exam tips
  • Know the central dogma DNA → RNA → protein: transcription makes mRNA, translation reads it in three-base codons at the ribosome.
  • Remember RNA is single-stranded and uses U instead of T; replication is semiconservative.
  • Every cell has the same DNA — cells differ because they express different genes (differential gene expression).
  • A mutation is a change in the DNA sequence and may be harmful, neutral, or beneficial (the raw material for selection).
  • Link biotech tools to their jobs: PCR copies DNA; gel electrophoresis separates fragments by size.

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