BIOL 202 Flashcards

Question

Why is induced polyploidy often done in plants?

Answer 1

A: Plant cells/tissues can be grown into mature plants (greater developmental “plasticity”), enabling propagation of the new polyploid

Answer 2

triploidy + dispermy: A giant invasive triploid crayfish spreading globally A: Likely dispermy (two sperm fertilizing one egg), possibly facilitated by crowded pet-shop/tank conditions. A: Bigger size (triploids tend to be bigger) + can reproduce without mating (all female, per lecture).

Answer 3

A: Polyploidy increases size (strawberries); triploidy causes sterility/seedlessness (banana, watermelon)

Answer 4

A: Mitosis: homologs do not pair; Meiosis I: homologs must pair to segregate correctly

Answer 5

A: The third unpaired homolog segregates randomly, producing aneuploid gametes

Answer 6

A: Aneuploid gametes: haploid set plus extra chromosomes from mis-segregation

Answer 7

A: Only if all extra chromosomes go to the same daughter cell during meiosis I

Answer 8

= change in number of entire chromosome sets (multiples of n). Examples: triploid banana (3n), octoploid strawberry (8n).

Answer 9

Extra sets from the same species.

Answer 10

Extra sets from different species (often ancestral species). After genome doubling → sometimes called amphidiploid (“double diploid”).

Answer 11

Fertility depends on meiosis pairing Homologous chromosomes must pair. If no pairing partner → missegregation → aneuploid gametes → sterility. Genome doubling restores pairing partners → restores fertility.

Answer 12

Commercial strawberry: 2n = 8x = 56 Contains remnants of four ancestral diploid subgenomes Origin largely natural over evolutionary time Not modern GMO — ancient hybridization events Sequence model: Diploid × diploid → 4 sets Hybrid × diploid → 6 sets Hybrid × diploid → 8 sets After doubling and stabilization → fertile octoploid lineage. Shows: Allopolyploidy can occur naturally in the wild.

Answer 13

Parents: Radish: 2n = 18 (n = 9) Cabbage: 2n = 18 (n = 9) Gametes: n = 9 each Fertilization → F1 hybrid: nR + nB (18 total) Chromosomes NOT homologous enough to pair Meiosis fails → aneuploid gametes → sterile If genome doubling occurs spontaneously: 2nR + 2nB = 36 Now each radish chromosome pairs with radish Each cabbage chromosome pairs with cabbage Meiosis restored → fertile Backcross to parent species: Allopolyploid gamete: nR + nB Parent gamete: nB F1 = nR + 2nB Radish set lacks pairing partner → sterile Cannot interbreed with parent species ➡️ Reproductive isolation = new species formed

Answer 14

Initial hybrid: 9 + 10 = 19 chromosomes Sterile (no pairing) Genome doubling required: 19 × 2 = 38 Now pairing possible Fertile allopolyploid Why still “2n”? Relative to itself, it has two copies of each of its chromosomes Diploid with respect to its own genome Even though chromosomes originated from two species

Answer 15

Aneuploidy = abnormal number of individual chromosomes Trisomy = 2n + 1 Monosomy = 2n − 1 In humans: Most trisomies → nonviable (die in utero) Viable examples: Trisomy 21 (Down syndrome) XXY (Klinefelter) XO (Turner — only viable monosomy) XXX and XYY often phenotypically mild Trisomy 13 & 18 → usually die in infancy

Answer 16

Problem = gene dosage imbalance In diploid: Gene A : Gene B = 1:1 ratio Balanced protein production In trisomy: 2 copies A : 3 copies B Altered ratio Disrupts pathways requiring precise stoichiometry Not all genes are dosage sensitive But dosage-sensitive genes → developmental defects Example: Partial trisomy 21 helped map which regions cause Down syndrome features

Answer 17

Jimsonweed: n = 12 Researchers created: 12 different trisomics (2n + 1) Each extra chromosome produces a different seed pod phenotype Shows: Plants tolerate aneuploidy better than animals Extra copy of different chromosomes → distinct phenotypes Strong visual evidence for gene dosage effects Trisomic leaf cell chromosome count: 2n + 1 = 2×12 + 1 = 25

Answer 18

Normal chromosome pairs → segregate normally. Trisomic chromosome: 3 homologs Multiple segregation possibilities 3 possible meiosis I configurations 6 possible gamete types Leads to: n n + 1 n − 1 gametes Often reduces fertility due to high proportion of aneuploid gametes

Answer 19

Nondisjunction = failure of chromosomes to separate properly Can occur in: Meiosis I (homologs fail to separate) Meiosis II (sister chromatids fail) If in Meiosis I: 100% gametes abnormal 2 n+1 2 n−1 If in Meiosis II: 50% abnormal 1 n+1 1 n−1 2 normal n Crossovers are important: Lack of recombination increases nondisjunction Mutations blocking crossing over increase nondisjunction frequency

Answer 20

n × (n+1) → 2n + 1 (trisomy) n × (n−1) → 2n − 1 (monosomy) Rare case: (n+1) × (n−1) → 2n May be normal if chromosomes complement Could also still be abnormal depending on which chromosomes involved Spontaneous nondisjunction is rare, so most abnormal gametes are fertilized by normal ones.

Answer 21

Cause: Deletion of part of chromosome 5 (short arm, 5p) Observations: Visible in karyotype Shortened chromosome arm Conclusions: Deleted region contains genes important for development Loss of one copy is NOT lethal (individuals are born) Genes in region are haploinsufficient One copy is not enough for normal development Indicates dosage-sensitive genes Key concept: We can infer biological properties (haploinsufficiency) even before knowing exact genes

Answer 22

A chromosome is one long DNA polymer. It can break: Randomly Due to DNA damage (e.g., radiation) Naturally during meiosis (crossing over) Cells normally repair break

Answer 23

Rearrangements occur when: Breaks are repaired incorrectly Ends that were not originally together are joined Homologous repetitive sequences mispair Key concept: Breakage + misrepair = structural rearrangement Rearrangements include: Deletions Duplications Inversions

Answer 24

Mechanism 1: Two breaks in one chromosome (cis misrepair) Mechanism 2: Unequal crossing over between repetitive DNA

Answer 25

If a chromosome breaks at two positions: Example: Original order: 1–2–3–4 If end of 1 joins directly to 4 → region 2 & 3 are lost Intervening sequence may: Form a circle Be degraded Be lost Important: A deletion is NOT a gap. It is: A shorter chromosome Still an intact DNA molecule Missing sequence

Answer 26

During meiosis: Homologous chromosomes may misalign due to repeated sequences. Instead of aligning correctly: PMS1 pairs with PMS2 (offset alignment) Crossing over occurs incorrectly → produces: One chromosome with deletion One chromosome with duplication This is called: Unequal crossing over

Answer 27

Williams syndrome: Frequency: ~1/10,000 people Traits: Strong musical ability Hyper-social personality Cause: 1.5 Mb deletion On one homolog of chromosome 7 Band 7q11.23 ~17 genes removed

Answer 28

Why it occurs: Region is flanked by repetitive PMS sequences. During meiosis, these repeats misalign. Unequal crossover removes intervening region Diagram explanation: Two repeats flank blue region Misalignment loops out the blue region Crossover produces: One chromosome missing blue region (deletion) One chromosome with two blue copies (duplication)

Answer 29

1️⃣ Cytology (karyotype analysis) 2️⃣ Pairing observation 3️⃣ DNA analysis 4️⃣ Complementation test (functional test)

Answer 30

Observe chromosome length Look for visibly shortened chromosome

Answer 31

During meiosis: Deleted chromosome pairs with normal homolog Missing region cannot pair Forms a “deletion loop” Diagram explanation: Normal chromosome: A–B–C–D–E–F Deleted chromosome: A–B–E–F During pairing: C–D region has no partner → loops out visibly

Answer 32

Genomic sequencing Molecular techniques

Answer 33

Concept: Cross two mutants to determine if mutations are in same gene. If both mutations affect: Same gene → fail to complement → mutant phenotype Different genes → complement → wild-type phenotype Applied to deletions: Cross: Unknown chromosome × Tester strain (Tester strain has multiple recessive mutations) For each locus: If unknown has wild-type allele → function restored If unknown lacks gene (deleted) → mutant phenotype appears Deletion “uncovers” recessive phenotype. Important limitations: Cannot be 100% sure chromosome has no deletion Failure to complement could mean: Deletion Two separate mutations Interpretation depends on probability

Answer 34

Flies homozygous for eya: → No eyes Goal: Find gene location. Method: Cross eya mutant to flies with known deletions. If: Deletion overlaps eya gene → no complementation → no eyes If: Deletion does not overlap → complementation → normal eyes Diagram explanation: Several deletion strains tested. Only one deletion fails to complement. Therefore: eya gene lies within that deleted region. Important evolutionary note: eyeless homolog exists in humans and mice Early master regulatory gene for eye development Conserved across evolution

Answer 35

Cross the question mark chromosome (unknown) to tester chromosome and look to see for each of the loci, if there’s a wild type allele - if the chromosome can contribute a wild type allele, the function should be restored - if there is no wild type, we will see complementation Wild type on loci, all + If chromosome has a deletion, the traits in the region in offspring will be missing in chromosome Deletion uncovers mutant phenotype of the loci - can we be sure that this chromosome has a deletion? - no, because its just a question mark, could have two mutations instead of a deletion

Answer 36

1️⃣ Unequal crossing over 2️⃣ Breaks on homologous chromosomes

Answer 37

Misalignment between homologous chromosomes. Produces: One deletion One duplication

Answer 38

Incorrect rejoining between homologs Duplication size: Can be small Can be large Large duplications are called: Segmental duplications

Answer 39

Definition: Large duplicated blocks of DNA found in multiple genome locations. Human chromosome 7 map shows: Blue lines connecting duplicated regions Same sequence appears in multiple places Diagram explanation: If region at one position matches region at another: A line connects them. Many overlapping lines: → Genome contains extensive duplication network. Purple areas: Hotspots — regions duplicated many times. Key insight: Genome has many repeated blocks.

Answer 40

1️⃣ Duplications contribute to genetic diversity 2️⃣ Duplications increase probability of further rearrangements

Answer 41

Extra gene copy allows: One copy keeps original function Other can mutate Potential new gene function evolves

Answer 42

Because: More repeated sequences → more mispairing during meiosis Therefore: More duplications → more unequal crossing over → more duplications & deletions This creates: A self-amplifying genomic instability pattern

Answer 43

Requires: Two chromosome breaks Region flips Ends rejoined incorrectly Original: 1–2–3–4 After inversion: 1–3–2–4 Two types: Paracentric inversion Pericentric inversion Important: Somatic and meiotic consequences differ.

Answer 44

Does NOT include centromere

Answer 45

Includes centromere

Answer 46

Unlike deletions/duplications: No change in DNA quantity Only order changes. Effect depends on breakpoint location.

Answer 47

No gene disrupted No phenotype

Answer 48

Gene disrupted Loss of function

Answer 49

Fusion gene created Example: A / D fusion Possible outcomes: Both genes defective Novel gene product Oncogenic fusion (tumorigenic potential)

Answer 50

DNA has two strands. If: Entire gene + promoter + regulatory elements are inside inversion Then: Orientation relative to promoter remains correct Gene transcribes normally If: Promoter outside inversion but coding region inside Then: Transcription disrupted Gene read from wrong strand Important: All regulatory elements must remain intact. Scale matters: Chromosome is huge. Genes are small units within it.

Answer 51

An inversion is a chromosomal rearrangement where a segment flips orientation after breakage and incorrect rejoining.

Answer 52

If both breakpoints fall between genes → no genes disrupted (often no detectable somatic effect), unless distant cis-regulatory sequences are affected. If one breakpoint lands within a gene → that gene can be disrupted. If both breakpoints land within genes → can create gene fusions (e.g., A/D fusion example).

Answer 53

Because transcription direction is driven by the promoter + gene as a unit: typically the inversion flips regulatory elements + gene together, so from the gene’s perspective, orientation is “internally consistent.” Only an unlikely breakpoint between promoter and gene would risk changing which strand is transcribed

Answer 54

Start with normal sequence → DNA breaks → inverted alignment of the broken segment → rejoining in the inverted orientation to complete the inversion

Answer 55

In prophase I, homologs undergo synapsis (pairing along their length) and recombination/crossing over. If one homolog carries an inversion, pairing requires an inversion loop, and crossover inside that loop can produce abnormal chromatids

Answer 56

In an inversion heterozygote, homologs try to align homologous sequences, so the inverted region must loop out so that matching segments can pair (normal vs inversion alignment)

Answer 57

Crossover in a paracentric inversion heterozygote produces a dicentric chromatid (2 centromeres) and an acentric fragment (0 centromeres)

Answer 58

A: The dicentric bridge gets pulled toward opposite poles → breaks randomly; the acentric fragment is lost, yielding chromatids with major deletions (only the non-recombinant chromatids remain “normal/inversion” intact).

Answer 59

A: Recombinant chromatids don’t become dicentric bridges the same way, but they still end up with large duplications and deletions (shown as duplication/deletion of different “arms”); these are typically nonviable products.

Answer 60

A: With a single crossover inside the inversion loop, 2 chromatids recombine → become abnormal (dup/del; dicentric/acentric outcomes depending on type), while the 2 non-recombinant chromatids remain intact (one “normal arrangement,” one “inversion arrangement”).

Answer 61

A: Often no somatic symptoms, but reduced fertility/infertility risk due to production of unbalanced gametes during meiosis (the “Think Break” prompt).

Answer 62

A: Normally, recombinant progeny frequency reflects distance (more distance → more recombinants). But if an inversion spans the interval, crossovers still occur yet recombinant chromatids often become inviable, so recombinants are under-counted in progeny.

Answer 63

A: dp and cn are 45 map units apart (expect ~45% recombinants), but the example shows very few recombinants (e.g., only small counts in single-trait classes) → consistent with an inversion causing recombinant gametes to be nonviable.

Answer 64

A: Crossovers can occur outside the inversion loop (outside the inverted region), producing viable recombinants. Also, inversion size matters: larger inversion → higher probability a crossover lands inside it → more “missing” recombinants in counts.

Answer 65

A: “50% defective” describes products of a single meiosis when a crossover lands in the inversion. “Observed recombination frequency” is based only on viable offspring you can count, so crossovers that yield inviable gametes still happened but don’t appear in progeny tables.

