pt 2 Flashcards

Question

how does RNAi work?

Answer 1

uses complementary RNA to silence gene expression

Answer 2

used successfully in cell culture but took over a decade to live up to its promise as a treatment for disease

Answer 3

using gene therapy to stimulate naturally occurring microRNAs (miRNAs)

Answer 4

- Can therapeutic gene expression be controlled? - Can we safely and efficiently target only cells that require the gene? - How can gene therapy be targeted to specific regions of the genome? - How long will therapy last? - Will immune system reject? - How many cells need to express the therapeutic gene to treat the condition effectively?

Answer 5

transcription activator-like effector nuclease. these are engineered proteins used for precise genome editing which combine: A DNA-binding domain from TALEs (proteins derived from plant pathogens like Xanthomonas) A DNA-cutting domain from the FokI endonuclease

Answer 6

Cas9 uses a guide RNA to locate and bind a specific DNA sequence next to a PAM site. It then creates a double-strand break, allowing targeted gene disruption or editing. Has been used successfully in a wide range of organisms

Answer 7

nuclease: - Cas9 single-guide RNA: - CRISPR RNA (crRNA) - trans-activating crRNA (tracrRNA) - optional: section of DNA repair template that is utilised in either non-homologous end joining (NHEJ) or homology directed repair (HDR)

Answer 8

the target sequence (20bp long) and the PAM

Answer 9

specificity - allows us to target and modify particular DNA sequences in the vast expanse of a genome

Answer 10

Goal: Use CRISPR-Cas9 to cut the mutated HBB gene and provide a corrected DNA template for homology-directed repair (HDR). - Extract patient hematopoietic stem cells (HSCs) from bone marrow or blood. - Use CRISPR-Cas9 + a DNA template to correct the point mutation in HBB ex vivo. - Re-infuse the edited HSCs into the patient → they produce healthy red blood cells.

Answer 11

A knockout deletes or disrupts a gene using error-prone NHEJ to inactivate it. A knockin inserts or replaces a gene sequence using precise HDR, often to fix or modify a gene.

Answer 12

- cancer is a genetic disease - different inheritance pattern than other genetic disorders

Answer 13

- mutations in genes that regulate cell division, or activate cell death - environmental factors

Answer 14

- inherited mutations can predispose to cancer (germline) - mutations causing cancer occur in somatic cells - mutations accumulate in clonal descendants of a single cell

Answer 15

1. protein that stimulates cell division starts working more 2. protein that blocks the cell from dividing stops working

Answer 16

- uncontrolled cell growth - genomic and karyotypic instability - potential for immortality - ability to invade and disrupt local and distant tissues

Answer 17

- autocrine stimulation - loss of contact inhibition - loss of apoptosis

Answer 18

- when autocrine stimulation is absent, the cell receptor responds to the binding of a ligand produced elsewhere - when autocrine stimulation is present, the cell receptor responds to the binding of a ligand produced in the cell

Answer 19

- contact inhibition present: normal cells stop dividing once they cover a surface and touch neighbouring cells. - contact inhibition absent: cancer cells continue to divide even after touching neighboring cells.

Answer 20

- following irradiation, most normal cells recognise the DNA damage that has occurred, leading to cell death - following irradiation, most cancer cells continue dividing despite the DNA damage that has occurred

Answer 21

1. replication occurs 2. mismatch is created by DNA polymerase - in a normal cell, this mismatch is corrected by mismatch repair and the following round of replication results in a normal DNA sequence - in a cancer cell, the mismatch is not correct and the following round of replication results in a mutated DNA sequence

Answer 22

- increased rate of chromosomal aberrations - this includes gains/losses of chromosomes, chromosomal rearrangements, and formation of abnormal structures (eg dicentric chromosomes)

Answer 23

number of passages (X-Axis): where you take a cell, grow it, move it to another plate, and repeat cumulative cell number (Y-axis) - in normal cells, as number of passages increases after a certain point, cumulative cell number plateaus - in cancer cells, they lose the ability to stop passaging so the line is straight

Answer 24

1. angiogenesis: the formation of blood vessels to distribute nutrients to all the cancer cells 2. metastasis: where cancer cells break through the basement membrane and go to another body part 3. evasion of immune surveillance

Answer 25

proposes that cancer arises through the accumulation of multiple genetic and epigenetic alterations in clonal copies of a single cell

Answer 26

- scientists explored this through X-inactivation in females - they looked at proteins on the X chromosome - in normal tissue, there was a mixture of cells with proteins from both X chromosomes - in tumour cells, there was a mixture of cells with proteins from only one X chromosome

Answer 27

1. lung cancer death rates: males started smoking around 1930 and lung cancer death rates peaked around 1990. females started smoking around 1950 and lung cancer death rates peaked around 2000. 2. incidents of cancer increase exponentially between age 0 and age 80

Answer 28

- retinoblastoma is caused by inherited mutations in the RB gene - individuals who inherit a single RB- mutant allele are more prone to this cancer as loss of heterozygosity through WT mutation is more likely

Answer 29

a) oncogenes b) tumour suppressor genes

Answer 30

- members of signal transduction systems that produce proteins promoting cell proliferation - cancer occurs when an oncogene obtains a gain-of-function mutation in ONE allele

Answer 31

- members of cell cycle checkpoint control and DNA repair mechanisms - produce protein that inhibits cell proliferation or protects the genome - cancer occurs when a tumour-suppressor gene obtains a loss-of-function mutation in BOTH alleles

Answer 32

- tumour-causing viruses - transform normal mouse cells with human tumour DNA - genome wide functional screens

Answer 33

- Retroviral DNA integrated near proto-oncogenes in cellular DNA. - The viral promoter overactivated these genes, turning them into activated oncogenes. - Deletion and fusion events also captured proto-oncogenes into the viral genome. - Studying these viral fusion genes revealed which host genes could drive cancer.

Answer 34

- human tumor DNA transformed normal mouse cells, causing them to grow uncontrollably. - By isolating the integrated human DNA, scientists identified specific mutated proto-oncogenes

Answer 35

Bcr: signaling molecule involved in regulating cell growth and cytoskeletal organization. Abl: regulating cell differentiation, division, adhesion, and response to DNA damage. - normal chromosome 9 (containing abl) - normal chromosome 22 (containing bcr) - both chromosomes break and swap segments - changed chromosome 9 now does not contain abl, whilst changes chromosome 22 (Philadelphia Chromosome) contains bcr-abl which is now constitutively active -> cancer

Answer 36

RAS is involved in the cell proliferation pathway - in a normal cell, the binding of a growth factor to a receptor activates RAS, causing proliferation - in a cancer cell, proliferation occurs even without the growth factor

Answer 37

Her2 regulates cell growth - in normal cells, each cell has 2 copies of Her2 - in breast cancer cells, each cell has more than 2 copies of Her2, leading to multiple copies of the Her2 receptor on the surface of the cell and the cell being able to activate itself for cell growth

Answer 38

through genetic analysis of families with inherited predisposition to cancer

Answer 39

- one normal allele is sufficient for normal cell proliferation in heterozygotes - however, the wild-type allele in somatic cells of the heterozygote can be lost or mutated, leading to abnormal cell proliferation

Answer 40

- nondisjunction (loss) - uniparental disomy - mitotic recombination - gene conversion - deletion - point mutation

Answer 41

1. negative regulators of cell proliferation 2. components of cell cycle checkpoints (before S phase, after G2) 3. involved in DNA damage repair 4. promote apoptosis when DNA damage is excessive

Answer 42

checkpoints monitor the genome and cell-cycle machinery before allowing progression to the next stage of the cell cycle

Answer 43

DNA synthesis can be delayed to allow time for repair of DNA that was damaged during G1

Answer 44

mitosis can be delayed to allow time for repair of DNA that was damaged during G2

Answer 45

monitors formation of mitotic spindle and engagement of all pairs of sister chromatids

Answer 46

1. p53 is a transcription factor that is activated by UV or ionising radiation 2. induces expression of CDK inhibitor, p21 3. p21 inhibits the activity of CDK4 - cyclinD complexes 4. if these complexes are inhibited, then Rb is not phosphorylated 5. Rb remains unphosphoryalted so it can inhibit E2F, preventing entry into the S phase of the cell cycle

Answer 47

E2F is no longer inhibited, so the genes necessary for DNA synthesis are activated

Answer 48

- inactive p53 means p21 is not induced - p21 cannot inactivated CDK4/cyclin D so this complex phosphorylates Rb - E2F is constitutively phosphorylated, causing the cell to enter the S phase - if mutations in the DNA have occurred, these are not repaired before mitosis

Answer 49

targets proliferating cells: - disrupts DNA replication - prevents formation of the mitotic spindle

Answer 50

drugs used in chemotherapy target rapidly dividing cells; however, they also affect normal dividing cells, leading to adverse side effects

Answer 51

- design drugs that specifically bind and inactivate oncogenic proteins - mobilise the immune system to eradicate cells that overexposes a particular oncogene

Answer 52

- typically the Bcr/c-Abl enzyme binds to ATP and a target protein, leading to chronic myelogenous leukaemia - Gleevec binds to the ATP active site, inhibiting it

Answer 53

- typically, abundant Her2 in cancer cells leads to dimerization, turning the signal transduction pathway on and leading to cell proliferation - Herceptin attracts the immune system cells to attack the cancer cells. this prevents dimerization and recruits killer T cells, turning the signal transduction pathway off and stopping cell proliferation

Answer 54

- many tumour suppressor gene products are involved in DNA repair/DNA damage checkpoints - in many cells excessive DNA damage induces programmed cell death strategy: utilise the ultra sensitivity of cancer cells to DNA damage to elicit self-destruction by inducing programmed cell death

Answer 55

PARP inhibitors block the action of poly(ADP-ribose) polymerase, needed to mend nicks in DNA without PARP, nicks in DNA -> double stranded breaks -> self-destruction by apoptosis

Answer 56

in a normal cell with a functional homologous recombination (HR) pathway. there is HR-mediated DNA repair of the double-strand breaks, enabling cell survival. some cancer cells do not have this

Answer 57

- they increase genomic instability - in cells with dysfunctional apoptotic pathways, this may simply lead to the accumulation of new cancer associated mutations

Answer 58

customising medicine by designing the most effective drug therapy and treatment strategies based on the specific genetic profile of a patient

Answer 59

- possibly most effective treatments would be directed against the common (early_ mutations to target all the cells - help find the driver mutations and 'druggable' targets

Answer 60

monoclonal antibody can prevent the interaction between tumour cells and T cells so that the T cell can no longer recognise the tumour cell

Answer 61

Car T cells have a chimeric antigen receptor which can recognise the cancer cell

Answer 62

- easier to implement for hematopoietic cancers (unique antigens) - can induce a cytokine storm (life-threatening)

Answer 63

- evolution has conserved basic strategies of development across multicellular eukaryotes - many homologous proteins have highly conserved functions

Answer 64

- eyeless (Drosophila), Pax-6 (mouse), and Aniridia (humans) genes are homologous - Pax-6 and Aniridia wild-type genes can direct eye development in Drosophila

Answer 65

Forward: Phenotype → Find gene - Mutagenize, screen for trait, map gene - Use when you know the trait but not the gene Reverse: Gene → Find phenotype - Knockout/mutate gene, observe effect - Use when you know the gene but not its function

Answer 66

- motor axon guidance mechanisms in Drosophila. - scientists mutagenise C. elegans, observe the phenotype, then try and identify the gene responsible for each phenotype

Answer 67

type of genetic screen used to find genes that influence the severity or expression of a known phenotype, rather than causing the phenotype on their own.

Answer 68

two or more genes perform overlapping or similar functions, so that loss of one gene has little or no effect on phenotype

Answer 69

- Wild type: Normal eye structure. - sev::Ras^G12V: Rough eye due to hyperactive Ras signaling. - Enhanced phenotype (E⁻/E⁺; sev::Ras^G12V): worsened mutant phenotype → E is a negative regulator of Ras. - Suppressed phenotype (S⁻/S⁺; sev::Ras^G12V): More normal eye → S is a positive regulator of Ras.

Answer 70

- we can design RNAs that can target specific genes and inactivate them, similar to a knockout - this produces a phenocopy of a loss of function mutation

Answer 71

a phenotype that mimics the effects of a gene knockout, caused by methods like RNAi or CRISPR.

Answer 72

Nature of the encoded protein: - we can infer amino acid sequence from nucleotide sequence and use computer searches to identify known motifs Location and timing of gene expression: - during development, where and when is the mRNA found? Location of the protein product: - during development, where and when is the protein found? Developmental phenotypes: - what cells or tissues are affected by loss-of-function?

Answer 73

- attach antibodies to the protein - tag the proteins with GFP

Answer 74

genes don't work in isolation, and many biological processes are complicated and require the coordinated action of many genes

Answer 75

analysis of effects of one gene on expression of another gene: - does a mutation in one gene affect the level or distribution of mRNA or protein from another gene? analysis of double mutants - epistatic interactions: - do mutations in two different genes define successive steps in a pathway?

Answer 76

- staining of Drosophila wing imaginal disks for wingless protein (Wg, green) and vestigial protein (Vg, red) - wild-type produces a different staining pattern than wingless mutant - this suggests that the wingless gene product is required for the expression of the vestigial gene

Answer 77

- phenotypes of the two mutants must differ - alleles must either be null or constitutive

Answer 78

A mutant allele that results in complete loss of function of the gene.

Answer 79

A mutant allele that causes the gene to be continuously active, regardless of normal regulatory signals.

Answer 80

- the phenotype of the double mutants reveals epistatic interactions - in a double mutant, the phenotypes the same as the Ras^G12V mutant, therefore seven less is upstream of Ras - the downstream gene is the one that is epistatic - thus, Ras is downstream and epistatic to sevenless ('masks' sevenless)

Answer 81

- eliminates unwanted or damaged cells - was understood by epistatic analysis screens

Answer 82

each embryonic segment defines a specific structure in the adult: 3 head segments, 3 thoracic segments, 8 abdominal segments

Answer 83

Christiane Nusslein-Volhard, Eric Wieschaus, Edward B. Lewis

Answer 84

1. Screening for abnormal embryos in homozygous mutant females - Identified recessive mutations in maternal-effect genes 2. Screening for abnormal homozygous mutant embryos - Identified recessive mutations in three classes of zygotic segmentation genes

Answer 85

Function in a hierarchy that progressively subdivides the embryo into successively smaller units - maternal genes - gap genes - pair-rule genes - segment polarity genes

Answer 86

- expressed by the mother - mRNAs deposited in egg but not translated until after fertilisation - products of maternal effect genes establish morphogenic gradients in the egg - these gradients ensure that gap genes are expressed only in certain broad regions of the embryo

Answer 87

- expression is controlled by maternal gene products - expression begins at syncytial blastoderm stage - activate pair-rule genes in a series of seven stripes

Answer 88

- seven zones of expression are controlled by gap and maternal gene products - levels of pair-rule gene products restrict the expression of segment polarity genes to a series of 14 stripes, one per segment

Answer 89

- expression in 14 segments is controlled by pair-rule gene products - this divides the embryo into 14 segment-sized units

Answer 90

two maternal-effect gene products which are morphogens: bicoid (bcd) and nanos (nos)

Answer 91

a substance that defines cell fate in a concentration-dependent manner

Answer 92

- bcd and nos are transcribed by the mother and their mRNAs are localized to opposite poles of the oocyte - bcd and nos mRNAs are not translated in the embryo until after fertilisation - each protein forms a gradient in the embryos (diffusion)

Answer 93

bcd is highest at anterior and lowest at the posterior

Answer 94

nos is lowest at anterior and highest at the posterior

Answer 95

- bcd mRNA localizes to the anterior pole of the oocyte - bcd protein diffuses from the anterior pole of the embryo to produce an anterior-to-posterior gradient

Answer 96

- dosage of maternal bcd gene determines how much of the embryo becomes head structures (the more protein in the anterior, the greater the size of the segment) - bicoid null mothers produce embryos that lack all head and thoracic structures

Answer 97

- bicoid protein activates the expression of hunchback gene and represses translation of caudal mRNA - this causes posterior-to-anterior gradient of caudal protein - nanos protein represses the translation of hunchback mRNA - this causes anterior-to-posterior gradient of hunchback protein

Answer 98

by zygotic genes: - bicoid, hunchback, and caudal proteins are transcription factors that control the spatial expression of zygotic genes - zygotic gene expression begins in the syncytial blastoderm stage

Answer 99

- gap genes - pair-rule genes - segment polarity genes

Answer 100

primary pair-rule genes and secondary pair-rule genes

Answer 101

- expression is controlled by transcription factors encoded by maternal genes and zygotic gap genes - upstream region of each pair-rule gene has multiple binding sites for transcription activation/repression

Answer 102

- expression is controlled by transcription factors encoded by other pair-rule genes

Answer 103

- in seven stripes - two segment periodicity: each stripe has two segments

Answer 104

eve transcription is activated by Bcd and Hb, but repressed by giant (Gt) and Kruppel (Kr) proteins

Answer 105

by combinations of maternal-effect and gap proteins (see slide 32_

Answer 106

- segment polarity genes are expressed in stripes that are repeated with single segment periodicity (one stripe per segment) - interactions between various polarity genes maintains the periodicity

Answer 107

transcription factors

Answer 108

in successive levels of the hierarchy, genes are expressed in narrower bands

Answer 109

- transcription of homeotic genes is controlled by gap, pair-rule, and segmentation genes - at the cellular blastoderm stage, each homeotic gene is expressed within a subset of body segments - homeotic genes are master regulators that control transcription of many genes responsible for development of segment-specific structures

Answer 110

when one body part develops as if it were another, due to mutations in homeotic (Hox) genes, which control body plan identity during development.