Answer 66

A: A translocation is breakage + improper rejoining between different chromosomes. A balanced translocation is an “even swap” with no net loss of genetic material (material exchanged reciprocally).

Answer 67

A: If there’s no loss of DNA and breakpoints don’t disrupt essential sequences, somatic phenotype may be normal. But breakpoints can land in genes and create fusion genes; historically, a translocation was the first chromosomal abnormality linked to a specific human cancer (breakpoint-driven oncogenic fusion concept).

Answer 68

A: Homologous regions align, producing a cross-shaped pairing configuration. The risk is segregation pattern: one pattern yields complete gametes; another yields duplications/deletions.

Answer 69

Alternate segregation → both products complete (balanced genetic content). Adjacent segregation → both products incomplete (duplications + deletions).

Answer 70

A: A Robertsonian translocation is a balanced translocation involving acrocentric chromosomes (13, 14, 15, 21, 22). Frequency ~1/1000 babies; phenotype usually normal

Answer 71

A: The long arms fuse into a translocation chromosome; the small reciprocal fragment can be lost. Carriers still retain a complete functional complement of essential genetic content (implies the lost short-arm material is not dosage-critical; also breakpoints didn’t disrupt essential genes).

Answer 72

A: They may carry the translocation chromosome + two normal homologs (effectively “3 chromosomes” in that set), but still have two copies of the important long-arm genetic material, so somatic phenotype is often normal

Answer 73

A: There are 3 relevant chromosomes (two normal + one translocation). Segregation can package chromosomes into gametes in multiple ways; the slide summarizes this as “Segregation during meiosis → 6 possible gametes.”

Answer 74

A: Some gamete combinations create 2N+1 for chromosome 21 material (a trisomy for 21 content) → Down syndrome. Unlike nondisjunction “random” cases, this can be familial/inherited because the parent is a translocation carrier producing recurrent unbalanced gametes.

Answer 75

A: Visible pigment traits; and maize reproduction includes double fertilization: each pollen grain has two identical haploid sperm—one fertilizes the egg (n) and one fertilizes the central cell (2n).

Answer 76

A: (1) Discovery of transposable elements via unstable alleles, (2) Impact of transposable elements on the genome, (3) Genomic surveillance mechanisms that limit mobility.

Answer 77

A: C promotes purple pigment in aleurone; c is recessive and can’t make pigment; C1 is a dominant inhibitor that represses pigment. Dominance: C1 > C > c.

Answer 78

A: Using pollen from a C1 (dominant inhibitor) strain to fertilize C/C ovules → predicted all colorless kernels, but one pollen parent produced kernels with blue pigment in sectors (“unstable” pattern).

Answer 79

A: Chromosome breakage causes loss of the C1 allele in some developing endosperm cells → C1 no longer inhibits → C can function locally → pigmented sectors. Evidence included high breakage rates in that strain.

Answer 80

A: Because breakage was observed to occur at the same place, the breakpoint was named Ds; observed genetic instability is due to chromosome breaks at specific locations.

Answer 81

A: Failure to complement (called out explicitly on the slide).

Answer 82

A: In related strains, breaks occur at a different location, implying Ds moved (new location inferred from sector phenotypes). Caveat: not all Ds insertion sites are associated with chromosome breaks.

Answer 83

A: A separate genetic element, Ac, is required for breaks to occur and for Ds movement/excision; Ds is non-autonomous (needs Ac), while Ac is autonomous (can “jump out” on its own).

Answer 84

A: “Jump out” = excise/transpose from a locus. In a described Ds type (c-Ds), excision can allow chromosome ends to be joined, restoring a functional dominant C allele.

Answer 85

A: Timing of Ds excision during endosperm development: Early jump-out → many descendant cells inherit restored C → large spot. Late jump-out → fewer descendants → small spot

Answer 86

in the human genome may harbor hints about the onset of a rare, inherited neurodegenerative disorder called ataxia- telangiectasia as well as other related diseases, a new Yale School of Medicine study finds. The findings also point to possible directions for treatment of the disease

Answer 87

through study of “unstable” alleles

Answer 88

Each pollen grain contains two identical haploid sperm cells

Answer 89

gene is important for the production of purple pigment in maize. c is a recessive allele that is unable to make pigment. C1 (dominant negative) is a dominant inhibitor allele that represses pigment production

Answer 90

Predicted result: all colorless kernels unexpected result from one particular “pollen parent”: some kernels have blue pigment! In sectors!

Answer 91

Careful observation of “unstable” pigment gene alleles led to the discovery of transposable elements Hypothesis 1: C1 allele is lost in some cells of the developing endosperm (“if these cells can produce pigment, then maybe C1 is no longer inhibiting pigment production, e.g. maybe it’s just not there”) Experiment: look for breakage in chromosomes of the pollen parent Result: high rate of breakage in this strain Interpretation: loss of the C1 allele is caused by chromosome breakage. (due to chromosome rearrangement getting end of the dominant negative allele)

Answer 92

chromosome breaks at specific locations MEMORIZE SLIDE

Answer 93

What term can we use to describe what’s happening in these sectors? FAILURE TO COMPLEMENT- >chromosomal deletions due to breakage - >failure to complement

Answer 94

Breaks occur in a different location in a related strain Not all Ds element insertion sites are associated with chromosome breaks! A genetic element called “Ac” is required for breaks to occur Interpretation: In some strains, Ds has moved to a new location. This new location can be inferred from the phenotype of the sectors. Observation: in these sectors, pigment is restored but other kernel traits (Sh, Bz, Wx) remain unchanged

Answer 95

Movement of the Ds element depends on another locus: the Ac element MEMORIZE SLIDE

Answer 96

Requires Ac to “jump out” (aka “excise” or “transpose”

Answer 97

Ac jumps out on its own *The type of Ds is found in the c-Ds allele has a different structure. When the Ds in c-Ds excises, the two ends of the chromosome are joined together, restoring a functional dominant C allele.

Answer 98

In the presence of an Ac element, Ds elements can mobilize. Jumping out: frequent Jumping in: rare In the example of the C pigment gene, Ds mobility can: * revert the c(Ds) mutant allele * generate a mutant allele of C

Answer 99

Transposable elements are everywhere. Not just a weird plant thing! Transposons, or “transposable elements (TEs)” are ancient, found in all organisms: bacteria, plants, yeast, multicellular animals (including humans). TEs have contributed to placental evolution by inserting themselves into genes, enabling new functions and providing alternative promoters or enhancers. At the individual level, there is evidence that these TE-influenced regulatory elements play a critical role in placental function. Studies in maize identified transposable elements genetically, but it was later studies in yeast and Drosophila that identified the actual pieces of DNA that are mobile and discovered how they move

Answer 100

Class 1: Retrotransposons (eukaryotes) * Class 2: DNA transposons (prokaryotes and eukaryotes)

Answer 101

There are multiple transposon families in maize, each with autonomous and non-autonomous elements. Ac and Ds are in the same family. Each has a transposase version that is specific for its own family.

Answer 102

Similar to retroviruses * First identified in yeast as a repeated sequence in the genome. * Found in many organisms

Answer 103

* LINE (long interspersed element) is a retrotransposon found in the human genome.

Answer 104

* LINE1 (L1), the only autonomous transposable element in humans, comprises almost 17% of the whole human genome.

Answer 105

Retrotransposons move via an RNA intermediate. * Transposition is mediated by reverse transcriptase (encoded by the retrotransposon pol gene), not transposase. * This is a replicative (“copy-paste”) mechanism. After transcription (step1) LINE1 RNAs are translated (step 2) to produce two proteins, ORF1p and ORF2p, which are essential for mobility * ORF1p is an RNA-binding protein with nucleic acid chaperone activity that binds to LINE1 transcript * ORF2p possesses endonuclease and reverse transcriptase activities

Answer 106

The lecture frames genomes as dynamic because they contain mobile DNA (transposable elements, TEs) that can move within genomes. The goals are: (1) how TEs were discovered via unstable alleles, (2) how TE mobility impacts genomes over evolution and within individuals, and (3) the genomic surveillance mechanisms that limit mobility. Slides also emphasize TE ubiquity across life + that TE insertions can contribute regulatory novelty (ex: placenta)

Answer 107

Transposable elements (TEs) are mobile DNA sequences found in all organisms (bacteria, plants, yeast, animals incl. humans). Early genetic discovery came from maize unstable alleles, but later work (notably yeast and Drosophila) identified the actual mobile DNA pieces and helped elucidate movement mechanisms. TE insertions can also create new regulatory elements (alternative promoters/enhancers), with noted relevance to placental evolution/function

Answer 108

Class 1 = Retrotransposons (common in eukaryotes): move through an RNA intermediate and use reverse transcriptase (from a pol gene), producing a replicative “copy-paste” outcome. Class 2 = DNA transposons (found in prokaryotes + eukaryotes): typically encode a transposase and move by DNA excision/integration, classic “cut-and-paste.”

Answer 109

Class 2 elements have terminal/flanking sequences recognized by transposase, which excises and reinserts the element (cut-and-paste). Maize contains multiple transposon families, each with autonomous (encode functional transposase) and non-autonomous elements (missing key coding capacity but can still move if transposase is provided in trans). Ac and Ds are in the same family; each family’s transposase is specific for its own family

Answer 110

Retrotransposons: 1. Cellular RNA pol transcribes TE → RNA intermediate 2. TE RNA translated to make proteins (incl. reverse transcriptase) 3. Reverse transcriptase makes DNA copy 4. DNA inserts at a new genomic location → copy-paste expansion. Key: transposition uses reverse transcriptase (pol), not transposase

Answer 111

After transcription (step 1), LINE1 RNAs are translated (step 2) to produce ORF1p and ORF2p, essential for mobility

Answer 112

RNA-binding protein with nucleic acid chaperone activity; binds the LINE1 transcript

Answer 113

has endonuclease + reverse transcriptase activities (cuts target DNA and synthesizes DNA from RNA)

Answer 114

LINE1 (L1) is a human retrotransposon; LINE1 (L1) is the only autonomous TE in humans and comprises ~17% of the human genome (slide value). Transcript also discussed other TE categories (SINEs, DNA transposons) and the broader idea that TE-derived DNA vastly exceeds protein-coding DNA; the slides echo that TE-derived sequence is a major genome fraction and frame TE accumulation as a defining feature of large genomes

Answer 115

A new insertion is inherited only if it occurs in the germline lineage (cells that become sperm/eggs). If it occurs in somatic cells, it can affect the individual (mutation/mosaicism) but is not passed to offspring. The lecture explicitly pauses to make you map the cartoon mechanism onto an organism: inheritance depends on cell type (germline vs somatic)

Answer 116

Plant genomes show that genome size differences often reflect TE content (slide theme: “Transposable elements in plants account for differences in genome size”). TEs are usually found in introns or intergenic regions (not typically in coding exons), consistent with selection against strongly disruptive insertions and/or insertion preferences

Answer 117

The Vvmyb1A gene is required for purple pigment (Cabernet). An insertion of Gret1 (an LTR retrotransposon) produced a loss-of-function allele (Chardonnay: loss of color). A later rearrangement in Gret1 generated revertant colored grapes (ex: Ruby Okuyama)

Answer 118

It illustrates that a TE insertion can create a new phenotype that becomes adaptive under environmental change, allowing the allele to rise in frequency via natural selection (classic industrial melanism story: differential camouflage → predator avoidance → allele frequency shift). The lecture uses this as a “relatable” case where TE-generated genetic variation can be filtered by selection

Answer 119

About 1/600 spontaneous mutations causing important human diseases result from LINE or SINE transposition. Two mechanistic routes highlighted: Exon/coding disruption: L1Hs insertion interrupts a coding exon of factor VIII (F8) → haemophilia A (loss of coding integrity). Splicing disruption via regulatory elements: AluYa5 insertion in Dent disease disrupts an exonic splice enhancer (ESE) → exon 11 skipping, introducing a stop codon in exon 12. (Your transcript also notes the prof corrected herself: this case is exonic and primarily a splicing-regulatory defect rather than “pure intron insertion.”)

Answer 120

Many cancers (lung/colon/pancreatic/ovarian) are associated with markers for LINE-1 mobility; a LINE-1 activity marker (brown) appears in cancerous tissue but not normal tissue. Hypothesis: cancer cells may have deregulated transposon control, leading to elevated TE transcription/retrotransposition, which can drive genome instability and accumulate mutations that contribute to tumorigenesis.

Answer 121

TE mobility can cause mutations + genome instability. Cells evolved mechanisms to repress TE mobility and protect genome integrity

Answer 122

P-elements were discovered by crossing lab strains with wild-caught strains; in the dysgenic direction, germline fails (atrophic gonads/sterility), with high mutation, abnormal chromosomes, nondisjunction—yet somatic development can look okay. The reciprocal cross yields fertile offspring, implying a protective mechanism exists that blocks P-element mobility in one direction

Answer 123

piRNAs repress transposon mobility in animal germlines. They are short single-stranded RNAs (26–30 nt in mammals) that load into a protein complex to destroy complementary target RNAs. Unlike siRNAs, piRNAs do not start as dsRNA; instead they originate from long transcripts from pi-clusters Directionality: P-strain (wild) females have P elements in their pi-cluster, so their eggs contain piRNAs that block P-element transposition in the embryo germline → F1 fertile. M-strain (lab) females lack P elements in pi-clusters → eggs lack piRNAs; if the male introduces P elements, they can mobilize → germline genome instability → sterility

Answer 124

Use a genetic screen: mutagenize randomly, then select mutants where Tc1 mobility is no longer repressed. If knocking out a gene allows Tc1 to move, infer that gene normally participates in TE silencing. Assay: Tc1 insertion in unc-22 disrupts unc-22 → twitching/uncoordinated worms. If Tc1 mobilizes (exits unc-22), unc-22 function is restored → smooth-gliding worms. This gives a clean phenotypic readout of “TE mobility ON.”