Answer 111

Wild type wing vs bithorax mutant vs postbithorax mutant

Answer 112

in spatially restricted domains

Answer 113

each on a different chromosome: HoxA HoxB HoxC HoxD

Answer 114

all animal genomes - even those of sponges - contain Hox genes

Answer 115

if you just mutagenise the wild type, you may have a whole bunch of different mutations, but if there's redundancy you won't see the phenotype you are studying

Answer 116

sev is a receptor tyrosine kinase that activates Ras signaling to specify the R7 photoreceptor.

Answer 117

When sev is activated by its ligand, it triggers a cascade that activates Ras, which in turn promotes cell differentiation (the specification of the R7 photoreceptor neuron).

Answer 118

gain of function mutation - this mimics constant "on" signaling, so R7 cells develop even without the proper upstream signal, or develop abnormally, causing things like a rough eye phenotype.

Answer 119

genes that are 100% inherited from the maternal side

Answer 120

genotype =/= phenotype - monozygotic twins are genotypical identical but not always phenotypically identical, and these differences can't always be explained by environment - mice with genotype A^vy/a genotype are also genetically identical, but some have brown fur and some have yellow fur

Answer 121

the study of mitotically and/or meiotically heritable changes in gene function that cannot be explained by changes in DNA sequence

Answer 122

1. DNA methylation on CpG islands 2. covalent modification of histone tails 3. noncovalent modification of histones 4. non-coding RNAs

Answer 123

CpG = Cytosine-phosphate-guanine nucleotides - some regions have many CpG islands, in which cytosine may be either methylated or unmethylated - typically, unmethylated promoters enable gene expression - typically, methylated promoters result in heterochromatin, repressing gene expression - the effects of methylation depend on the region; sometimes DNAm leads to silencing, sometimes it leads to increased expression

Answer 124

- acetylation: loosens chromatin by neutralizing lysine charges, promoting gene expression - methylation: can activate or repress gene expression, depending on the site and number of methyl groups. - ubiquitination: tags histones for degradation or signals chromatin remodeling, affecting transcriptional activity

Answer 125

reposition nucleosomes to make DNA more or less accessible.

Answer 126

Transcriptional Silencing: Small RNAs (like siRNAs) guide chromatin modifiers to DNA, causing heterochromatin formation and blocking transcription.

Answer 127

- promoters can be hidden when wrapped in nucleosomes, leading to lowered gene expression - chromatin remodelling complexes can expose gene promoters, allowing RNA polymerase to bind - nucleosomes in heterochromatin can be tightly packed, generating silenced heterochromatin

Answer 128

the environment

Answer 129

- the agouti gene is normally expressed in a regulated pattern, giving banded hairs (black-yellow-black) and brown/gray fur. - the Aᵛʸ allele contains an IAP (Intracisternal A Particle) retrotransposon upstream of the Agouti gene. - the IAP has a promoter that can drive ectopic (abnormal) Agouti expression. - Unmethylated IAP → promoter is active → overexpression of Agouti → yellow coat. - Methylated IAP → promoter silenced → Agouti expressed normally (or not at all) → brown or mottled coat.

Answer 130

an allele whose expression is not fixed but can vary in a stable way between cells or individuals due to epigenetic states established early in development.

Answer 131

its expression is controlled by DNA methylation at a retrotransposon promoter, leading to variable, heritable gene expression without DNA sequence changes.

Answer 132

1. Maternal diet can directly affect the epigenetic regulation of genes in offspring. 2. These changes can be stable enough to affect not just the immediate offspring, but also subsequent generations.

Answer 133

- When pregnant mothers were fed a diet rich in methyl donors, their offspring showed a higher proportion of the pseudoagouti phenotype (brown fur). - This was because methylation silenced the Agouti gene, preventing ectopic expression and resulting in darker coat color. - When mothers were fed a normal diet, their offspring showed more of the yellow or mottled phenotypes. - This was due to lower methylation of the Agouti gene, allowing it to be expressed and leading to lighter coat color.

Answer 134

- F0 mothers were supplemented with methyl donors during pregnancy. - Her F1 offspring showed increased methylation at the Agouti locus. - The F2 generation also exhibited more pseudoagouti phenotypes, despite not being directly exposed to the diet. - This suggested that the epigenetic marks were inherited through the germline. - F0 mothers were supplemented with no methyl supplementation. - the F1 and F2 offspring showed more yellow or mottled phenotypes, indicating low methylation.

Answer 135

specific functions of different cell types are generated through differential gene regulation

Answer 136

- random X-chromosome inactivation in females occurs early during development for dosage compensation - these X-chromosomes are reactivated in germ cells - inactivation of the paternal/maternal X chromosome is random but persists in the subsequent cells produced - this means that females express X^m in some cells and X^p in others, leading to clonal patches

Answer 137

in the calico cat, X-inactivation leads to a mosaic of fur colours

Answer 138

- process begins at the X-inactivation centre (XIC), which activates the gene Xist. - Xist makes a long non-coding RNA (lncRNA) that spreads across the X chromosome, coating it and starting the silencing process. - Hypoacetylation of a Lys of histones (H3/H4), methylation of histone H3 and underlying DNA shut down gene expression - the inactive X becomes tightly packed into a structure called a Barr body. - most genes on this X are turned off, but a few, called escapees, remain active.

Answer 139

- histone tail usually positively charged, DNA negatively charged - when the tail is acetylated, the charge is neutralised and DNA becomes more loosely packed

Answer 140

Hypoacetylation = regions that are silenced Hyperacetylation = transcriptionally active regions

Answer 141

autosome + one copy of Xist -> autosome becomes a heterochromatic Barr body thus, we could add Xist to one of the 3 copies of chromosome 21 in children with Down syndrome

Answer 142

genomic imprinting

Answer 143

it results from transcriptional silencing

Answer 144

when the expression of a gene depends upon its parental source (ie whether it is inherited from the maternal or paternal side)

Answer 145

- paternally silenced - only the maternal allele is expressed

Answer 146

- maternally silenced - only the paternal allele is expressed

Answer 147

- methylated by special methylases - demethylated by demethylases

Answer 148

through the action of DNA methyltransferases, slide 22

Answer 149

long term repression through DNA methylation

Answer 150

- in the early primordial germ cells, existing methylation marks are removed. - later in gametogenesis, new methylation marks are added. - female places maternal imprints on all eggs - male places paternal imprints on all sperm

Answer 151

- single enhancer downstream of H19 gene controls the expression of both genes on maternal chromosome: genes and regulatory sequences are not methylated, so CTCF binds to insulator blocking activation of IGF2, but allowing for the activation of H19 on paternal chromosome: insulator and promoter region of H19 gene is methylated. CTCF cannot bind to insulator, and activator is able to active transcription of IGF2. - thus, due to mthe methylation of its promoter, the activator cannot active H19

Answer 152

- In regular genes, a mutation in one allele can often be compensated by the other normal allele. - In imprinted genes, since one allele is silenced, a mutation in the active allele acts like a dominant mutation - A mutation in the silenced allele is not expressed despite being present in the genome

Answer 153

obese, small hands and feet, eats uncontrollably, does not mature sexually, short stature, mental retardation

Answer 154

developmental delays, severe mental and motor retardation, prominent jaw, happy disposition

Answer 155

A deletion of the paternal copy of 15q11–q13. The same region on the maternal chromosome is silenced by imprinting.

Answer 156

A deletion of the maternal copy of 15q11–q13. The paternal allele of a key gene, UBE3A, is silenced in neurons due to imprinting. No active UBE3A in neurons → leads to AS.