Answer 125

Screens identified 25 genes blocking Tc1 mobility; many are RNAi silencing pathway genes. RNAi logic: short (~21–25 nt) RNAs derived from dsRNA precursors; processing includes cleavage by Dicer. Mature small RNAs load into RISC, which recognizes mRNA targets by complementarity and either promotes degradation or represses translation. miRNA is endogenous gene-encoded; siRNA typically targets foreign nucleic acids

Answer 126

If a Tc1 inserts near a cellular gene, “read-through transcription” can transcribe both ends of Tc1, including 54-bp terminal inverted repeats (TIRs). The Tc1 RNA can fold into dsRNA, generating a silencing trigger. This is presented as an example of genome surveillance: detect TE sequences and prevent their mobility.

Answer 127

Transposable elements (TEs) are mobile DNA sequences that can excise or copy themselves and insert elsewhere, so TE mobility can cause mutations and genome instability (breaks, rearrangements, disrupted genes/regulatory regions). Cells therefore evolved mechanisms that repress TE mobility to protect genome integrity. Key framing: understand not only the mechanisms but also the experiments, results, and interpretations used to discover them.

Answer 128

Crossing lab strain flies with wild-caught flies (both D. melanogaster) could destabilize hybrid progeny genomes because transposable elements moved. Crucially, instability occurred in one cross direction: when the lab strain parent was the female and wild-caught parent was the male; the reciprocal cross did not produce instability. Interpretation: something in the egg cytoplasm (maternal contribution) can silence P-element mobility in the embryonic germ line, producing a normal germ line and fertility when present

Answer 129

C. elegans has endogenous TEs including Tc1, an autonomous DNA transposon analogous to the maize Ac element. Tc1 contains a transposase gene (enzyme enabling excision/insertion) and terminal repeats at both ends; the lab strain genome contains ~32 Tc1 copies. Key observation: Tc1 transposes in somatic cells but not in germline cells. This implies Tc1 is capable of moving, but germline cells prevent it

Answer 130

A: Interpretation: the transposase gene in each Tc1 copy is silenced in germline cells, so Tc1 cannot mobilize there. Hypothesis: germline cells contain silencing machinery that represses Tc1 mobility; components of this machinery are encoded by genes. Important insight: if genes encode the machinery, then mutating/knocking out those genes should cause silencing failure and allow Tc1 mobility

Answer 131

A: Rationale: If you knock out a gene required to repress TE mobility, TEs will be able to move. But if you don’t know which genes are involved, you can’t do targeted knockouts. So you mutate genes at random (chemical mutagen), then screen for mutants where Tc1 becomes mobile. Inference: the mutation likely knocked out a host silencing gene (not Tc1 itself and not unc-22 itself). This is a genetic screen.

Answer 132

A: Starting strain: a nonfunctional unc-22 allele caused by a Tc1 insertion inside unc-22, disrupting unc-22 function and producing an uncoordinated/twitching worm. Normally Tc1 mobility is repressed in the germ line, so progeny remain twitching. If silencing is blocked, Tc1 can excise (“jump out”) in the germ line; the unc-22 gene becomes re-joined/functional and worms become smooth-gliding (wild-type swimming). This is an easy visual readout on plates: most worms twitch; rare revertants glide smoothly and are easy to isolate and breed into families for analysis

Answer 133

The smooth-swimming revertants arise because the mutagen induced a mutation in another gene (a host factor) required for Tc1 silencing, allowing Tc1 to excise from unc-22 and restore unc-22 function. It is not primarily interpreted as the mutagen directly “fixing” unc-22, nor as “mutating Tc1” to cause reversion; instead, the screen is designed to find loss of repression.

Answer 134

A: Not necessarily; multiple genes can contribute to the silencing pathway, so different revertants may carry mutations in different genes. You test this with a complementation test (classic genetics): cross mutants and see whether the phenotype complements. Researchers identified ~25 different genes whose mutation allowed Tc1 excision/mobility in the germ line; many map to the RNAi silencing pathway

Answer 135

A: The pathway uses short ~21–25 nt RNAs derived from double-stranded RNA (dsRNA) precursors. dsRNA is processed by Dicer into small dsRNA fragments; one strand becomes the guide and loads into RISC (RNA-induced silencing complex). Guide RNA provides sequence complementarity so RISC targets matching mRNAs to promote degradation or repress translation. Framing: miRNA derives from endogenous genes; siRNA often derives from foreign sequences (transgenes, viruses, TEs). In Tc1 control, loss of any key component can break silencing and allow mobility.

Answer 136

Model: with ~32 Tc1 copies, a few Tc1 elements near cellular genes can be transcribed by read-through transcription from a neighboring gene. In that case, both ends of Tc1 get included in the RNA. Tc1 ends contain 54-bp terminal inverted repeats (TIRs); when both TIRs are present in the same RNA, they can base-pair and the Tc1 RNA spontaneously forms dsRNA (hairpin with intervening region looping out). dsRNA triggers Dicer → siRNAs → RISC loaded with Tc1 sequence, which then recognizes and chops up complementary Tc1/transposase transcripts from anywhere in the genome. Outcome: no transposase protein → no excision → no mobility in germ line. Hypothesized nuance: Tc1 can still transpose in somatic cells because RNAi is less efficient there, allowing some transposase production. Key trigger: read-through transcription that includes both inverted repeats (often when Tc1 inserts next to a cellular gene).

Answer 137

A: In Drosophila (and other animals), germline TE repression uses piRNAs (Piwi-interacting RNAs), a small-RNA mechanism conceptually similar to RNAi but distinct in origin. piRNAs are short single-stranded RNAs (not produced from dsRNA precursors like siRNAs; mammals noted as ~26–30 nt in slides). They originate from long transcripts from specialized genomic regions called pi-clusters (often heterochromatic) that contain clusters of inserted transposable element fragments. These long RNAs are processed into piRNAs, which associate with a piwi-Argonaute–containing complex that recognizes complementary TE mRNAs by base pairing and degrades them, thereby silencing TE mobility across the genome. Genome surveillance logic: an active TE is “recognized” effectively when it happens to insert into a pi-cluster, which then generates piRNAs against it.

Answer 138

Wild-caught female (P strain) has P elements present in her pi-cluster, so she produces piRNAs and piRNA destruction complexes that are deposited into the egg cytoplasm. When crossed, the embryo’s germ line is protected: P-element transcripts are targeted and degraded, TE mobility stays silenced, and F1 are fertile. Lab strain female (M strain) lacks P elements in her pi-cluster, so her eggs do not contain anti-P piRNAs/complexes. If P elements arrive via the male genome, the embryo’s germ line lacks the maternally provided silencing program, allowing P-element mobility → genome instability → atrophic gonads and sterility (hybrid dysgenesis)

Answer 139

Hypothesis: p53 represses retrotransposon mobility; when p53 is lost/mutated, retrotransposons become deregulated, increasing transcription/mobility, which contributes to genome instability. Genome instability can drive cellular responses including cancer (and potentially inflammation/other sporadic disease processes). p53 is a major tumor suppressor: senses cellular stress/damage, halts division or triggers cell death; inherited loss of one functional copy predisposes to cancer (e.g., Li-Fraumeni syndrome), and p53 mutations are common across tumor types

Answer 140

In Wilms (kidney) tumor samples, researchers compared tumors with mutant p53 vs intact p53. Staining: blue marks DNA; green detects LINE-1 ORF1p (ORF1 protein from an autonomous LINE-1 retrotransposon). Tumors mutant for p53 showed dramatically elevated LINE-1 ORF1p expression relative to p53-wild-type tumors. This supports correlation consistent with the hypothesis that loss of p53 is associated with derepressed retrotransposon activity, but by itself does not prove causation

Answer 141

RNA transcripts from the Drosophila retrotransposon TAHRE. In wild type (WT) flies, transcript levels are low. In p53−/− flies, TAHRE transcript levels are higher (loss of repression). Adding back a transgenic wild-type Drosophila p53 in p53−/− flies (Dp53 rescue) restores repression (transcripts reduced). Adding back wild-type human p53 also represses TAHRE in p53−/− flies (functional substitution; shown with replicate rescues). In contrast, introducing human p53 alleles from cancer patients (tumor-associated mutants; shown as black bars) fails to repress transposon activity. Interpretation: p53 activity is required for silencing retrotransposon activity in flies; human p53 can perform this role, but many cancer-associated p53 variants cannot, supporting the idea that loss of this repression function could contribute to genome instability in tumors

Answer 142

A mutation is a change in DNA sequence; more broadly, any unpredictable change in the structure or amount of DNA can be called a mutation. Mutations can be detrimental, neutral, or beneficial; can affect coding or regulatory sequences; and are a key source of genetic variation

Answer 143

Mutations occur on different scales: chromosomal mutations (gain/loss of all/part of a chromosome), point mutations (single nucleotide changes or small indels), and insertional mutations (large insertions such as transposable elements)

Answer 144

Most mutations occur in somatic cells and are not inherited; only mutations occurring during formation of gametes (germ line) can be inherited

Answer 145

DNA double helix: sugar-phosphate backbones on the outside; purines/pyrimidines form rungs. Strands are antiparallel. Strand directionality defined by sugar carbon numbering (1–5): the 5′ end has carbon #5 not bound to another nucleotide; the 3′ end has carbon #3 not bound to another nucleotide. Purines: A, G. Pyrimidines: C, T

Answer 146

A: An ORF is a stretch of DNA that does not contain a stop codon. A codon is a trinucleotide in DNA/RNA encoding an amino acid or a termination signal.

Answer 147

A new germline mutation typically occurs on one homolog (rare event; not the same site on both homologs). The individual is then heterozygous, so ~50% of gametes carry the mutant homolog, giving ~50% of offspring inheriting it (phenotype depends on dominance/recessiveness).

Answer 148

Transition: purine→purine or pyrimidine→pyrimidine (example GC→AT). Transversion: pyrimidine→purine or purine→pyrimidine (example CG→TA).

Answer 149

Indels are small base additions/deletions. If they’re only 1–2 nucleotides, they’re still considered point mutations (even though large insertions/deletions exist at bigger scales)

Answer 150

Synonymous (silent): changes codon but same amino acid. Nonsynonymous (missense): codon changes to encode a different amino acid. Nonsense: codon becomes stop codon (UAA/UAG/UGA); effect depends on distance from the 3′ end of ORF and can trigger NMD (mRNA degradation). Frameshift: indel shifts reading frame for all downstream codons.

Answer 151

Conservative = chemically similar amino acid substitution (example Lys→Arg; both charged, more similar). Non-conservative = chemically different substitution (example Lys→Thr; higher chance of altering structure/function).

Answer 152

Because the ribosome reads triplets; an indel not in multiples of 3 shifts the reading frame, changing every downstream codon, often creating/eliminating stop codons and producing a completely different (often truncated) protein

Answer 153

Wild type function: no effect on gene function.

Answer 154

Hypomorphic: partial/weak; protein retains some function or is produced at reduced level. Null: protein nonfunctional or not produced.

Answer 155

Hypermorphic: protein hyperactive. Ectopic: more protein or made in wrong time/place (often regulatory). Neomorphic: new function gained.

Answer 156

A: G12V “locks” Ras in its active conformation, making it constitutively active, driving uncontrolled proliferation that can lead to cancer. Approximately 30% of human cancers are driven by Ras mutations (per slide note).

Answer 157

Cave crayfish bright colors: proposed to have evolved purely by chance, not necessarily for attracting mates/warning predators (possible byproduct/neutral). Lactose tolerance: mutation arose randomly, then conferred advantage; lactase “on” mutations are common in modern Europeans but not ancestors; as late as 5000 BC many ancient Europeans likely couldn’t digest milk as adults.

Answer 158

A: Noncoding mutations can disrupt transcription, splicing, stability, translation, function, etc.

Answer 159

Mutations in promoters/enhancers can alter binding of general or specific transcription factors, changing transcription levels; severe cases can yield reduced or absent gene expression (problem at transcription, not translation).

Answer 160

A mutation can create a new splice donor site (a sequence that looks like a splice site) so splicing occurs at the wrong location. A mutation can inactivate/destroy a splice donor, causing intron retention (effectively inserting intronic sequence into mRNA), potentially triggering frameshifts.

Answer 161

A: While canonical intron ends are key, additional sequences in introns and exons also recruit/position the spliceosome (example mentioned: exonic splicing enhancers), so mutations in coding or noncoding regions can still disrupt splicing.

Answer 162

Observation: a kidney cancer patient with PBRM1 mutation had a much worse outcome than expected. Hypothesis: they also had another mutation worsening outcome. They found a synonymous mutation in BAP1, a tumor suppressor.

Answer 163

A: The synonymous mutation occurred in an exonic splicing enhancer, causing exon 11 skipping. Exon 10 spliced directly to exon 12; if the skipped exon length is not a multiple of 3, the ORF becomes out of register, producing a frameshift and early stop, and the resulting mRNA/protein become unstable/degraded. Net effect: the “silent” change behaved like a loss-of-function / effectively null allele and associated with worse prognosis.

Answer 164

If the skipped exon is, e.g., 30 nt, that’s a multiple of 3, so the frame can remain intact when adjacent exons join. If it’s 29 nt, you’re missing 1 nt relative to triplet reading, so splicing exon 10 to exon 12 shifts the reading frame downstream.

Answer 165

A: Northern (N) measures RNA (mRNA); Western (W) measures protein. Electrophoretic mobility/position reflects size (smaller runs farther/faster), while band intensity reflects amount present.

Answer 166

Missense: usually same size mRNA/protein; may still show normal bands (function can still be altered). Nonsense: often smaller protein (runs farther), but can also trigger NMD → little/no mRNA and no protein. Frameshift: size change is hard to predict (could be shorter via early stop or different length if stop lost); migration can vary. Regulatory/splicing/promoter: can yield reduced/absent RNA and thus reduced/absent protein.

Answer 167

A: DNA replication errors, spontaneous chemical changes to DNA, and mutations induced by mutagens.

Answer 168

Bases can transiently shift into alternative chemical forms (and/or ionized states) that change H-bonding so they mispair during replication

Answer 169

are isomers differing in atom positions/bonds and exist in equilibrium. If proofreading doesn’t fix mismatches, they can produce transition mutations

Answer 170

A: During replication (especially in repetitive sequences, like runs of the same base), strands can misalign and form a loop. If the newly synthesized strand loops out, replication can incorporate extra bases → insertion. If the template strand loops out, polymerase skips bases → deletion. Which strand loops out determines insertion vs deletion.