Answer 157

1. rearrangement breakpoint may acquire new patterns of gene expression and create new gene functions by fusion of two separate genes 2. some rearrangements contribute to the process of speciation 3. duplications provide extra gene copies that can acquire new functions

Answer 158

- deletion - inversion (180 degree rotation of a piece of DNA) - deletion in one chromosome, duplication in another - translocation of a piece of DNA into another chromosome

Answer 159

- deletion - inversion - deletion in one chromosome, duplication in another - reciprocal translocation of a piece of DNA into another chromosome

Answer 160

1. intragenic: small deletion within a single gene 2. multigenic: many genes deleted

Answer 161

- short for deletion (deficiency) homozygote, is an individual that has both copies of the same chromosomal region deleted - usually inviable

Answer 162

- gene imbalance - might result in haploinsufficiency

Answer 163

a DNA loop formed during meiosis when one homologous chromosome has a segment deleted. The extra DNA on the normal homolog that has nothing to pair with loops out.

Answer 164

when a recessive allele is expressed in a heterozygous individual because the dominant allele has been deleted or is missing.

Answer 165

- deletions may uncover recessive mutations - they can be used to locate genes for mapping

Answer 166

1. Start with a mutant strain that has a recessive mutation (e.g. mut) causing a known phenotype. 2. Cross it with a strain that carries a known deletion of part of the chromosome (e.g. Df1). 3. Examine the phenotype of the offspring (heterozygotes: mut / Df1): - If the offspring show the mutant phenotype, then the deletion likely removes the same gene as the mutation → No complementation. - If the offspring show the wild-type phenotype, the mutation must lie outside the deleted region → Complementation occurs.

Answer 167

tandem duplications: the duplicated segment is inserted right next to the original. non tandem (dispersed) duplications: the duplicated segment is inserted elsewhere in the genome, not adjacent to the original.

Answer 168

- less likely to affect phenotype - in some cases causes a dosage effect/genetic imbalance - genes may be placed in a new location that modifies their expression

Answer 169

1. X-ray breaks/any other cause of breaks: - X-rays break one chromosome in two places - X-rays break homologous chromosome in one place - during repair, the freed segment from the first chromosome is mistakenly inserted at the break site on the homolog -> non tandem duplication 2. Unequal crossing over: - Homologous chromosomes misalign during meiosis. - Crossing over occurs at these misaligned points. - One chromosome gains extra DNA (duplication), the other loses it (deletion).

Answer 170

unequal crossing over, causing increase and reciprocal decrease in the number of copies (eg Bar-eye in drosophila)

Answer 171

- most inversions to not alter phenotype unless breakpoints occur within genes - but genes may be placed in a new location that modifies their expression (eg Antennapedia)

Answer 172

1. pericentric inversion - includes the centromere 2. paracentric inversion - does not include the centromere

Answer 173

- Genes remain intact. - Order of genes is reversed in the inverted segment. - Usually no gene disruption or loss of function. - May affect gene expression if regulatory regions are affected.

Answer 174

- The gene is disrupted (split) and mutated - Usually causes loss of gene function or creates a truncated protein. - Can lead to a nonfunctional or altered gene product.

Answer 175

- Both genes are disrupted at the breakpoints. - May create fusion genes by joining parts of two genes. - Can produce novel or dysfunctional proteins. - Often causes loss of function or gain of abnormal function.

Answer 176

- form in inversion heterozygotes - enables pairing of homologous regions despite the reversed gene order. - produces abnormal recombinant chromosomes.

Answer 177

Normal chromosome + Inversion chromosome. Inversion loop forms outside the centromere: - if crossing over occurs inside this loop, it produces one dicentric chromosome (with two centromeres) and one acentric fragment (without a centromere) - the acentric fragment is lost - there is a random break in the dicentric bridge of the dicentric fragment Results in: one normal product, two deletion products, and one inversion product with all genes present. Reduced number of viable gametes

Answer 178

Normal chromosome + inversion chromosome. Inversion loop includes the centromere: - if crossing over occurs inside this loop, it results in gene imbalance - one normal product, two different inviable deletion/duplication products, one viable inversion product (all genes present). Reduced number of viable gametes

Answer 179

- engineered chromosomes used in genetics that carry multiple inversions and sometimes other rearrangements. - these prevent crossing over from happening during meiosis

Answer 180

- most translocations do not alter phenotype unless breakpoints occur within genes - but genes may be placed in a new location that modifies their expression

Answer 181

a) nonreciprocal intrachromosomal translocation b) nonreciprocal interchromosomal translocation c) reciprocal interchromosomal translocation

Answer 182

- arises from a specific reciprocal translocation between chromosomes 9 and 22 - this translocation fuses the BCR gene from chromosome 22 with the ABL gene from chromosome 9, creating the BCR-ABL fusion gene. - the BCR-ABL fusion produces a constitutively active tyrosine kinase protein that drives uncontrolled cell division and the malignant proliferation of white blood cells seen in CML.

Answer 183

- alternate - adjacent-1 - adjacent-2

Answer 184

- semiysterility since <50% of the time there are viable gametes - pseudo linkage since genes can't independently assort - only alternate segregation produces viable progeny

Answer 185

reciprocal exchange between acrocentric chromosomes generates a large metacentric chromosome and a small chromosome (which may be lost)

Answer 186

- a parent carries a Robertsonian translocation involving chromosome 21 and another acrocentric chromosome (commonly chromosome 14). - this carrier has 45 chromosomes but is usually healthy because they have all the essential genetic material. - however, during gamete formation, abnormal segregation can lead to a child inheriting two normal chromosome 21s plus the translocated chromosome containing an extra copy of 21 material. - this causes trisomy 21, the genetic cause of Down syndrome. table

Answer 187

Fluorescent in situ hybridisation (FISH) - FISH Karyotype - Multicolour banding

Answer 188

- uses fluorescent probes that bind to specific DNA sequences or regions on chromosomes. - detects presence, absence, or location of specific genes or chromosomal regions. - useful for identifying known abnormalities (e.g., deletions, duplications, translocations). - shows bright fluorescent signals on chromosomes under a microscope.

Answer 189

- specialized form of FISH that uses multiple probes along a single chromosome to create a unique banding pattern with different colors. - allows high-resolution analysis of structural chromosome rearrangements. - can distinguish small intrachromosomal changes like inversions or complex rearrangements. - provides a detailed “barcode”-like pattern for precise mapping.

Answer 190

chromosomes from tumour cells may be present in larger copies

Answer 191

- fast - inexpensive - highly sensitive

Answer 192

- house mice in the island of Madeira - different Madeira mouse populations have unique Robertsonian translocations, forming distinct chromosomal races. - reduced fertility of heterozygotes for translocations can contribute to reproductive isolation and promote speciation

Answer 193

1. karyotypes generally remain constant within a species - most genetic imbalances result in a selective disadvantage 2. related species usually have different karyotypes - closely-related species differ by only a few rearrangements - distantly-related species differ by many rearrangements - correlation between karyotypic rearrangements and speciation

Answer 194

number of different/unique chromosomes that make up a single complete set eg 23 in humans

Answer 195

number of chromosomes in a gamete eg 23 in humans

Answer 196

n = x - each gamete contains a single complete set of chromosomes

Answer 197

A condition where a cell has a complete set(s) of chromosomes.

Answer 198

A condition where a cell has an abnormal number of individual chromosomes, not involving whole sets.

Answer 199

monoploidy (x) diploidy (2x) triploidy (3x) tetraploidy (4x)

Answer 200

euploidy: 2n nullisomy: 2n-2 monosomy: 2n-1 trisomy: 2n+1

Answer 201

- male bees, wasps, and ants - they undergo parthenogenesis - usually lethal in other organisms

Answer 202

development of unfertilised egg into an embryo (with no fertilisation): - single set of chromosomes - produce gametes by mitosis

Answer 203

- unmasks recessive lethal alleles (eg X-linked diseases) - if an individual survives to adulthood, this most likely leads to sterility

Answer 204

- visualize recessive traits directly - introduction of mutations

Answer 205

1. take a diploid plant 2. haploid pollen grains are treated and plated onto agar 3. growth of haploid embryoids 4. embryoids treated with plant hormones 4. monoploid plant (usually sterile)

Answer 206

- in meiosis, chromosomes are supposed to pair up and separate. - in monoploids, there's nothing to pair with, so meiosis is disrupted, leading to infertile gametes.

Answer 207

- colchicine inhibits the formation of the mitotic spindle - thus the plant cells become diploid and meiosis is able to occur normally

Answer 208

alfalfa, coffee, peanuts, large apples, pears, grapes

Answer 209

large strawberries

Answer 210

- associated with origin of new species - may positively correlate with size and vigor

Answer 211

- autopolyploids - allopolyploids

Answer 212

- originate within a species - all polyploids with an odd number of chromosome sets are sterile because they cannot produce balanced gametes, producing aneuploid gametes

Answer 213

- if x is small - balanced gametes are only produced if two copies of each chromosome always segregate to the same daughter cell and the third to the other

Answer 214

- when the 2x genome of a diploid is doubled to 4x, with all four sets coming from the same species - could be spontaneous or induced by a drug such as colchicine - often the source of a new species

Answer 215

- each chromosome has three homologs to choose from - in order to form balanced gametes, the four copies of each group of homologues must form two bivalents - successful tetraploids produce balanced 2x gametes and are fertile

Answer 216

- hybrid of two or more closely-related species - partially homologous chromosomes

Answer 217

doubled diploid: contains two different diploid genomes

Answer 218

F1 hybrid of wheat and rye (triticale) is sterile because there are no pairing partners for the rye chromosomes

Answer 219

- some combine high yield of wheat with ability of rye to grow in unfavourable environments - some combine high level of protein from what with high level of lysine from rye

Answer 220

XXY, XXX, XO, XYY

Answer 221

can result in a mosaic - am individual has two or more populations of genetically different cells in their body.

Answer 222

an organism that has both male and female physical characteristics — often with distinct regions of male and female tissues

Answer 223

1. mitotic nondisjunction 2. mitotic chromosome loss

Answer 224

offspring of fertile aneuploids have an extremely high chance of aneuploidy because of production of unbalanced gametes

Answer 225

- incidence of abnormal phenotypes caused by aberrant chromosome organisation or number is 0.4% - half of spontaneously aborted foetuses have chromosome abnormalities - incidence of abnormal phenotypes caused by single-gene mutations is 0.01%

Answer 226

2n-1; usually lethal in utero in humans, but there are a few exceptions: - Monosomy 21: born with severe multiple abnormalities but die shortly after birth - Turner Syndrome (XO): 99% of affected foetuses are not born. Those who are born have developmental abnormalities

Answer 227

short stature, sterile due to rudimentary ovaries and lack of menstruation

Answer 228

- embryogenesis: X inactivation occurs after several rounds of cell division, so it is thought that the 2nd X chromosome has important function within the first hundred divisions - some of the genes on the 'inactivated' X chromosome are expressed - germ-line: X chromosome reactivation

Answer 229

2n+1: often lethal in animals owing to chromosome imbalance

Answer 230

Down syndrome - females can be fertile - males infertile - average life expectancy is 40-60 years due to congenital heart disease

Answer 231

Edward syndrome - severe physical and mental abnormalities - heart defects, growth retardation, small jaw, kidney abnormalities, narrow pelvis, rocker bottom feet, clenched fists - average life expectancy of a few weeks - 1/6000 to 1/10000 live births

Answer 232

patau syndrome - severe physical and mental abnormalities - major abnormalities of heart kidneys, brain, face and limbs, small or absent eyes, harelip, small malformed head - average life expectancy of 130 days - 1/12500 to 1/21700 live births

Answer 233

XYY XXX XXY (Klinefelter syndrome)

Answer 234

1/1000 male births - humans tolerate X chromosome aneuploidy because of X inactivation - sterile

Answer 235

some X-linked genes escape inactivation — especially those in the pseudoautosomal regions (PARs). One key gene, SHOX, is essential for bone growth. Loss (like in Turner syndrome) or gain (like in XXX or XXY) of SHOX copy number leads to short or tall stature, respectively.

Answer 236

- look for abnormal karyotypes - possible to screen for biochemical and genetic disorders - tests are done in combination with blood tests for certain maternal and fetal proteins, and with ultrasound tests

Answer 237

first trimester screening test (11 to 13 weeks) ultrasound (nuchal translucency) and maternal blood test [pregnancy associated plasma protein-A (PAPP-A) and β-human chorionic gonadotrophin (β-hCG)]

Answer 238

- chorionic Villi Sampling, CVS (10 to 13 weeks) - amniocentesis (16+ weeks), less invasive

Answer 239

screening tests - if there are abnormalities, we then conduct diagnostic tests.

Answer 240

movement of small segments of DNA called transposable elements from one position to another in the genome

Answer 241

Barbara McClintock in the late 1940s, awarded Nobel prize in 1983

Answer 242

any segment of DNA that evolves the ability to move within a genome

Answer 243

from genetic studies of corn

Answer 244

in all organisms

Answer 245

- previously considered to be selfish DNA carrying no genetic information useful to the host - now it is known that some TEs have evolved functions that are beneficial to the host sometimes

Answer 246

50bp to 10kb

Answer 247

TEs can be present in hundreds of thousands of copies per genome

Answer 248

- some kernels showed a mottled coloration, rather than a uniform color - the mutant phenotype was colourless, whereas the wild type phenotype was red - this pattern could not be explained by standard Mendelian inheritance, since all cells in a kernel should theoretically have the same genotype and produce the same pigment. - this suggested that gene expression was being turned on and off during development, pointing to the movement of genetic elements within the genome.

Answer 249

an unstable version of the C gene responsible for kernel pigmentation. It causes mottled color due to the insertion of a transposable element that disrupts gene function.

Answer 250

It causes colorless kernels with red spots, because the pigment gene is inactivated by an inserted Ds element. If Ac is present, it enables Ds to jump out, restoring gene function in some cells and producing spots of pigment.

Answer 251

Ac is required for Ds to transpose. If Ac is present, it allows the Ds element to move out of the C gene, restoring pigment production in some cells.

Answer 252

Ac (Activator) is autonomous and can transpose on its own. Ds (Dissociator) is non-autonomous and needs Ac to move.

Answer 253

Kernel color is set early in development, when cells are still dividing. Ac must be present in the same cell as Ds during this time to make Ds jump out of the C gene and restore pigment. Once the kernel matures, cells stop dividing and pigment production is fixed, so even if Ac moves later, it can’t change the color anymore.

Answer 254

several types, inserted several times

Answer 255

approximately 12.5% of the genome

Answer 256

44% of the genome

Answer 257

DNA transposons Retrotransposons

Answer 258

move directly without being transcribed into RNA (eg TEs studied by McClintock in corn, P elements in Drosophila)

Answer 259

move via reverse transcription of an RNA intermediate (eg copra elements in Drosophila, L1 and Alu in humans)

Answer 260

LINEs, SINEs, HERVs

Answer 261

encode their own reverse transcriptase and endonuclease, allowing them to copy and insert themselves into new genomic locations.

Answer 262

non-autonomous retrotransposons that rely on LINE enzymes to move - they do not encode any proteins

Answer 263

ancient LTR retrotransposons that originally encoded enzymatic components but are mostly now inactive due to mutations

Answer 264

an enzyme called transposase that helps them cut and move to different places in the DNA.

Answer 265

An intron was added to a yeast Ty1 retrotransposon (located on a plasmid and flanked by LTRs). After splicing and reverse transcription, the DNA inserted into the yeast's genomic DNA lacked the intron, proving the RNA (not DNA) was the template.

Answer 266

Target-site duplication: - they are transcribed into RNA, which is reverse-transcribed into cDNA by reverse transcriptase. This cDNA is inserted into a new genomic site via endonuclease cleavage, generating target-site duplications flanking the new insertion.

Answer 267

most transposons contain two components: - inverted repeats (IRs) of 10-200 bp long at each end - gene encoding transposase, which catalyses the transposition through recognition of the IRs (cuts at border between the IR and genomic DNA)

Answer 268

P element excised from its original location and is transposed to new location by transposase enzyme

Answer 269

in both cases, exonucleases first widen the gap. Then there is: - repair of gap using a sister chromatid or homologous chromosome containing a P element, resulting in the transposon remaining in its original position OR - repair of gap using a homologous chromosome lacking a P element, resulting in the transposon no longer being at the original position

Answer 270

- these are usually inserted into introns - often defective and unable to transpose again due to mutations (ie lack of repeats or active transposase) - epigenetic changes from heterochromatin, resulting in no transcription/transposition - inserted into safe havens (non-essential/repetitive regions)

Answer 271

- production of transposon repressors through alternate splicing - piRNAs (kiwi-interacting RNAs) that block TE transcription and translation (both DNA transposons and retrotransposons)

Answer 272

- unequal crossing over can occur between TEs found in slightly different locations on homologous chromosomes due to misalignment - recombination between them can lead to duplications, deletions, or inversions of the DNA between them.