Answer 171

A: An incorporated error (mismatch after one replication event) is still repairable. If it is not repaired before the next round of replication, it can become a replicated error, which is permanent (the incorrect base becomes the template for “matching” incorporation).

Answer 172

A: In repeats like (CAG)_n, slippage can cause a segment of the template to be copied twice, expanding repeat number in the daughter strand. Greater repeat number (greater n) increases the chance of greater slippage, promoting further expansion.

Answer 173

* DNA replication errors * Spontaneous chemical changes to DNA * Mutations induce by mutagens

Answer 174

Base mispairing Strand slippage

Answer 175

Bases are chemicals that can exist in different forms, such as tautomers or ionized forms. These altered forms can have different base-pairing properties, so if DNA polymerase copies DNA while a base is in one of these altered forms, it may insert the wrong nucleotide Tautomerization and ionization of bases leads to mismatched base pairing, which can generate a permanent mutation

Answer 176

A base such as guanine can transiently exist in an alternative form, such as an enol form, that pairs incorrectly. DNA polymerase then adds the wrong base, and after further rounds of replication, that error can become a stable permanent mutation.

Answer 177

Strand slippage is a replication error in which the template strand and the newly synthesized daughter strand become misaligned during DNA replication. If one strand loops out during replication, DNA polymerase may either: copy extra nucleotides, causing an insertion, or fail to copy some nucleotides properly, causing a deletion repeated DNA sequences provide multiple similar complementary pairing possibilities, making misalignment easier during replication

Answer 178

They are repeated DNA sequences made of three nucleotides repeated in tandem, such as CAG CAG CAG...

Answer 179

base mispairing and strand slippage. In base mispairing, a base transiently exists in an altered chemical form, such as through tautomerization or ionization, which changes its pairing behavior, causing DNA polymerase to insert the wrong nucleotide. In strand slippage, the newly synthesized strand or the template strand misaligns with the other during replication, allowing a loop to form; this can cause DNA polymerase to insert extra nucleotides or omit some, leading to small insertions or deletions.

Answer 180

Repeated sequences are especially prone to slippage because repeated motifs provide multiple equivalent pairing options, making it easier for the template and daughter strands to become misaligned during replication. In a trinucleotide repeat region such as (CAG)n, part of the template can loop out. DNA polymerase then fails to recognize that it has already copied that section and recopies the looped region, increasing repeat number in the daughter strand. This creates a self-reinforcing cycle: more repeats → more opportunity for slippage → more repeat expansion

Answer 181

Trinucleotide repeat diseases are disorders caused by the expansion of repeated three-base motifs in DNA, often through strand slippage during replication. The lecture highlighted several disorders caused by expanded CAG tracts, including Huntington’s disease, spinobulbar muscular atrophy (Kennedy disease), DRPLA, and several spinocerebellar ataxias

Answer 182

Since CAG encodes glutamine (Q), proteins produced from these expanded coding repeats contain abnormally long polyglutamine (polyQ) tracts. These expanded polyQ tracts tend to fold abnormally, form aggregates, and promote neural dysfunction and degeneration, explaining why many of these disorders are neurological

Answer 183

Trinucleotide repeat expansion can cause disease even if it occurs outside the coding region. In Fragile X syndrome, expansion of CGG repeats in the 5′ UTR / non-coding regulatory region of FMR1 alters gene regulation rather than protein sequence. The slide showed that increasing repeat number increases instability and the likelihood of further expansion, and once the repeat reaches a large size, the surrounding region becomes methylated, especially in the promoter/CpG island region. This methylation blocks transcription of FMR1, so the mutation acts as a loss-of-function mutation due to transcriptional silencing, not because of a nonsense mutation or coding-sequence disruption

Answer 184

Huntington’s disease is a classic example of repeat expansion in a coding region, specifically in the HTT gene, where the disease is autosomal dominant. The key genotype-phenotype relationship is that more repeats correlate with earlier age of onset and faster progression of symptoms. The lecture described the classical pedigree observation called anticipation, where the phenotype becomes worse over generations—symptoms appear earlier and are more severe. This makes sense molecularly because repeat expansion tends to increase across generations, especially once repeat number is already elevated. The slide categorized HTT alleles approximately as normal (<26 repeats), mutable normal (27–35), reduced penetrance (36–39), and disease-associated (>40)

Answer 185

Deamination is the spontaneous removal of an amino group from a base. The key example emphasized was cytosine deamination, which converts cytosine into uracil

Answer 186

Because uracil pairs like thymine, it pairs with adenine instead of guanine. So if cytosine is deaminated and not repaired before replication, DNA polymerase inserts A opposite the uracil. At this stage, the incorporated mismatch is still potentially fixable. But once that A-containing strand is used as a template in the next round, DNA polymerase inserts T opposite A, converting the original C·G pair into a T·A pair. This is therefore a C·G → T·A transition

Answer 187

Can lead to sequence errors during replication. 10,000 purines lost per cell in a day! (usually repaired) Depurination is the spontaneous loss of an entire purine base from DNA, leaving behind an apurinic (AP) site

Answer 188

This is mutagenic because during replication there is no base available for proper pairing, so DNA polymerase may either stall or insert a nucleotide more or less arbitrarily—commonly A. The lecture emphasized that this can lead to sequence errors during replication and that depurination is surprisingly frequent, with about 10,000 purines lost per cell per day, though most are normally repaired. If the polymerase inserts an incorrect base opposite the AP site and that base is later copied in the next replication round, the result becomes a permanent mutation in both descendant strands. Again, the mutation comes not from the loss alone, but from the replication of the damaged site

Answer 189

Mutagens induce mutations by different mechanisms. * alter a base so that it mispairs with another base (e.g. alkylating agents) * alter a base, blocking DNA replication * replace a base in DNA, i.e. get incorporated instead of an actual base (base analogs) * damage a base so that it can no longer base pair with any base (e.g. UV light, ionizing radiation

Answer 190

Mutagens can induce mispairing or damage/non-recognition

Answer 191

* Can be incorporated into DNA in place of normal bases * Can be less stable, shifting their base-pairing affinities and leading to nucleotide changes. * Example: 2-aminopurine is an analog of A, but when protonated can mispair with C, leading to transitions

Answer 192

* UVB light can induce covalent interactions between adjacent pyrimidines * If not repaired, this can block DNA replication UVB light can induce covalent interactions between adjacent pyrimidines, especially adjacent thymines, producing lesions such as a cyclobutane pyrimidine dimer (thymine dimer) or a 6-4 photoproduct. These lesions distort DNA structure and can block DNA replication and transcription if not repaired. Unlike a simple altered base that predictably mispairs, UV-induced lesions often create a more severe structural problem, so the mutational outcome may depend on how the lesion is repaired or bypassed

Answer 193

Alkylating agents add alkyl groups to bases, changing their pairing properties so they mispair and produce base substitutions, often transitions. Base analogs are chemicals that resemble normal bases closely enough to be incorporated by DNA polymerase during replication. Once incorporated, they may be chemically unstable or able to adopt alternative forms, shifting their base-pairing affinities. The lecture’s example was 2-aminopurine (2-AP), an analog of adenine. It can be incorporated in place of A, but in altered form, especially when protonated, it can mispair with C, causing transition mutations

Answer 194

Intercalating agents are planar molecules that mimic the shape of base pairs and slide between stacked bases in the DNA double helix

Answer 195

Because they insert into the helix, they disrupt normal spacing, which can trick DNA polymerase during replication. The polymerase may treat the intercalated molecule as if an extra base pair were present and insert an extra nucleotide, or it may skip one, leading to single-nucleotide insertions or deletions rather than ordinary base substitutions

Answer 196

The Ames test is a functional assay used to determine whether a compound is mutagenic. It starts with a his− strain of Salmonella, which cannot grow without histidine because it has a defective histidine biosynthesis gene. When plated on medium containing only a very low amount of histidine, most cells cannot continue growing. However, a few rare cells undergo spontaneous mutations that restore histidine function; these survivors form visible colonies called revertants. The test’s logic is that if exposure to a chemical increases the number of revertant colonies above background, that chemical must be causing mutations. It therefore detects mutagenicity functionally rather than by directly sequencing DNA

Answer 197

Even without an added mutagen, a few colonies still appear because spontaneous mutation occurs at a low background rate. In a large bacterial population, rare random events can restore the function of the mutant histidine gene or otherwise compensate for it, allowing a few cells to grow on minimal histidine medium. These are the natural or spontaneous revertants. This background is crucial because mutagenicity is not judged by whether any colonies appear, but by whether the treated condition yields more revertants than the spontaneous baseline.

Answer 198

different tester strains are needed because different mutagens produce different mutation classes. If the original his− mutation in Salmonella is a base substitution, then a mutagen that induces point mutations may restore function and yield revertants. But if the original defect is a frameshift, then a simple point mutation usually will not fix it, because restoring protein function typically requires the reading frame to be shifted back by another insertion or deletion. Therefore the revised Ames strategy uses two different Salmonella strains: one His− transition mutant and one His− frameshift mutant. If revertants appear mainly in the substitution strain, the compound likely causes base substitutions; if they appear in the frameshift strain, the compound likely causes frameshift-inducing indels

Answer 199

A chemical may appear harmless if tested only in a strain whose mutation type it cannot revert. For example, a compound that mainly causes base substitutions would generate revertants in a strain carrying a substitution mutation, but might produce almost none in a strain carrying a frameshift mutation. If only the frameshift tester were used, the chemical could be falsely judged non-mutagenic. This is why the professor emphasized the elegance of the Ames test: it ties together knowledge of mutation type, reversion logic, frameshifts, and point mutations, and it shows why experimental design must match the biology of the mutation being tested

Answer 200

The original Ames test uses bacteria, but bacteria are not multicellular organisms and do not metabolize compounds the same way humans or mice do. A chemical might be harmless when applied directly to Salmonella but become mutagenic only after metabolic processing, especially by enzymes in the liver

Answer 201

To address this, the revised Ames test adds a liver extract containing metabolic enzymes to the bacterial assay. This mimics the conversion that might occur in an animal body and allows detection of compounds that become mutagenic only after metabolism. The setup still uses both the transition-mutant and frameshift-mutant Salmonella strains, plus proper controls without the test compound, so researchers can detect mutagenicity above background and distinguish mutation classes

Answer 202

forward genetics: phenotype --> genotype reverse genetics: genotype --> phenotype

Answer 203

Genome analysis has accelerated because sequencing capacity has risen sharply while sequencing costs have fallen, leading to enormous amounts of genome sequence data. Public resources now contain millions of genomes, including reference and annotated genomes, and databases let users browse genome tables, taxonomy pages, genomic data, and gene data

Answer 204

Researchers start with what we understand best: protein-encoding genes. A typical gene contains recognizable features such as transcriptional regulatory elements, promoter, 5′ UTR, introns, exons, codons, 3′ UTR, splice sites, translation start/stop information, and polyadenylation-related signals. Among these, the most informative sequence feature is the coding region, because a long stretch of codons contains much more identifiable information than short motifs like splice sites or transcription factor binding sites. So the first annotation strategy is to scan DNA for open reading frames (ORFs) as evidence of possible exons/genes

Answer 205

An open reading frame (ORF) is a stretch of codons that can potentially encode protein because it continues without interruption by a stop codon for a sufficient length

Answer 206

the same nucleotide sequence can be read in three different frames, so a start codon in one frame may disappear in another. It also reminds you that codons refer to translation, so stop codons stop translation, not transcription. Because DNA is double-stranded, ORFs can exist on either strand, giving six possible reading frames total: three frames on one strand and three on the other. Computational tools therefore predict translation in all six frames and mark putative ORFs above some minimum length, with stop codons shown explicitly

Answer 207

A raw ORF is only a prediction: random DNA can sometimes contain a long stretch lacking a stop codon, so not every ORF is functional. To strengthen the case, researchers ask whether the ORF is conserved across species and whether there is evidence it is actually transcribed. The lecture shows that when a predicted ORF lines up with a peak of sequence conservation across 100+ insect species, that supports the idea that it corresponds to a real exon of a real gene. Conservation suggests that natural selection preserved the sequence because it matters functionally

Answer 208

Identification of a cDNA that maps to a predicted ORF provides the best evidence that the ORF is part of a real Because cDNA is made from mRNA, and mRNA only exists if the region was actually transcribed in cells. The lecture states directly that identification of a cDNA that maps to a predicted ORF provides the best evidence that the ORF is part of a real gene. The logic is: cells transcribe genes to make mRNA, mRNA is isolated, reverse transcription generates cDNA copies, those cDNAs are sequenced, and the sequences are aligned back to the genome to determine where the mRNA came from. If a cDNA maps to a predicted ORF, that means the sequence is not just a theoretical coding stretch — it was actually expressed. An EST is just a partially sequenced cDNA gene

Answer 209

Cells transcribe genes to make mRNA mRNA is isolated, reverse transcription generates cDNA copies of mRNAs, cDNAs are sequenced Sequences are aligned to the genome sequence to determine where the mRNA came from

Answer 210

When cDNA is aligned back to genomic DNA, the sequences that match correspond to exons, because those are the parts retained in the mature mRNA. The intervening genomic regions that are absent from the cDNA correspond to introns, which were removed during RNA splicing. So cDNA mapping does more than confirm that a gene is real: it also helps identify gene structure, including exon boundaries, transcript ends, and alternative splice forms

Answer 211

Even after extensive genome analysis, the exact number of human genes has been debated. The lecture shows that estimates have narrowed dramatically over time and places the current value at just under 20,000 protein-coding genes as of 2023. This matters because it shows that even something that seems simple — “count the genes” — is actually conceptually messy in genomics. Defining what truly counts as a gene is harder than textbook diagrams make it seem

Answer 212

~3% of the human genome is made up of exons of protein-encoding genes. (exons plus introns plus regulatory sequences: = ~28%) ~1% encodes protein-coding sequences. ~45% is made up of repetitive sequences

Answer 213

Because not all exonic sequence is translated into amino acids. Exons include untranslated regions (UTRs) in addition to coding sequence. So mature transcripts can contain exonic DNA that is retained in the RNA but not translated into protein. This explains why exon content is larger than protein-coding content

Answer 214

Pseudogenes Functional RNA transcripts * microRNAs * siRNAs * piRNAs * long non-coding RNAs Gene regulatory elements * enhancers * transcription factor binding site

Answer 215

ENCODE stands for the Encyclopedia of DNA Elements. Its goal is to identify all functional elements within the human genome, including all transcribed regions, regions controlling transcription, and tools for functional analysis such as knockdowns. In this lecture, ENCODE represents the broader shift from simply sequencing genomes to functionally annotating them

Answer 216

Comparison between organisms: to compare types of genes and their arrangement in the genome, to infer information about genome structure and evolutionary processe

Answer 217

Comparison within an organism: to identify gene families and gene duplication

Answer 218

Comparison between individuals of the same species: to identify differences associated with phenotypes or disease

Answer 219

Because sequence conservation suggests functional importance. If a sequence stays similar across many species over evolutionary time, that implies natural selection has preserved it, usually because it performs an important role. This logic is used first for coding genes and then extended to noncoding regions

Answer 220

Regulatory regions such as enhancers do not have obvious long sequence signatures like codon-based ORFs. They may contain short transcription factor binding motifs, but those motifs are much less information-rich and harder to distinguish from background sequence. The lecture therefore says it is harder to identify important regulatory regions, and uses conservation of noncoding DNA as one strategy for finding candidate functional regulatory element

Answer 221

Comparison of human, mouse and rat genomes has found more than 5000 sequences of >100bp and 481 sequences of <200bp that are absolutely conserved. Many are conserved in other species too. Many are found in non- coding regions. Hypothesis: these sequences might regulate regulate gene

Answer 222

By comparing human, mouse, and rat genomes, researchers found more than 5000 sequences longer than 100 bp and 481 sequences longer than 200 bp that are absolutely conserved. Many are also conserved in other species, and many lie in noncoding regions. Since they do not simply look like protein-coding exons, the lecture proposes the hypothesis that these highly conserved noncoding sequences might regulate gene expression

Answer 223

They create a reporter construct: clone the candidate conserved sequence upstream of a reporter gene such as lacZ or GFP, insert the reporter into a host genome as a transgene, and then test where the reporter protein is expressed. If the candidate sequence can drive reporter expression in a particular tissue or developmental pattern, that provides evidence that it has enhancer activity.