Answer 273

- bar gene product limits eye growth - its enhancers increase transcription of the gene Bar (one duplication of gene + enhancer): reduced expression leading to bar eye Double Bar (two duplications of gene + enhancer): even more reduced expression

Answer 274

- two transposons can form a large, composite transposon - composite transposon can move to new location - the level of expression of the gene between them may be altered

Answer 275

1. insertion sequence (IS) element: IR - transposase - IR 2. complex transposon (IS + genes): IR - transposase - gene - IR 3. composite transposon (two transposons) - IR - transposase - IR - gene A - gene B - IR - transposase - IR

Answer 276

- wrinkled pea mutation likely resulted from insertion of a TE near the sbe1 gene (starch branching enzyme) - ~100 human genetic diseases i.e. forms of Hemophilia A, Hemophilia B, cystic fibrosis, neurifibromatosis, muscular dystrophy - many spontaneous mutations in the white eye gene (Drosophila)

Answer 277

change genes from one allelic form to another, sometimes leading to the creation of entirely new alleles

Answer 278

genes mutate randomly, at any time and in any cell of an organism

Answer 279

very rarely; on average 1.2x10^-8 mutations/gene/gamete in humans

Answer 280

spontaneous: - arise in absence of known mutagen (polymerase errors, reactive oxygen species etc) - provide 'background rate' of mutation induced (by geneticist -> mutagenesis) - action of mutagen alters nucleotide sequence

Answer 281

only mutations in germline cells can be transmitted to progeny; somatic mutations cannot be transmitted

Answer 282

- mutations affect phenotype rarely - different genes mutate at different rates (mutation rates range from less than 10^-9 to more than 10^-3/gene/gamete) - mutation rate can increase after exposure to a mutagen (eg UV light, certain chemicals)

Answer 283

a base is replaced by one of the other 3 bases

Answer 284

deletions - block of one or more nucleotide (base) pairs is lost insertion - block of one or more nucleotide (base pairs is added)

Answer 285

transition: purine to purine (A<->G) or pyrimidine to pyrimidine (C<->T) transversioin: purine to pyrimidine or pyrimidine to purine

Answer 286

guanine, adenine

Answer 287

thymine, cytosine

Answer 288

- they act as markers for genes - mutations can disrupt gene function. this allows for the study of how the wild-type gene works

Answer 289

the form found in nature (or in a standard laboratory stock); an allele whose frequency is 1% or more of the population

Answer 290

the form that has changed due to a mutation; an allele whose frequency is less than 1% of the population

Answer 291

changes wild-type allele to a different allele

Answer 292

causes novel mutation to revert back to wild-type allele

Answer 293

rate of forward mutation is almost always higher than rate of reverse mutation (except TEs)

Answer 294

the hydrolysis of a purine base from the deoxyribose-phosphate backbone, leading to an apurinic site

Answer 295

100 times/hour in every human cell

Answer 296

the removal of an amino group from a cytosine, leading to the conversion of cytosine to uracil

Answer 297

by uracil-DNA glycosylase - most of the time, it is easy for cells to recognise deamination as uracil is only meant to be in RNA, not DNA

Answer 298

- after replication, one of the strands will be normal (CG) and the other will be mutant (UA) as adenine base pairs with uracil - after another round of replication, one of the strands will be (TA) and the other will be (UA) overall, there is a C-G to T-A transition mutation

Answer 299

naturally occurring radiation such as cosmic rays and x rays often lead to mutations like deletions

Answer 300

- UV light (especially UV-B and UV-C) primarily causes pyrimidine dimers - induces covalent bonds between adjacent pyrimidines (especially thymine-thymine dimers, but also C-T and C-C).

Answer 301

- oxidative damage can occur to any of the 4 base pairs - ROS oxidize guanine into 8-oxoG. 8-oxoG mispairs with adenine during replication. This causes a G:C → T:A transversion in the next round of replication.

Answer 302

- DNA polymerases may lead to DNA replication mistakes - this is caused by the incorporation of an incorrect during replication

Answer 303

DNA polymerase has very high fidelity, so such errors are exceedingly rare

Answer 304

in regions of repeated bases during replication or crossing over in meiosis

Answer 305

DNA polymerase can slip or misalign on the template strand, forming a loop of extra repeats on the new strand, leading to expansion

Answer 306

DNA polymerase can slip or misalign on the template strand, forming a loop of extra repeats on the template strand, leading to contraction

Answer 307

Fragile X-syndrome, Huntington and other disorders of the nervous system

Answer 308

- dictionary used to translate nucleic acids to a amino acids - codon language - redundant (multiple codons per amino acid)

Answer 309

this protects against the negative impacts of mutations

Answer 310

altered codon resulting from the mutation specifies the same amino acid

Answer 311

altered codon resulting from the mutation specifies a different amino acid

Answer 312

identifies specific DNA sequences

Answer 313

identifies (m)RNA sequences

Answer 314

identifies protein sequences

Answer 315

conservative: - substitutes chemically similar amino acid - less likely to alter function or structure of protein non-conservative: - substitutes chemically different amino acid - more likely to alter function or structure of protein

Answer 316

introduction of a (early) stop codon (TAG, TAA, TGA)

Answer 317

- the insertion or deletion of nucleotides in a DNA sequence not in multiples of three - shifts the reading frame used during translation, drastically altering the resulting protein

Answer 318

a second mutation that occurs within the same gene as a previous mutation and restores, at least partially, the original function that was lost or disrupted by the first mutation

Answer 319

- mutations in promoter or termination signal sites - splice donor/acceptor site mutations

Answer 320

- disrupts splice donor/acceptor site, resulting in incorrect retention/excision of intron - often leads to large additions or deletions that may cause frameshift

Answer 321

result in reduced or abolished protein activity

Answer 322

loss-of-function mutations are usually recessive

Answer 323

- null (amorphic) mutations - completely block function of gene product (eg deletion of an entire gene) - hypomorphic mutations - gene product has weak, but detectable, activity

Answer 324

- when one WT allele reaches threshold for the WT phenotype, the WT allele is haplosufficient - even with one mutant allele, we still see the WT phenotype - recessive mutations

Answer 325

- when one WT allele does not reach the threshold for the WT phenotype, the WT allele is haploinsufficient - heterozygotes are affected - dominant mutations

Answer 326

polydactyly - phenotype is seen in individuals with only one mutant copy

Answer 327

phenotype varies with the amount of functional gene product

Answer 328

enhance a function or confer a new activity

Answer 329

they are typically dominant

Answer 330

hypermorphic mutations: - generate excessive gene product or more potent gene product neomorphic mutations: - generate gene product with new functions or ectopically expressed at inappropriate time or place, generating a novel phenotype

Answer 331

achondroplasia - caused by a hypermorphic allele of the FGFR3 gene, which leads to increased receptor activity. - this overactive FGFR3 inhibits bone growth, resulting in the characteristic short stature.

Answer 332

mutation in Antennapedia gene of drosophila causes ectopic expression of a leg-determining gene in structures that normally produce antennae

Answer 333

- usually occur in genes that encode multimeric proteins - produce a mutant gene product that interferes with the function of the normal (wild-type) protein

Answer 334

- proposed that genes control metabolism by encoding specific enzymes. - In PKU, a mutation in the gene for phenylalanine hydroxylase leads to a loss of enzyme function, causing a buildup of phenylalanine. - This showed that mutations can block metabolic pathways, linking gene defects to biochemical function

Answer 335

normal pathway: phenylalanine -> tyrosine -> p-hydroxyphenylpyruvate -> homogentisic aid (HA) -> maleylacetoacetic acid -> CO + H2O in alkaptonuria: - HA oxidase nonfunctional - HA accumulates (toxic) - turns urine black in air - pathway stops

Answer 336

- compounds that are used latest in the pathway will support the growth of the most mutants - compounds that are used earliest in the pathway will support the growth of the fewest mutants

Answer 337

the total digital information contained within the DNA sequences of an organism's chromosomes

Answer 338

the branch of biology dedicated to the study of the whole genomes (both human and not human)

Answer 339

the science of using computational methods to analyse biological information

Answer 340

harnessing the function of newly discovered genes/identifying more targets for therapy

Answer 341

- allows genes and mutations of interest to be identified much more rapidly and easily - opens up prospects for larger-scale, more complete understanding of how genes interact for biological functions (systems biology)

Answer 342

through the publicly funded human genome project (HGP)

Answer 343

- Francis Collins 'Hierarchical Shotgun' - J. Craig Venter Celera Genomics Parallel (whole-genome shotgun)

Answer 344

- first determine the physical location of large pieces of DNA (large-insert clones) - sequence these pieces

Answer 345

- Fragment the genome into many small pieces. - Sequence all fragments to get short reads. - Assemble overlapping reads into contigs. - Use paired-end reads to link contigs across gaps. - Build scaffolds from connected contigs. - Use paired reads to help resolve repeats and place contigs in the correct order and orientation.