Answer 224

One ultraconserved element near the ISL1 locus in the mouse genome was tested using a reporter assay. A reporter driven by the conserved sequence showed lacZ expression in mouse embryos in a pattern that closely matched the endogenous mouse ISL1 gene, especially in the head and spinal cord. The conclusion is that conserved noncoding sequences can identify functional elements, and specifically that ultraconserved noncoding sequence from the human ISL1 locus can drive reporter expression in mouse embryos in the same pattern as endogenous ISL1

Answer 225

The experiment supports the hypothesis that the conserved noncoding region can function in gene regulation and shows that the element is sufficient to drive reporter expression in the tested pattern. It also shows that the human version can function in mouse

Answer 226

it does not prove the sequence is necessary for normal ISL1 expression in the mouse, because necessity would require deleting or disrupting the endogenous element and seeing whether ISL1 expression is lost. So the result shows capability and functional compatibility, but not full endogenous necessity

Answer 227

Look for blocks of DNA sequence that are conserved between chimps and macaques but not found in the human genome. Might loss of these regions explain some of the differences between chimps and humans?

Answer 228

Another approach is to look for DNA blocks that are conserved in chimps and macaques but missing from humans. The lecture asks whether loss of such regions could help explain human-specific traits. One tested example involved a conserved enhancer deleted in humans. In reporter assays, this enhancer could drive expression in two structures present in mice and chimps but lost in humans: sensory vibrissae (whiskers) and penile spines. The interpretation given is that loss of this ancestral enhancer in humans is correlated with loss of those structures. The key word is correlated: the result is consistent with the hypothesis, but not by itself absolute proof of causation

Answer 229

Synteny is the conserved relative order of genes between genomes. The lecture states that 99% of mouse genes have a homolog in humans, and 99% of human genes have a homolog in mouse. Beyond shared genes, the overall genome organization, including the relative order of genes and noncoding regions, is also highly conserved. Thus, mouse and human genomes contain large blocks of genes in common, with genes in the same order along chromosomes

Answer 230

Because the locations and orientations of syntenic blocks have been reshaped by chromosomal rearrangements during evolution. The lecture interprets the different locations of shared blocks as reflecting rearrangements that occurred since the last common ancestor. So synteny captures both conservation and historical change: the same genes are often retained together, but inversions, translocations, and other rearrangements move those blocks around over evolutionary time.

Answer 231

Observation: Mouse and human genomes have large blocks of genes in common, and the genes within these blocks are in the same order along the chromosome. Interpretation: The different locations of these blocks reflects the rearrangements that have occurred relative to their last common ancestor

Answer 232

Homologs are genes that are related by common ancestry.

Answer 233

Orthologs are homologous genes found at the same genetic locus in different species, inherited from a common ancestor. For example, gene A in frog, human, and mouse would be orthologs if they all descended from the same ancestral gene A.

Answer 234

Paralogs are homologous genes found at different loci within the same species, arising by gene duplication. For example, gene A and gene B in the mouse genome may be homologous to each other because an ancestral gene duplicated and produced two related gene copies. Thus, orthology reflects speciation, while paralogy reflects duplication

Answer 235

If a similar gene is found across multiple species, you can infer that the common ancestor also had that gene. If two related genes are found within one species, you can infer that an ancestral gene likely duplicated before the descendant lineages split. By comparing which genes are present, absent, or duplicated, researchers can reconstruct likely evolutionary histories. The lecture’s diagram uses genes A and B across frog, mouse, and human to show that conserved copies across species indicate ancestral presence, while related A/B pairs within species suggest an older duplication event in a deeper ancestor

Answer 236

The key question was: did monotremes like the platypus gain a vitellogenin gene, or did other mammals lose it? Most eutherian mammals such as humans, chimpanzees, mice, and dogs do not lay eggs and do not have a functional vitellogenin gene, whereas platypus, an egg-laying mammal, does. Since presence of the vitellogenin gene correlates with egg laying, comparative genomics asks whether the gene newly appeared in monotremes or whether it was present in a common ancestor and later lost in other mammals. To answer that, the lecture looks at an evolutionary outgroup: birds The presence of the vitellogenin gene in the genome correlates with whether or not the animal lays eggs. Did monotremes GAIN the gene, or did the eutherians LOSE the gene? How do we tell? Ask if it was present in their last common ancestor

Answer 237

Birds such as chickens have three vitellogenin genes, platypus has one functional vitellogenin gene, and humans contain only recognizable mutated/fragmented remnants rather than functional yolk genes. This strongly supports the model that vitellogenin genes were present in the common ancestor of mammals and birds, and that these genes were then progressively lost in the mammalian lineage. Specifically, the lecture suggests that mammals originally inherited these yolk genes, but two were lost earlier in synapsid/mammalian evolution and the final one was later lost in eutherian mammals, while monotremes retained one. So the platypus did not gain a new yolk gene; rather, most other mammals lost theirs

Answer 238

It shows that sequence comparison does more than identify genes — it lets us build a model of evolutionary history. By looking at gene presence, gene loss, and mutated remnants, we can infer what the last common ancestor likely contained and how different lineages changed over time. In other words, comparative genomics can reveal not just what genomes contain now, but the steps evolution likely took to produce present-day organisms * Related genes within an organism that encode proteins of similar amino acid sequence * Can contain 2 to >100 members (called PARALOGS) * Members can be functionally redundant or have independent functions

Answer 239

It shows that sequence comparison does more than identify genes — it lets us build a model of evolutionary history. By looking at gene presence, gene loss, and mutated remnants, we can infer what the last common ancestor likely contained and how different lineages changed over time. In other words, comparative genomics can reveal not just what genomes contain now, but the steps evolution likely took to produce present-day organisms

Answer 240

* Related genes within an organism that encode proteins of similar amino acid sequence * Can contain 2 to >100 members (called PARALOGS) * Members can be functionally redundant or have independent functions. * Human keratin-associated protein 1 family. These proteins are involved in the structure of human hair fibers, and this family is thought to have arisen through gene duplication during evolution

Answer 241

Comparison within one genome can reveal families of related genes, especially genes that arose through duplication. These duplicated homologous genes within a species are paralogs. Paralogs may be just two genes or a very large family with more than 100 members. They can remain functionally redundant, meaning different copies still do similar jobs, or they can evolve independent/specialized functions over time. This kind of within-genome comparison therefore tells us how genomes expand and diversify functionally.

Answer 242

comparing genomes from different dog breeds to understand the genetic basis for their differences

Answer 243

human keratin-associated protein 1 family as an example. These proteins contribute to the structure of human hair fibers, and the family is thought to have arisen through gene duplication during evolution. Sequence comparison shows that multiple related genes in the genome encode proteins of similar amino acid sequence, consistent with a duplicated gene family. This illustrates how a genome can contain clusters of related genes that arose from a shared ancestral gene and then either retained overlapping roles or diversified

Answer 244

The lecture explicitly ties gene families back to earlier topics like chromosome rearrangements, unequal crossing over, deletions, and duplications. If chromosomes break or recombine improperly during meiosis, duplicated segments can form. Over evolutionary time, such duplications can be retained and diverge into paralogous gene families. So gene family evolution is not an isolated genomics idea — it connects directly to the chromosome-level mutation mechanisms discussed earlier in the course.

Answer 245

The Dog Genome Project compares genomes from different dog breeds to understand the genetic basis of their differences. Dogs are all the same species, but breeds vary enormously in size, coat type, skull/face shape, limb length, and behavior, making them an excellent system for connecting phenotype to genotype. Humans have selectively bred dogs for many generations for desired traits, so genomics can now ask: when people selected for small size, short legs, curly fur, pointy ears, or particular behaviors, what parts of the genome were they actually selecting? It is essentially a modern genomic version of linking artificial selection on phenotype to underlying genotype

Answer 246

that humans and dogs share many similar genes and disorders, so studying canine genetics can inform human health research as well. The NHGRI Dog Genome Project specifically focuses not only on dog morphology but also on canine cancer and health traits, with the idea that understanding gene-trait relationships in dogs can also provide insight into disease biology in their “two-legged caregivers.” So dog genomics is useful both for understanding breed diversity and as a comparative medical model

Answer 247

Mutations in the MC4R gene usually lead to obesity by preventing a sense of fullness. But one such mutation leaves people uninterested in eating, scientists report. The study subjects had been thin all their lives and just did not care much about food. Now, a group of researchers in Britain may have found the reason. The people carry a mutation that mutes appetite, and greatly reduces their chances of getting diabetes or heart diseas

Answer 248

A case study will be used to examine specific issues of bioethics and forensic science that occur in forensic investigative genealogical searching, which include genetic privacy, discrimination and public safety concerns. The forensic investigative process and various investigative DNA tools will also be described. The Golden State Killer Case (1) will be examined to highlight and discuss forensic ethical issues to develop an ethical framework, as well as provide recommended solutions to pressing public safety and privacy issues facing crime laboratories and criminal investigators. Use of the ethical concept of proportionality (2) will be utilized to contrast and balance competing ethical concern

Answer 249

A man has a rare genetic mutation that essentially guaranteed he would develop Alzheimer’s disease by his early 50s. Somehow, he has escaped Alzheimer’s for at least 25 years longer than anyone expected. Scientists are searching for the secret in his DNA that has protected him from dementia, hoping it could lead to ways to treat or prevent Alzheimer’s for many other people.

Answer 250

Some volunteers for biobanks, which collect genetic information to study health across populations, have been surprised to be informed they carry risky mutation

Answer 251

Several examples were given. Researchers have identified 14 different genes affecting dog size. Breeds with short legs often have an extra copy of the FGF4 gene. At least three genes influence the length, texture, and curl of the coat. Some breeds show a high incidence of rare cancers, motivating efforts to understand why genetically. The lecture also mentions a locus associated with sociability in dogs that is absent in wolves and corresponds to the human chromosomal region deleted in Williams syndrome, a condition associated with hyper-sociability in humans. These examples show that within-species genome comparison can uncover the genetic basis of morphology, disease susceptibility, and even behavioral traits

Answer 252

Comparing genomes within a species can reveal the genetic basis of complex visible traits, disease predispositions, and selected behaviors. It shows how population structure, breeding, and selection expose genotype–phenotype relationships that would be much harder to detect in less structured populations. In a way, dog breeds provide a naturally organized experiment in artificial selection that genomics can decode

Answer 253

Human genome comparison can be used to ask why some people have a certain trait, why some have a disease, why some people carrying a high-risk mutation stay healthy, and what variants might be protective or modifying. The lecture gives broad examples from the literature and news, including appetite-related mutations, Alzheimer’s-protective backgrounds, unexpected discovery of risky mutations in biobank participants, and bioethical issues around forensic genealogy and privacy. The main point is that comparing human genomes can reveal not just causes of disease but also resilience, modifier genes, and ethical challenges

Answer 254

If genome databases such as direct-to-consumer genetic datasets identify people carrying a mutation believed to be lethal, yet those people are obviously alive, it suggests that the original mutation’s effect is being modified or suppressed — possibly by second-site mutations, genetic background, compensatory mechanisms, or incomplete penetrance. This again emphasizes that genotype is often interpreted in the context of the rest of the genome, not as an isolated deterministic event

Answer 255

The final example was uniparental disomy (UPD), presented as a striking case where large-scale genome comparison uncovers unexpected inheritance patterns. This example ties together earlier course content on chromosome segregation errors with modern genome databases and bioinformatics. It shows how comparative genomics can reveal rare but important events in human inheritance by analyzing very large datasets

Answer 256

Researchers screened 4.4 million individuals in the 23andMe database and used identity by descent (IBD) patterns to identify parent-child duos. The principle is that each chromosome in the child should match one of the parent’s homologs because a child normally inherits one copy of each chromosome from each parent. Using this logic, they identified 916,712 parent-child duos. Then they searched for cases where that expected pattern failed

Answer 257

UPD occurs when an individual has the normal disomic number of a chromosome — that is, two copies, not one and not three — but both copies came from the same parent, with no copy from the other parent for that chromosome. In the study discussed, among the 916,712 parent-child duos, researchers found 199 individuals with this pattern, representing 205 incidences of UPD total. The lecture notes this was around 0.005%, or about 1 in 20,000, making it more common than previously thought

Answer 258

UPD is surprising because under normal Mendelian inheritance, one homolog comes from the mother and one from the father. So if a child has two chromosomes from the same parent, the obvious question is: how did both copies get into the same F1 individual? This points directly back to meiosis and segregation errors

Answer 259

The key starting point is nondisjunction. If nondisjunction occurs in meiosis I or meiosis II, a gamete can end up with two copies of one chromosome from the same parent instead of one. The lecture explicitly calls back to earlier course material on nondisjunction, because this is the essential first mistake needed to produce a gamete capable of generating UPD

Answer 260

A gamete carrying two copies of one chromosome would normally fertilize or be fertilized by a normal gamete carrying one copy, producing a trisomic zygote. But most trisomies are severe or lethal, whereas the people in the database were clearly alive and functioning. So nondisjunction alone would produce trisomy, not the observed normal diploid state with UPD. Therefore, another event must have happened afterward

Answer 261

The lecture explains that UPD is generated by nondisjunction in meiosis followed by “trisomy rescue” in the embryo. First, nondisjunction creates a disomic gamete containing two copies of one chromosome from one parent. This gamete is fertilized by a normal gamete, creating a trisomic zygote. Then, in an additional rare event, one chromosome is lost during early embryonic development. If the chromosome that is lost is the one from the other parent, the embryo is restored to a normal diploid chromosome number, but both remaining copies are from the same parent — producing UPD

Answer 262

In heterodisomy, the child inherits both homologs from one parent; this usually reflects meiosis I nondisjunction

Answer 263

the child inherits two copies of the same homolog from one parent; this usually reflects meiosis II nondisjunction. If crossing over occurred during meiosis, the resulting disomy can be a mixture rather than perfectly one type or the other. So the exact UPD pattern can reveal something about where in meiosis the original error happened.