Answer 346

1. sequencing the human genome 2. advancements to sequencing technologies

Answer 347

computing power doubles every 18 months

Answer 348

advances in sequencing technology (resulting in lowered cost) outpaced Moore's Law

Answer 349

- the human genome consists of approximately 3.1 billion base pairs - the genome is approximately 99.9% the same between individuals of all nationalities - single nucleotide polymorphisms (SNPs) and copy number variations (CNVs) account for much of the genome diversity identified between humans - less than 2% of the genome codes for genes - the vast majority of our DNA is non-protein coding, and repetitive DNA sequences account for at least 50% of the noncoding DNA - the genome contains ~20,000 protein-coding genes - nearly 50% genes do not yet have a function - high degree of conservation between species

Answer 350

because it means we can use other animals as model organisms

Answer 351

one can do association studies without paying attention to pedigrees, just by treating groups of individuals that have a trait as though they are related for that trait

Answer 352

the application of SNP genotyping to large populations of people for the purpose of discovering genetic associations between particular SNPs and traits

Answer 353

SNPs being identified as tightly linked to, or playing causative roles in, a range of common diseases such as breast cancer, diabetes, Crohn's etc

Answer 354

- GWAS studies are more broadly applicable and provide greater power and resolution than traditional pedigree analysis - GWAS studies do not depend on the analysis of closely related family members. there is no limit to the number of humans that can be included in a GWAS test population - direct comparative studies between affected and unaffected can be performed. we can also map and identify traits' associated genes that follow any pattern of inheritance, simple or complex

Answer 355

IGF1 encodes a hormone involved in juvenile growth in mammals and is the major contributor to the difference in size between small and large breeds of dogs

Answer 356

- 7255 studies - 808, 580 unique SNP-trait associations

Answer 357

fragments of genomes carried by our distant ancestors can be observed as blocks of DNA called haplotypes that are shared between many 'unrelated' people who are in fact distant relatives

Answer 358

we have the power to develop personalised drugs for someone with a particular genetic disease or drugs that alter the genome

Answer 359

1. take a group of patients with a disease, and a group of non-patients who don't have the disease 2. take their DNA and detect disease-specific SNPs by comparing the differences within the two groups 3. use linkage disequilibrium (higher frequency of co-segregation of marker and trait to identify locus of interest)

Answer 360

- incomplete penetrance - phenocopy - locus heterogeneity - polygenic determination

Answer 361

an individual carrying the disease genotype may not express the disease phenotype

Answer 362

only 66% of individuals with a mutant BRCA1 allele will develop cancer by the age of 55

Answer 363

disease phenotype can be expressed by an individual who does not have the disease genotype, caused by environmental conditions

Answer 364

3% of women still develop breast cancer by age 55 from somatic mutations in the breast cells themselves

Answer 365

mutations in more than one locus cause the same phenotype in different families: different disease genotypes are responsible for the same phenotype

Answer 366

BRCA1 (chromosome 17) and BRCA2 (chromosome 13) both predispose to breast cancer

Answer 367

mutant alleles at more than one locus influence expression of the disease phenotype in a single individual

Answer 368

loci that influence quantitative traits

Answer 369

2 or more important genes

Answer 370

- using genetic mapping and association of molecular markers with the trait - statistical methods allow the identification of more than one QTL at a time

Answer 371

1. make an informative cross between individuals which differ at the trait(s) of interest. This step creates a population with a range of phenotypes that can be studied. 2. determine the frequency distribution in the F2. This step shows how variable the trait is and suggests how many genes may be involved (more genes = smoother distribution). 3. use molecular markers to genotype the individuals, attempting to find markers that co-segregate within the trait. 4. use a statistical method to determine if markers are co-segregating (associating) with the trait or not. 5. plot the degree of association (LOD score) on a linkage map. This tells you where in the genome the important genes affecting the trait are located.

Answer 372

RFLP - Restriction Fragment Length Polymorphism - a type of molecular marker based on SNPs - if one allele has a restriction site, the enzyme cuts there - if the other allele lacks the site, it produces fragments of different lengths. - these fragment patterns are detected using gel electrophoresis and probes.

Answer 373

STRs - Short Tandem Repeats - alleles vary in the number of STRs that they contain - electrophoresis can be used to separate different alleles based on size

Answer 374

- longest/heaviest alleles move the least, whereas shorter/lighter alleles move the most - when a sequence of DNA is cut by a restriction enzyme, the resulting fragments will result in different bands based on their size

Answer 375

we will notice that the phenotype does not change with different genotypes - different sized parents overlap in terms of the alleles being studied

Answer 376

we will notice that the phenotype does change with different genotypes - different sized parents have different (combinations of) alleles

Answer 377

probability of linkage/probability of no linkage

Answer 378

the log of the probability/likelihood of co-segregation of trait and alleles at a locus being due to chance

Answer 379

p = 0.001 likelihood of association due to chance (ie there is a 99.9% chance that the association is REAL)

Answer 380

The threshold value represents the minimum LOD score required to declare a statistically significant association between the marker and the trait.

Answer 381

molecular techniques - fine mapping - can be employed to narrow the QTL down to candidate genes

Answer 382

Recombinant Inbred Lines: - Generated by crossing two genetically distinct parents (that differ in the trait), then selfing or sibling-mating for many generations. - Each line becomes homozygous but carries a unique combination of parental segments. - This allows for mapping of QTLs with increased precision by comparing many recombinant genotypes with their phenotype.

Answer 383

Near-Isogenic Lines: - Backcrossing a trait donor (carrying the QTL) into a recurrent parent (without the QTL) for many generations, selecting for the QTL region each time. - Until you get an organism that is differs from the recurrent pattern in only the QTL trait

Answer 384

genetic engineering can be used to test the function of the candidate gene, proving its role in determining the genotype

Answer 385

- there are candidate genes in the fw2.2 QTL interval - these are thought to determine tomato size - the predicted fw2.2 is related to the human proto-oncogene RAS, which is involved in cell cycle control - genetic engineering: take a large fruit NIL. take another, same large fruit NIL with the small fruit allele introduced as a trans gene by genetic engineering. If there is a fruit size decrease in the second tomato, then the gene is involved in determining this trait; if not, then it is not

Answer 386

- requires either extensive pedigree data or crosses: this is difficult in some taxa like humans - only maps effect in single cross/pedigree - no guarantee of widespread importance in population

Answer 387

otherwise we will not know which allele came from which parent in the offspring

Answer 388

Positional cloning is a technique used to identify the specific gene responsible for a trait based on its location on the chromosome

Answer 389

we can separate the genetic effects from the environmental effects by quantifying one variable, while controlling the other

Answer 390

environmental variance - the portion of phenotypic variation in a population that is due to environmental factors, rather than genetic factors

Answer 391

genetic variance - a measure of the variation in a trait due to differences in genotype within a population

Answer 392

the extent of phenotypic variation that is attributable to genetic variation H^2 = Vg/Vp

Answer 393

H^2 = 1 - all the phenotypic variation is attributable to genetic variation H^2 = 0 - all the phenotypic variation is attributable to environmental effects

Answer 394

- heritability of a trait is always defined for a specific population or specific family in a specific set of environmental conditions - this is because the amounts of genetic, environmental, and phenotypic variation may differ among traits, families, populations and among different environments

Answer 395

Tells Us: If H^2 is high, the phenotype of an individual is likely to be attributable to its genotype IN THAT FAMILY Does not tell us: 1. What phenotype an individual will have based on their parents' phenotypes. Even if H^2 is high, an individual's precise phenotype cannot be predicted based on its parents' phenotypes (parents pass on their alleles, not their genotype) 2. What is going on in other families. H^2 is family specific and varies among different families, populations, or environments

Answer 396

Vp = Vg + Ve however, Vg = Va + Vd + Vi where - Va is variation due to additive effects - Vd is variation due to dominance effects - Vi is variation due to epistatic effects

Answer 397

dominance variance, which is the portion of genetic variance attributable to interactions between alleles at the same locus

Answer 398

epistatic interaction variation is not transmitted from parents to offspring -> new genotypes and thus new epistatic relationships are formed with each generation

Answer 399

the extent of phenotypic variation that is attributable to additive genetic variation

Answer 400

if all the phenotypic variation is attributable to additive variation, h^2 = 1 (the maximum it can be) if all the phenotypic variation is attributable to other genetic and environmental effects, h^2 approaches 0

Answer 401

~ 18% of variation is attributed to the environment and or dominant or epistatic genetic variation slope = 0.82 ~ 82% of variation is attributed to additive genetic variation

Answer 402

tells us: if h^2 is high, the phenotype of an individual is predictable based on the phenotype of its parent in that family does not tell us: what is happening in other families

Answer 403

1. monozygotic twins: single ovulated egg fertilised by one sperm, then the embryo splits into two. 100% of alleles are shared, so we can study the effect of different environments on the same genotype 2. dizygotic twins: two ovulated eggs fertilised by different sperm, so 50% of alleles are shared, so we can study the effect of different genotypes on the same environment