Answer 264

Not all chromosomes are equally likely to show UPD. The final graph shows that some chromosomes appear much more frequently in maternal or paternal UPD than others. Some are common, some rare. The professor notes that this is a genuinely interesting biological observation and that people do not really know why some chromosomes are represented much more often than others. This is presented as an example of a new biological question that emerged directly from large-scale genome comparison

Answer 265

Breast cancer metastasized to liver (not primary liver cancer) Visible as nodular masses in liver tissue

Answer 266

Makes treatment much harder Tumor spreads → disrupts function of other organs → Early detection is crucial

Answer 267

~45% (~1 in 2 people) Similar in males and females

Answer 268

~22% overall (~1 in 4–5 people) Males: ~24% Females: ~21%

Answer 269

Lung cancer (≈ 1 in 4 cancer deaths)

Answer 270

High incidence + mortality Major public health burden Leads to large funding + dedicated research centers (e.g., McGill Goodman Cancer Center)

Answer 271

Sustaining proliferative signaling Evading growth suppressors Resisting cell death (apoptosis) Enabling replicative immortality Inducing angiogenesis Activating invasion + metastasis

Answer 272

Respond to growth signals (e.g., insulin) Stop dividing at confluency (contact inhibition) Undergo apoptosis if abnormal Limited divisions due to telomere shortening

Answer 273

Ignore growth signals → self-sustain proliferation Evade apoptosis Bypass telomere limits → immortality Lose contact inhibition → pile up Can induce blood vessel formation

Answer 274

“Normal” cultured cells Divide 3 times in 3 days Exhibit: Anchorage dependence Contact inhibition

Answer 275

Cells become transformed: Lose contact inhibition Grow on top of each other Become rounded morphology Grow without serum signals → Demonstrates single gene → cancer-like phenotype

Answer 276

Familial inheritance patterns (pedigrees) Example: Li-Fraumeni syndrome Can appear dominant OR recessive depending on context Due to underlying mutation + gene function complexities

Answer 277

Hereditary breast cancer and ovarian cancer syndrome Genes: BRCA1, BRCA2 Li-Fraumeni syndrome Gene: P53 Familial adenomatous polyposis Gene: APC Retinoblastoma Gene: RB1 ^^^^^^^^^ Genetic testing available Allows risk prediction + preventative decisions Example: BRCA mutation carriers

Answer 278

positive regulators driving tumorigenesis

Answer 279

negative regulators that are inactivated in cancer

Answer 280

prevents mutations, maintaining DNA integrity

Answer 281

Filtered tumor extract (no bacteria) → still causes cancer Demonstrated virus causes tumors → Discovery of RNA tumor viruses 1910: Peyton Rous discovered that particles smaller than bacteria could reproducibly induce sarcomas in chickens. This led to discovery of tumor virus (RNA tumor virus). Virus must carry a cancer-causing gene!!!!

Answer 282

viral oncogene (causes cancer) First oncogene src (v-Src) was isolated from Rous sarcoma virus in 1970

Answer 283

normal cellular gene (proto-oncogene) In late 1970s, search for related sequences in normal cells discovered proto-oncogenes (c-Src)

Answer 284

are required for normal cellular function, but when mutated, become oncogenes and promote cancer formation

Answer 285

They are retroviruses: RNA → reverse transcribed → DNA DNA integrates into host genome Retroviruses: Random insertion into genome High mutation rate Can alter host genes

Answer 286

Steps: 1. Virus integrates near the proto-oncogene 2. During replication → captures the host gene 3. Gene becomes mutated/rearranged 4. Forms viral oncogene (v-onc) Example: v-Src Loss of C-terminal inhibitory domain → Constitutively active kinase → Continuous growth signaling

Answer 287

Viral DNA inserts near gene Viral strong promoter/enhancer drives expression → Overexpression of proto-oncogene → Cancer development

Answer 288

In human, most proto-oncogenes are activated in the absence of viral infection: ¨ Mutations in coding sequences ¨ Chromosome abnormalities: increased expression, fusion protein (cancer specific form)

Answer 289

RTK: cell surface proteins that bind to extra cellular signaling molecules like growth factors. Binding of the signaling molecule results in phosphorylation on tyrosine residues

Answer 290

Small GTPase (NOT a kinase) GTP-bound Ras → activates Raf kinase → phosphorylation cascade → transcription factors activated → cell growth GTP bound Ras (not kinase) binds Raf (kinase), which initiates a cascade of phosphorylation events. Activation of TFs that promote cellular growth

Answer 291

Controlled by Sos (GEF) Tight regulation in normal cells Sos is a guanine nucleotide exchange factor

Answer 292

Single point mutation → constitutive activation → continuous growth signaling → tumorigenesis

Answer 293

Translocations (most important for leukemia) Gene amplification Deletions Fusion genes → Can cause: Overexpression Constitutive activation New protein functions

Answer 294

Exchange of DNA between non-homologous chromosomes → Can: Place gene under new promoter Create fusion gene

Answer 295

Proto-oncogene moved near strong promoter/enhancer Often immune cell promoters (e.g., Ig genes) → Uncontrolled expression 👉 Example (classical, even if not emphasized heavily): MYC near Ig promoter (Burkitt lymphoma)

Answer 296

Reciprocal translocation between Ch 8 and Ch 14, t(8:14), observed in Burkitt lymphoma (B-cell) c-Myc: Transcription factor whose normal function is to promote cellular growth/proliferation (proto-oncogene

Answer 297

The coding sequence of Myc is NOT changed. However, the Myc protein is OVEREXPRESSED by IG regulatory element

Answer 298

Philadelphia chromosome t(9:22) is observed in 95% of people with CML c-Abl: protein kinase in involved in many cellular processes (proto-oncogene

Answer 299

… A chimeric protein at the molecular level Bcr1-Abl becomes a hyperactivated kinase

Answer 300

Leukocoria = white reflex in pupil (instead of normal red reflex) Seen in flash photography due to abnormal retinal reflection Can indicate retinoblastoma (eye tumor) Normal red eye = reflection from retinal blood vessels White reflex = pathological sign

Answer 301

Retinoblastoma is a rare (1 / 15,000) childhood cancer of the eye. Pedigree analyses show that 40 % of the cases involve an inherited (familial) mutation. Retinoblastoma may be unilateral (affecting only one eye) or bilateral (affecting both eyes)

Answer 302

In 1971, Alfred Knudson observed a difference in the kinetics of retinoblastoma development between unilateral and bilateral cases Different kinetics of tumor development: Bilateral (inherited) → early, rapid Unilateral (sporadic) → later, slower Graph (log scale): Bilateral = 1st-order kinetics (1 event) Unilateral = 2nd-order kinetics (2 events) Cancer requires 2 mutations (“hits”) in same gene These are rate-limiting events

Answer 303

Knudson proposed that retinoblastoma results from two separate genetic defects (mutations), two hits -In sporadic cases, “two hits” must happen to a single cell. Such event is rarer and takes longer to develop

Answer 304

In inherited cases, children are born with the first hit. Chances of having the second hit is high, and it could happen to more than one cell. The hypothesis is consistent with the idea of having recessive mutations in both alleles of a single gene

Answer 305

Sporadic: probability = (mutation₁ × mutation₂ in same cell) Inherited: only need mutation₂ → Much higher probability → earlier onset

Answer 306

Tumor suppressor genes are recessive at the cellular level Both alleles must be inactivated

Answer 307

Chromosomal abnormality frequently found in Retinoblastoma 1986: The RB1 gene is cloned, discovery of the first tumor suppressor gene Frequent chromosome 13 deletion (13q14) in tumors Loss-of-function mutation → suggests tumor suppressor RB1 gene cloned in 1986 (first tumor suppressor)

Answer 308

RNA expression present in normal tissue Absent in tumor samples (Northern blot)

Answer 309

Oncogenes → gain-of-function → dominant (1 allele enough) Tumor suppressors → loss-of-function → recessive (need both alleles lost)

Answer 310

One normal allele produces enough protein to maintain function Cancer requires loss of both copies

Answer 311

Tumor suppressor genes are often inherited as a dominant trait -The chance of acquiring the second mutation in a somatic cell is high. In theory, a single cell acquiring the second mutations is sufficient to cause cancer. -Not all carriers will develop cancer (penetrance) Individuals inherit one defective allele (carrier) High probability of acquiring second mutation → Phenotype appears dominant at organism level

Answer 312

Not all carriers develop cancer Due to: environment immune surveillance stochastic mutations

Answer 313

Controls G1 → S checkpoint Prevents premature DNA replication

Answer 314

RB binds E2F transcription factor → inhibits it E2F normally activates DNA synthesis genes

Answer 315

Cyclin D/E + CDKs phosphorylate RB RB releases E2F → transcription of S-phase genes

Answer 316

E2F is constitutively active Continuous DNA synthesis → uncontrolled proliferation

Answer 317

Cyclin-CDK hyperactivation (oncogene) → constant RB phosphorylation → same effect

Answer 318

Responds to DNA damage Maintains genomic integrity activated by: DNA damage (DSBs, replication stress) Kinases: ATM, ATR, CHK1/2

Answer 319

p53 is mutated in > 50% of all human tumors Unable to bind DNA but able to form tetramer!

Answer 320

Cell cycle arrest (repair time) Senescence (permanent arrest) Apoptosis (if damage severe)

Answer 321

TAD : transcription activation domain DBD : DNA-binding domain OD : oligomerization domain

Answer 322

Many p53 mutant alleles in cancer function as a dominant negative P53 is a transcription factor that binds DNA as a homotetramer Mutation in the DNA biding domain in one of the two alleles

Answer 323

Missense mutations allow protein expression Mutant protein interferes with WT (dominant negative)

Answer 324

Viral oncoproteins from DNA tumor virus bind and inactivate Rb and p53 E6 → inhibits p53 E7 → inhibits RB → forces cells into S phase inactivated rb = To create S-phase environment Needed for viral DNA replication

Answer 325

HPV → cervical (~95% cases) HBV → liver cancer EBV → Burkitt lymphoma HTLV-1 → adult T-cell leukemia

Answer 326

A single cell acquiring a mutation that provides proliferative advantage gives rise to clones of mutated cells. A single cell from the clones may acquire a secondary mutation that provides additional proliferative advantage. Eventually, cycle of acquiring mutations and clonal expansion gives rise to fully transformed cancer cells

Answer 327

1. First mutation → clonal expansion 2. Second mutation → increased proliferation 3. Additional mutations → malignancy 4. Further changes → metastasis

Answer 328

loss of normal tumor suppressor gene APC = 1. polyp forms on colon wall 2. a benign pre-cancerous tumor grows --> activation of oncogene ras 3. an adenoma (benign tumor grows) --> loss of tumor suppressor gene p53 4. carcinoma develops (malignant develops) --> loss of antimetastisis gene 5. cancer spreads via bloodstream

Answer 329

Loss-of-function Require two hits Often inherited as dominant (phenotypically) Function: inhibit proliferation / maintain genome

Answer 330

PD-1 activation by tumor cells suppresses T cell function

Answer 331

-The U.S. ban on in-flight smoking began with domestic flights of two hours or less in April 1988. All planes became smoke-free by the end of the 1990s.

Answer 332

Antibodies (Y-shaped) bind specific regions (epitopes) on spike protein Recovered patients produce antibodies targeting: Receptor-binding domain Intermediate domain S2 domain Demonstrates antigen-specific adaptive immunity

Answer 333

Infections produced antibodies circulating in their blood that target three different parts of the coronavirus’s spike protein

Answer 334

Spike protein has many mutations (~37) Mutations occur at antibody binding sites Prevent antibody recognition → immune evasion RNA viruses mutate rapidly → evolutionary “arms race”

Answer 335

molecule (usually protein/polysaccharide) that elicits immune response

Answer 336

Protein in blood/body fluids Binds antigen → marks for destruction / neutralizes function

Answer 337

Immune system attacks self-antigens Caused by failure of elimination of autoreactive lymphocytes Increases with age

Answer 338

Innate: Non-specific Fast Present in all organisms (e.g., macrophages)

Answer 339

Adaptive: Antigen-specific Slower initial response Present in vertebrates only Includes B and T cells

Answer 340

production ad secretion of antibodies by a specialized lymphocytes (white cells) called B cells

Answer 341

specialized lymphocytes called T cells produce T-cell receptors that recognize and bind antigens found only on the surface of the body's own cells

Answer 342

Origin of immune cells:' From hematopoietic stem cells (HSCs) Found in bone marrow Differentiate into: Common lymphoid progenitor (→ B, T, NK) Common myeloid progenitor

Answer 343

Where do B and T cells mature? B cells: bone marrow T cells: thymus Each B cell and T cell recognize a unique non-self antigen that can elicit immune response (diversity in antibody and T cell receptor

Answer 344

Autoreactive B and T cells are eliminated Only cells recognizing non-self antigens survive

Answer 345

Body doesn’t know future pathogens Needs large pool of randomly generated receptors Ensures at least one cell can recognize any antigen

Answer 346

1. Clonal selection (specific B cell activated) 2. Clonal expansion (proliferation) Differentiation into: - Plasma cells (antibody secretion) - Memory B cells

Answer 347

Persist in circulation Enable faster secondary response Basis of vaccination

Answer 348

Trigger primary immune response Generate memory cells Enable faster response upon real infection

Answer 349

Inactivated pathogen Attenuated pathogen Purified antigen RNA vaccines (host produces antigen)

Answer 350

Only require nucleotide sequence Host cells produce protein Easier than synthesizing proteins (4 bases vs 20 amino acids)

Answer 351

2 identical heavy chains + 2 identical light chains Y-shaped Regions: Variable (V) → antigen binding Joining (J) Constant (C) → not involved in specificity

Answer 352

two types (kappa and lambda) encoded on different chromosomes, segments V,J,C (kappa/kappa or lambda/lambda)

Answer 353

5 types (alpha, delta, gamma, epsilon, or mu) segments V, D, J, and C - each type of light chain can potentially combine with each type of heavy chain - B cells produce a unique light and heavy chain combination that can recognize a specific antigen

Answer 354

It is estimated that the human genome has 20,000 to 25,000 protein coding genes However, humans have the capacity to produce 1011 antibodies with different specificities

Answer 355

- somatic recombination - junctional diversity - somatic hypermutation

Answer 356

RAG1, RAG2 and DNA repair enzymes introduce dsDNA breaks and joint random V and J segments. This occurs at the level of DNA

Answer 357

The heavy chain locus has multiple D (diversity) gene segments in addition to V J and C segments - immunoglobulin generated in a given B cell will always be the same -

Answer 358

few random nucleotides are lost or gain - in many cases, it will result in a frameshift that produces a nonfunctional gene

Answer 359

the immunoglobulin genes are subject to a high mutation rate deamination of cytosine which becomes uracil - uracil is replaced by the DNA repair mechanisms by another base, thus generating a point mutation

Answer 360

t-cell receptors are composed of alpha and beta chains that have variable regions alpha chain: 44-46 V gene segments, 50 J gene segments and a single C segment beta chain is similar to alpha chain, but contains D gene segments somatic recombination and junction diversity, but no hypermutation

Answer 361

Cell fusion generates hybrids containing genomes from the two different cell lines The genomes of fused cells are unstable, randomly keeping or losing chromosomes. In hybrids between rodent and human cells, rodent chromosomes are predominantly kept and most human chromosomes are lost. (This phenomenon is used to map a GOI in human) Cell fusion is highly inefficient. Therefore, a selectable marker is necessary to identify hybrids!! - genes on chromosomes are still being expressed --> thats the key

Answer 362

synthesis of nucleotide: Novo pathway or Salvage pathway Novo = major DNA precurosors, blocked by aminopetrin Salvage pathway: minor precursors, HPRT for purine synthesis makes dCTP without HAT medium, cells will die, cant make sufficient amount of dNTP for DNA synthesis HPRT and TK are usually not needed for cell viability

Answer 363

converted to guanine by Hypoxanthine-guanine phosphoribosyltransferase

Answer 364

phosphorylated by Thymidine kinase

Answer 365

has hypoxanthine and thymidine

Answer 366

Hat medium makes sure that only a few hybrid cell lines survive and the others die to identify gene of interest

Answer 367

make many hybrids between primate and rodent cells (some human chromosomes will be kept and many will be lost) --> identify the hybrid cell lines that can be infected by measles --> determine which human chromosomes are kept in those cells (structure of human chromosomes are different from rodent chromosomes) - make hybrids, isolate colonies

Answer 368

first three chromosome cell lines affected - common between those 3 is chromosome 1 - cell line 4 contains part of chromosome 1, which wasn't affected by measles virus -

Answer 369

A cell fusion experiment between human TK-deficient and rodent HPRT- deficient cell lines was carried out to identify the location of the human UMPK gene. Eight different cell lines are recovered from the experiment. The table below shows the presence or absence of the human UMPK activity (+ means present and – means absent) and the human chromosomes retained in each hybrid cell line - in order for the cells to survive they must have HPRT from human, so the one in common with all the cell lines is chromosome X

Answer 370

*Gene therapy involves adding a normal (wild-type) copy of a gene to the genome of an individual carrying defective (mutated) copies of the gene, often recessive mutations. *Only few of the 4000 inherited human diseases are currently treatable

Answer 371

1. derived from adenovirus 2. derived from retrovirus

Answer 372

* Will infect all cells even non-dividing ones * The vectors will not be integrated in the genome, thus the transgene is diluted and eventually loss change isn't permanent

Answer 373

* Many will infect only dividing cells, but others (HIV) can infect without host cell division. * The transgene and the viral vector will be incorporated into the genome of the infected cells (more stable) - if works then permanent

Answer 374

1. somatic cell line 2. germ line

Answer 375

gene therapy refers to the transfer of a gene in somatic cells.

Answer 376

gene therapy refers to the transfer of a gene in all cells of an organism through the germline transmission

Answer 377

rare autosomal disease of immune system (bubble boy disease) - affected individuals dont have an immune system - in abscence of ADA, deoxyadenosine accumulated in T lymphocytes and eventually kill these cells (t lymphocutes are responsible for the stimulation of the antibody producing B lymphocytes) - first disease treated with gene therapy - most common treatment for SCID is bone marrow transplantation

Answer 378

- LTR needed for vector DNA integration in the cells - SV40 promoter will allow transgene expression in human cells - the NEO is needed for the selection of human cells that will produce the virus in vitro

Answer 379

In 2007, four of the ten patients have developed leukemia due to the integration of the retroviral vector near a proto-oncogene -LTR acts as a promoter

Answer 380

* Integration sites cannot be controlled. * Expression level of the rescue gene may not be optimal. * Ex vivo experiment is limited to certain types of cells. Potential solution: directly edit genomic sequences of stem cells from the patient

Answer 381

The clustered regularly interspaced short palindromic repeats (CRISPR) system uses a RNA molecule to target specific DNA sequences To target different sequences in the genome, one can simply change the crRNA sequences

Answer 382

Overexpression of 4 proteins, Oct3/4, Sox2, c-Myc, and Klf4, in fibroblasts was sufficient to reprogram cells to become pluripotent stem cells. 2012 Nobel Prize for the discovery that mature cells can be reprogrammed to become pluripotent. (Unlike embryonic stem cells, there is no moral implication of using iPS)

Answer 383

Almost every cell in our body has essentially the same DNA sequence. What makes each cell type unique is largely determined by the expression of specific subsets of tissue-specific genes (mostly controlled at the level of transcription).

Answer 384

Simple system, relatively few levels at which “gene activity” can be regulated Transcription initiation (to transcribe or not…?) is one of the main mechanisms for regulating genes in prokaryotes Many prokaryotic metabolic genes involved in the same process are organized into operons (one transcript unit coding for multiple proteins under a single promoter) - multiple protein coding sequences

Answer 385

Sigma factor subunit of RNA polymerase binds to -35/-10 promoter sequences to properly position the holoenzyme at transcription start site o In prokaryotes, RNA polymerase easily binds the promoter – default state of genes is “ON” (a bit…)

Answer 386

activators = proteins bringing RNA polymerase to promoter repressors = either overlaps promotor sequence or sits between where transcription begins and promotor binds Binding of sequence-specific DNA binding proteins around the promoter either “activates” or “represses” gene transcription (Operator = binding sites for repressors)

Answer 387

o Effector called an “inducer” if its presence leads to increased gene expression - inducers can act via either positive or negative regulatory mechanism - no longer binding DNA - inducer = quickest and easiest way for regulation - can act with repressor or activator

Answer 388

o investigated regulation of the lac operon in E. coli, required for lactose metabolism o Used a purely genetic approach to understand gene regulation, later confirmed by molecular techniques (e.g. biochemical characterization of repressor protein) - must be DNA sequence and diffusable element controlling gene expression

Answer 389

o To uncover the regulatory mechanism of lac operon Jacob & Monod used a mutagenesis approach: treated E. coli with a chemical mutagen used biochemical assays to measure ßgalactosidase and/or Lac permease activity in presence and absence of a synthetic inducer (IPTG) - similar to lactose in repressing lac repressor selected mutant strains where one or both of these activities was altered

Answer 390

o Three main classes of lac mutants: 1. Structural gene mutations (affect function of just one enzyme – the other is INDUCIBLE) 2. UNINDUCIBLE mutants – can’t make LacZ and LacY in presence of inducer (IPTG) single variant, altering expression of both 3. CONSTITUTIVE mutants – make both LacZ and LacY even in absence of inducer

Answer 391

o J+M used “partial diploid” bacteria harbouring a plasmid that contains a second copy of the lac operon allowed them to test dominance / recessiveness critical in determination of “cis” vs “trans”-acting factors - dominant allele can function and doesn't have to be on same gene cis diploid gene - make diploid and can use as mendelian genetics to see if recessive/dominant to wildtype

Answer 392

bacteria grown in medium containing glycerol (no glucose) - P- only affects transcription of genes physically attached to it on the same DNA molecule (i.e. chromosome or plasmid) – promoter is a cis-acting element. P- affects inducibility of both β–Galactosidase and Permease - failure to “turn on”

Answer 393

Oc (= O-): constitutive… both β–Galactosidase and Permease are constitutively active - failure to “keep off” similar to P, O only affects transcription of genes physically attached to it on the same DNA molecule – operator is a cis-acting element.

Answer 394

The functional copy of I does not have to be on same sequence; can be on different strand I- : similar to O, since both β–Galactosidase and Permease are constitutively active However, I doesn’t have to be on the same DNA molecule – can act in trans… diffusible factor

Answer 395

IS is different from I- ; it cannot be complemented by I+ (i.e. it’s dominant to I+ )

Answer 396

Z is uninducible, Y is inducible Y+ = structure gene will act as wildtype because of cis element OC = mutation of operator the polymerase will bind to P - operator controls P because thats where repressor binds

Answer 397

o CAP: Catabolite Activator Protein coded by the crp gene o CAP-cAMP binds P just upstream of RNA pol o facilitates RNA pol binding to P to increase transcription - binds to promoter consensus sequence matches

Answer 398

when glucose present no lactose, no transcription

Answer 399

Promoter elements: core and proximal; enhancers/silencers ▪ example – yeast Gal system ▪ Transcriptional regulation by changes to chromatin structure ▪ Position-effect variegation in Drosophila -

Answer 400

= chromatin

Answer 401

heterochromatin and euchromatin Any molecular processes that need to access chromosomal DNA (eg. transcription, replication, and repair) require that nucleosomes be “unpacked -

Answer 402

* tightly packed nucleosomes * “closed” DNA * associated with “structural” DNAs & non-transcribed regions * e.g. centromere, telomeres, repetitive elements

Answer 403

* loosely packed nucleosomes * “open” DNA * associated with transcribed regions of chromosomes -

Answer 404

ground state of gene expression is on, very accessible so polymerase can bind as long as sequence is there -- many genes are negatively regulated - if ground state is off, the DNA sequences is part of nucleosome and can't easily access, therefore there are many things that have to happen for gene expression/transcription to occur - transfactors are important for activating transcription

Answer 405

o like in prokaryotes, it involves cis-acting DNA sequences and trans-acting proteins

Answer 406

* core promoter * promoter-proximal elements * enhancers/silencers

Answer 407

* general transcription factors * common transcription factors * cell/tissue-specific transcription factors * transcription cofactors

Answer 408

Efficient transcription of such genes requires binding of specific transcription factors to enhancer sequences distance-independent cis-acting DNA sequences can be found near to or far from the transcription start site (TSS), situated upstream, downstream, or even within the gene itself (introns) * enhancers – promote transcription * silencers - prevent transcription *

Answer 409

Efficient transcription of such genes requires binding of specific transcription factors to enhancer sequences * bind DNA in a sequence-specific manner through DNA- binding domain * influence transcription by: * interacting with transcription machinery (e.g. RNApol) directly or indirectly OR * influence chromatin structure directly or indirectly * expressed in cell-type / tissue / time-specific manner

Answer 410

* there are many different families of transcription factors; all have a sequence-specific DNA binding domain * all have an activation or repression repression domain that mediates interactions with other components of the transcription machinery * can have other domains – dimerization, ligand-binding domains (= sensors) * transcription co-factors can have similar domains, but LACK A DNA BINDING DOMAIN * function to regulate gene transcription - think of transcription factors like a toolkit and cofactors are the tools themselves

Answer 411

“ENHANCEOSOME” Efficient transcriptional activation often requires the combined and coordinated activities of several transcription factors bound to different enhancers to recruit RNApol II Though they may be distant in the linear DNA sequence, the DNA is bent and looped such that transcription factor-bound enhancers are physically located near the promoter important for transcriptional regulation -targets to the correct position so transcription occurs - enhancers don't need to be near the promoter, can be downstream because DNA is flexible and can bend to form a loop so the enhancer is brought closer

Answer 412

o Molecular formula of glucose and galactose are identical. The difference between them is the orientation of the hydroxyl group at the 4th Carbon. o Many species including yeast can convert galactose to glucose-1-P for energy and carbon metabolism

Answer 413

o Like the lac genes of E. coli, enzymes that convert galactose to glucose-1-P are only produced when galactose is present. o Four galactose-responsive enzymes are required for this process (GAL1, -2, -7 and -10). o Three regulatory proteins (GAL3, - 4 and -80) are also required

Answer 414

o GAL4 binds to enhancers called “Upstream Activator Sequences” (UAS) located upstream of each of the GAL enzymes - each Gal gene has its own promoter - not organized into an operon - UAS sites are located at a distance from the promoters, not immediately adjacent - when binds, promotes sequence of GAL 7 abd GAL 2

Answer 415

Mutational analysis: Gal4- cells – Gal enzymes uninducible because transcription factor Gal80- cells – Gal enzymes constitutive Gal3- cells – Gal enzymes uninducible - o Gal80 binds to Gal4 activation domain – blocks activation o binding of galactose and ATP changes structure of Gal3, causing Gal3 to bind to Gal80 and remove it from Gal4 (i.e. Gal3 acts as a sensor and inducer).