Answer 404

clear-cut, 'either-or' phenotypes between alternative alleles eg. all of the traits Mendel studied in peas

Answer 405

the phenotype varies continuously with levels of protein function/the amount of functional gene product

Answer 406

A1: functional enzyme A2: nonfunctional enzyme A1A1: red (100% pigment production) A1A2: pink (50% pigment production) A2:A2: white (0% pigment production)

Answer 407

the heterozygous phenotype (250-500mg/dl) is distinct from and intermediate compared to either homozygous phenotype (<250mg/dl and >500mg/dl)

Answer 408

- blood and serum types, enzyme defects - simple Mendelian (monogene) - little environmental effect

Answer 409

- stature height, intelligence, milk yield - complex (polygene) - moderate to great environmental effect

Answer 410

- hare lib, many diseases - complex (polygene) - moderate to great environmental effect

Answer 411

a normal distribution, which is bell-shaped

Answer 412

alleles are incompletely dominant and have additive effects. the more genes or alleles, the more possible phenotypic classes, the greater the similarity to continuous variation

Answer 413

genetic and environmental influences

Answer 414

additional variation might arise

Answer 415

contribute to traits in a cumulative manner. the more of a particular allele an individual has, the stronger the expression of that trait

Answer 416

- a handful of ultra rare inherited mutations likely shave years off a person's life - each of these DNA variants can reduce lifespan by as much as 6 months - and different combinations dictate how early age-related diseases such as cancer, diabetes, and dementia will develop - thus, each variant is an additive allele

Answer 417

are determined by segregating alleles of many genes that interact together and with the environment eg height, weight, skin colour

Answer 418

quantitative traits

Answer 419

meristic (counting traits): takes on a range of discrete values. these traits are quantitative, but restricted to certain discrete values (eg salamander spots) continuous traits: takes on a potentially infinite number of states or values over a continuous range (eg height) threshold trait: individuals who have a certain number of risk factors will exceed a threshold and develop the disease.

Answer 420

no, they do not follow simple Mendelian rules or produce Mendelian ratios in pedigrees

Answer 421

siamese cats

Answer 422

region of the genome that correlates with the quantitative trait and may contain the genes affecting the quantitative trait

Answer 423

frequency distributions associated with each genotype at the QTL overlap. we cannot determine genotype by simply looking at an individual's phenotype as we can with genes that segregate in Mendelian ratios

Answer 424

at either end (low or high trait value) we can guess the genotype from the phenotype. however, at intermediate trait values, this is very hard

Answer 425

genotype + environment

Answer 426

the study of the genetics of a population and how the alleles vary with time

Answer 427

an interbreeding group of the same species within a given geographical area

Answer 428

the collection of all alleles in the members of the population

Answer 429

alleles can move between populations when individuals migrate and mate

Answer 430

phenylketonuria (PKU), a heritable metabolic disorder, autosomal recessive trait

Answer 431

the relation between genotype and allele frequency within a generation and from one generation to the next

Answer 432

- infinitely large population - individuals mate at random - no new mutations appear in gene pool - no migration into or out of population - no genotype-dependent differences in ability to survive and reproduce

Answer 433

no; all populations violate one or more assumptions of Hardy-Weinberg law

Answer 434

- equations derived based on assumptions are remarkably robust - HW law can be used as a null model

Answer 435

allele frequency of one allele

Answer 436

allele frequency of a second allele

Answer 437

1 all of the allele frequencies together equals 1 or the whole collection of alleles

Answer 438

p^2 + 2pq + q^2 = 1 all of the genotype frequencies together equals 1 p^2 and q^2 = genotype frequencies for each homozygote 2pq = genotype frequency for heterozygotes

Answer 439

of particular allele/total # of alleles in the population - count both chromosomes of each individual - allele frequencies affect the genotype frequencies

Answer 440

- allele frequencies do not change from generation to generation in a population at HW equilibrium - a HW population achieves the genotype frequencies of p2, 2pq, q2 in just one generation and once at equilibrium maintains them in subsequent generations - only 1 generation is required to reach HW equilibrium

Answer 441

- natural populations are undergoing microevolution: genetic change due to changing allelic frequencies in populations - allelic frequencies can change with: nonrandom mating, gene flow, genetic drift, natural selection (unequal reproductive success)

Answer 442

- chance random fluctuations in allele frequency that have a neutral effect on fitness - random allele frequency fluctuations might make certain disease alleles more common in particular populations

Answer 443

smaller populations are more affected than larger populations

Answer 444

acts on differences in fitness to alter allele frequencies

Answer 445

individual's relative ability to survive and transmit genes to the next generation (viability and reproductive success)

Answer 446

individuals with higher fitness strive and reproductive more than individuals with lower fitness

Answer 447

- homozygous recessive genotype (rr) has decreased fitness - fitness of RR and Rr is the same

Answer 448

- when q is small, the frequency of homozygous recessive individuals is low - most copies of the q allele are in heterozygotes, who do not have reduced fitness

Answer 449

although allele frequency of lethal alleles should continue to decline, some stay in population at stable frequencies

Answer 450

- In heterozygotes, the presence of some sickled red blood cells interferes with the life cycle of the malaria parasite. - As a result, HbA/HbS individuals are more resistant to severe malaria compared to HbA/HbA individuals.

Answer 451

- if onset is in middle or late age, after reproduction, then the trait will not affect fitness - sustain little or no negative selection

Answer 452

a balance between mutation to a new allele and selection against the allele

Answer 453

mutations: - when new allele has effect on fitness, selection will drive frequency towards an equilibrium with wild-type allele - if new allele has no effect on fitness, genetic drift will determine its frequency

Answer 454

short for haploid genotype - a group of alleles or DNA variations that are inherited together from a single parent because they are located close to each other on the same chromosome.

Answer 455

1. DNA collected from evidence at crime scene 2. DNA collected from suspected 3. five specific DNA sequences from different chromosomes are labeled and separated by size - blood (victim), blood (suspect), skin (evidence), blood (evidence) 4. use HW to calculate the probability that another person in the suspect's population group has the same pattern of these alleles

Answer 456

enables the correlation of occurrence of trait with occurrence of a particular allele

Answer 457

genotype many individuals and correlate with phenotype

Answer 458

distance between two loci

Answer 459

perform a chi-square test, which assesses the likelihood that a deviation from expectation is due to chance

Answer 460

- deviations from 1:1:1:1 ratios can represent chance events OR linkage - the chi-square test accounts for sample size

Answer 461

null hypothesis - observed values are not different from the expected values. for linkage studies, this is no linkage where you expect a 1:1:1:1 ratio of gametes alternative hypothesis - observed values are different from expected values. for linkage studies, this is that the genes are linked and you expect a significant deviation from 1:1:1:1 ratio

Answer 462

1. formulate null hypothesis: genes are not linked (1:1:1:1 ratio of phenotypes 2. compute: x^2: Σ (O-E)^2/E 3. determine degrees of freedom (number of independent measurements) 4. consult x^2 chart of critical values

Answer 463

if p<0.05, then we reject the null hypothesis and accept the hypothesis that genes are linked to

Answer 464

less than 5% of the time, we are incorrectly rejecting the null

Answer 465

he postulated that the frequency of crossovers between two genes is a function of their distance apart on the chromosome; he also created the first genetic linkage map

Answer 466

number of recombinants/total number of progeny

Answer 467

one centiMorgan (cM) = 1% recombination - 1 product of meiosis out of 100 is recombinant

Answer 468

50% (=unlinked)

Answer 469

1. anticipate and identify 8 kinds of meiotic products 2. identify pairs of reciprocal products 3. identify parental types as the most frequent pair of products 4. identify double crossover products as least frequent pair of products 5. compare parental and double crossover to deduce locus order 6. compute map distances (distances between loci) by breaking down the results for each interval - RF = SXO+DXO/total

Answer 470

crossover in one region interferes with simultaneous crossing over in adjacent regions

Answer 471

1. expected frequency of DCO = product of frequency crossovers in two regions 2. coefficient of coincidence = observed DCO/expected DCO 3. interference = 1- coefficient of coincidence

Answer 472

- recombination and physical maps are co-linear (syntenic) but are frequently misaligned as recombination distances (cM) are frequently not the same as physical distances (bp of DNA) - that is, one cannot say that X bp of DNA sequence always equals Y cM on a linkage map - this is because recombination varies across the length of a chromosome

Answer 473

no - we can also use molecular markers - any polymorphism can be mapped

Answer 474

In male Drosophila melanogaster, meiosis does not involve recombination, a phenomenon known as achiasmy

pt 2 Flashcards

(524 cards)