Answer 416

The activation domain of Gal4 can bind to proteins of the transcriptional machinery: * helps recruit RNA Pol II * also helps recruit chromatin modifying proteins

Answer 417

reporter activated is retina because contains promoter and Pax 6 enhancer

Answer 418

Different enhancers regulate PAX6 expression in specific tissues, an example of transcriptional complexity in eukaryote

Answer 419

Different enhancers regulate PAX6 expression in specific tissues, an example of transcriptional complexity in eukaryote

Answer 420

Different enhancers regulate PAX6 expression in specific tissues, an example of transcriptional complexity in eukaryote

Answer 421

Insulators: - DNA sequences that are specifically bound by "insulator proteins" like CTCF - help to organize chromosomal DNA into specific sub- domains (TADs) essential for proper gene regulation

Answer 422

- transcription factor binding sites are often inaccessible due to their incorporation in nucleosomes - nucleosome remodelling protein complexes like Swi/Snf use energy from ATP to reposition or remove single nucleosomes, exposing binding sites - Swi/Snf is a coactivator - does not bind DNA itself, but gets recruited by enhancer-bound transcription factors

Answer 423

- the amino-terminal ends of core histones protrude from the nucleosome - variety of covalent modifications of these tails can have profound effects on gene regulation: - acetylation - methylation - phosphorylation - ubiquitination - can alter charge to affect affinity for DNA - can also affect the binding of other regulatory proteins (histone code)

Answer 424

- histone modifications are added by histone code writers - histone modifications are removed by histone code erasers - histone modifications are recognized and bound by histone code readers - readers are often coupled with writer or erasers (same protein or by interaction of 2 proteins)

Answer 425

o Histone tail acetylation on lysines neutralizes positive charge, loosens interactions with DNA (negative charge) chromatin relaxation increased accessibility and transcription o Also creates binding sites for histone code readers that help promote activation o Acetylation catalyzed by histone acetyltransferase (HAT) enzymes (“writers”); transcriptional (co)activators often have this activity o Removed by histone deacetylase (HDAC) enzymes (“erasers”); transcriptional (co)repressors often have this activit

Answer 426

o Histone methylation occurs at lysine or arginine residues, catalyzed by histone methyltransferase (HMTase) enzymes o Doesn’t change the charge of histone o Many histone methylation marks (e.g. H3K9Me) are generally associated with gene silencing and act as a signal to recruit specific reader can be unmethylated, monomethylated, dimethylated, or trimethylated

Answer 427

HP-1 (Heterochromatin Protein 1) binds methylated histones, H3K9Me HP-1 promotes heterochromatin formation and recruits additional HMTase HMTase methylates neighbouring nucleosomes etc. etc

Answer 428

heterochromatin spreading is counteracted by HATS bound to barrier insulators = DNA sequence - can stop things from spreading to other regions

Answer 429

Two classes of pigments, brown and red pigments, contribute to the Drosophila eye color. The “white” (w) gene codes an ATP-binding cassette (ABC) transporter that carries the precursors of the red and brown eye color pigments into the developing eyes

Answer 430

Epigenetic silencing by heterochromatin spreading Mutagenize flies with X-rays Screen for offspring with unusual phenotyp

Answer 431

Examples of mutation identified (LOF): -HAT: Histone acetyltransferase Examples of mutation identified (LOF): -Su(var)3-9: Histone methyltransferase -Su(var)205: HP1

Answer 432

Methylation (H3K9Me) promotes heterochromatin formation and tends to spread on chromatin - HP-1 (Heterochromatin Protein 1) binds methylated histones, H3K9Me HP-1 promotes heterochromatin formation and recruits additional HMTase HMTase methylates neighbouring nucleosomes etc

Answer 433

Epigenetics: study of heritable traits that cannot be explained by changes in DNA sequence. o Many modifications to DNA and chromatin structure are inherited: passed on to daughter cells or future generations During DNA replication, existing nucleosomes are disassembled and new nucleosomes on daughter strands are reassembled on the daughter strands from a mix of old and new (unmodified) histones o Old histones with modifications (methylation) direct modification of new histones by recruiting readers and writers Hipfner

Answer 434

heterochromatin spreading is counteracted by HATs bound to barrier insulators = DNA sequence

Answer 435

The difference in the number of X chromosomes between males and females is compensated at the level of transcription in different organisms

Answer 436

* “Dosage compensation” - so that males and females have equal expression levels of the ~1000 genes located on the X chromosome * In early female mammalian embryo (~8-32 cell stage), each cell randomly “inactivates" one of the two X- chromosomes - forms heterochromatic “Barr body” * The same X-chromosome is inactive in all descendants of a given cell - “mitotic” epigenetic inheritance this “mosaicism” of females can lead to unexpected genetic consequences….

Answer 437

Question: which allele is dominant - black fur or orange fur?

Answer 438

* red-green colour blindness is often due to mutation of the OPN1-LW gene on X * mut/Y males are colour blind, but mut/+ females are……… While X-linked mutations mainly affect males, some female carriers can display “partial” symptoms, with variable penetrance and expressivity due to “skewed X inactivation" some X-linked diseases like Rett syndrome are only seen in females…. because mut/Y males are not viable (recessive lethal)

Answer 439

catalyzed by DNA methyltransferase (DNMT) enzymes o occurs primarily on cytosine in CpG dinucleotide (MCpG) o 60 ~80% of CpG are methylated genome-wide in vertebrate

Answer 440

o CpG methylation mostly associated with intergenic regions correlated with repressed chromatin state (heterochromatin) o CpG islands – CpG-rich clusters located near promoters (60% of genes) mostly (>95%) not methylated and transcriptionally active

Answer 441

o Can be direct effect: DNA methylation blocks transcription factor binding o Can also be indirect effect, due to recruitment of HDACs and HMTs that lead to repressive histone modifications

Answer 442

Like DNA sequences and histone tail modifications, DNA methylation patterns are passed on through mitosis o DNMTs have high affinity for hemimethylated sites ▪ guided by methylation pattern on parental strand

Answer 443

Unusual pattern of inheritance of the Igf2 gene involved in embryonic growth o “Insulin-like growth factor 2” o Member of insulin gene family o Binds to receptor on cell surface, stimulates cells to grow o autosomal gene Mutation of Igf2 gene in mice – heterozygous pups born small, but only if mutant allele was inherited from father

Answer 444

o up to ~200 genes in human and mouse genomes where only the paternal or maternal copy (but not both) is expressed – as if there were only one copy of the gene in the cell o Which allele gets expressed depends only on biological sex of parent from which it came - sex-specific gene silencing o The non-expressed allele is said to be IMPRINTED o imprinted copy is inactivated by a mechanism involving DNA methylation in the maternal or paternal germline o methylation imprint is maintained throughout life of the progeny (in somatic cells) - “monoallelic inheritance”

Answer 445

Epigenetic gene regulation through DNA methylation: Genomic imprinting In mice (and humans), Igf2 and H19 genes are adjacent

Answer 446

maternally imprinted only the Igf2 allele inherited from the father is expressed maternal copy is imprinted and inactive

Answer 447

H19 gene: paternally imprinted only the H19 allele inherited from the mother is expressed paternal copy is imprinted and inactive

Answer 448

Alternative methylation at an imprinting control region determines gene expression o sex-specific CpG methylation of Imprinting Control Region (ICR) – only in paternal gametes (sperm) o ICR DNA methylation prevents CTCF binding; me- thylation in H19 promoter silences transcription o unmethylated ICR binds CTCF, acts as an enhancer- blocking insulator

Answer 449

DNA methylation re- moved in germ cells Methylation of ICRs in males (X-inactivation is also erased No methylation of ICRs in females X-inactivation in female is established during embryogenesis

Answer 450

Silver-Russell syndrome - dwarfism o ~50% of cases - hypomethylation of paternal allele (“epimutation”) leads to repression of paternal Igf2 expression

Answer 451

Maternal deposition of piRNAs into the embryo cytoplasm provides resistance against P-elements in the progeny

Answer 452

* Size of a human egg (oocyte) 100 μm, the largest cell in human * Sperm head 5.1 μm by 3.1 μm and a tail 50 μm long Oocyte cytoplasm contains important cytoplasmic factors (eg. Proteins, RNAs, etc)

Answer 453

-Multiple organelles in each cells. -Organelles contain circular genome (mtDNA and cpDNA). -Organelle DNA replication and division is stochastic, not couple to the cell cycle

Answer 454

o 16,569 nt, circular DNA molecule, coding 37 genes (13 polypeptides). o dozens to hundreds of mitochondria/cell, each with multiple mtDNA copies/mt - 1% of cellular DNA! Gene sequences most similar to α- proteobacteria – endosymbiotic theory BUT… over 1000 proteins in a mt… HOW?

Answer 455

o 99% of mt proteins are encoded in the nuclear genome o The 13 mtDNA-encoded proteins all involved in electron transport chain o Other proteins produced in cytoplasm, imported o Nuclear mt genes are also similar to bacterial genes – must have moved to nucleus over time

Answer 456

Oocyte and sperm both contain mitochondria, but far more in oocyte Upon fertilization, the few sperm mitochondria are destroyed (or consumed) by the oocyte All mitochondria are inherited from the mother (maternal inheritance) mtDNA can be used as a tool to trace maternal heritage (For both males and females)

Answer 457

can lead to heteroplasmy o mt genes ~1000-fold higher rate of mutation than nuclear genes § more frequent DNA replication § no DNA repair o Spontaneous mtDNA mutations can lead to two distinct mt populations within a single cell - heteroplasmy o Random segregation of organelles at mitosis (or meiosis) can lead to homoplasmic cells for mtDNA mutation

Answer 458

Chloroplasts are responsible for green pigmentation (Chlorophyll) in plant homoplasmic: Mainly contains mutant chloroplasts heteroplasmic: In heteroplasmic plants, some branches can be homoplasmic homoplasmic: Mainly contains wild- type chloroplasts If variegated plants are fertilized by wild-type pollens (male germ cells), what will be the phenotype of progeny?

Answer 459

- Children of an affected mother all have the disease. -Whether the grand children have the disease or not is determined by the sex of the child. -In some cases where a mother has a mixture of affected and unaffected mitochondria (heteroplasmic), her children may be norma

Answer 460

o Affect at least 1/5000 people o Progressive o Multi-system disease (high energy-demand tissues)

Answer 461

-Three-person In vitro fertilization: resulting embryo has three parents

Answer 462

o Mid-1970s - Eric Wieschaus and Christine Nusslein-Volhard carried out a forward genetic screen to identify genes required for organizing the Drosophila embryo o First systematic mutagenesis screen in a multicellular organism

Answer 463

Screen 1: for genes required in the mother for normal development of the embryo § maternal mRNA deposited into egg, required for early development before transcription of zygotic genes begins § Maternal-effect genes - phenotype determined by genotype of MOTHER Two screens: Maternal vs zygotic genes 5 Stage 10 Drosophila egg chamber

Answer 464

Screen 2: for genes required in the embryo’s genome § zygotic genes – phenotype determined by genotype of the embryo § inheritance shows “normal” Mendelian pattern 6

Answer 465

Two maternal-effect genes specify the anterior-posterior axis: Anterior: bicoid mutants lack anterior structures (head, thoracic segments) Posterior: nanos mutants lack posterior (abdominal) segments

Answer 466

o bicoid (bcd) mutants lack anterior segments § role in establishment of basic positional information along anterior- posterior axis Fig 13-15a o encodes a transcription factor § protein in A à P gradient

Answer 467

bicoid mRNA (from mother) localizes to anterior of embryo Bicoid protein translated in anterior from maternal mRNA Bicoid proteins diffuse toward the posterior region, making a concentration gradient in syncytium – providing information about distance from anterior

Answer 468

Bicoid is required to form anterior structures… is it sufficient? YES – Injected bicoid mRNA directs formation of anterior structures

Answer 469

* maternal nanos mRNA is localized to the posterior end of the embryo, translated protein forms gradient from P to A * Nanos inhibits translation of uniformly-distributed maternal hunchback mRNA * creates an A to P gradient of the transcription factor Hunchbac

Answer 470

o Gap genes (9 identified) e.g. zygotic hunchback, kruppel, knirps, giant § all are transcription factors Gap genes translate maternal A-P gradients into broad subdomains Fig 13-14 § mutants lose several consecutive segments Fig 12-15b – Kruppel protein localization § transcriptionally activated or repressed over broad domains by maternal-effect gene products

Answer 471

o e.g. - zygotic hunchback (hb-z) § 3 binding sites for Bcd in hb promoter o presence of multiple binding sites allows for: § More sensitive detection of gradient of Bcd protein. § Activation of a specific gene could depend on the concentration of Bcd and the number of binding sites

Answer 472

The pattern of Gap gene expression is controlled by previously established gradients of maternal-effect gene products (activated or repressed they also feed back to control each other’s expression

Answer 473

o Pair-rule genes (8 identified) * all transcription factors * mutants missing every other segment *expressed in alternating segments

Answer 474

o Even-skipped (eve) § stripe 2 will have specific concentration of maternally loaded and zygotic factors * High [Hb-z] * Medium [Bcd] * Low [Gt], [Kr]

Answer 475

o even-skipped (eve) Each enhancer has different arrangements of binding sites for maternally loaded factors and Gap genes. This allows enhancers to be activated at a specific stripe.

Answer 476

Segment polarity genes § encode components of two cell-cell signaling pathways (Hedgehog, Wingless); includes secreted proteins, membrane receptors, transcription factors… activated/repressed by pair-rule genes § Function to define A and P within each segment… mutants have mirroring of one half of each segment

Answer 477

§ Mutant animals lack a particular structure, which is replaced by another structure normally found other body segments (homeotic transformation). § Ultrabithorax (Ubx) – discovered in 1915; second thorax and set of wings in place of halteres § Antennapedia (Antp) – leg in place of antenna

BIOL 202 Flashcards

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