L1: Synthetic Genomes Flashcards

Question

What is the effect of removing all DNA methylation?

Answer 1

met1, drm1/2 and cmt2/3 mutants lack methylation and full mutants show clear developmental defects - more genes are upregulated - genomic instability due to movement of Transposable Elements (TEs)

Answer 2

- MBD and SRA domain recognises DNA methylation - BROMO domain recognises acyl groups on HISTONES - BAH and CHROMO domain recognises methylated histones

Answer 3

SET and RING associated domain (SRA) - recognises methylation of cytosine residues When binding leads to flipping out of the methylated cytosine allowing interaction with proteins such as KRYPTONITE for methylation of nearby histones @ the replication fork, hemi-methylated strands form and VIM is important in recognising this and recruits MET1 = met1 and vim mutant have the same methylation pattern, suggesting they work together

Answer 4

Feedback loop that couples DNA methylation with H3K9me2 methylation. KRYPTONITE(SUVH4) works with SRA-domain containing proteins to detect DNA methylation, and methylate histones CMT3 is capable of interacting with H3K9 methylation marks via chromo and BAH domains. CMT3 function is to methylate DNA = Mutation in DNA methylation (drm1/2, cmt2/3) eliminates CHG and CHH methylation which results in reduced H3K9 methylation also

Answer 5

Superman and clark kent are epialleles capable of switching between phenotypes in different generations, suggesting epigenetic differences instead and genetic mutations associated with floral development SUPERMAN - prevents excessive number of male organs (stamens) In mutant superman or Clark Kent epiallele, SUP gene is hypermethylated, silenced and so increase in stamen number - Mutagenesis found that a mutation in KRYPTONITE (SUVH4) could cause would lead to reversion to WT - Similarly mutation in cmt3 leads to WT phenotype as methylation to generate SUPERMAN phenotype is not possible

Answer 6

SET domain = Histone methylatransferases SRA domain = reader of DNA methylation

Answer 7

BAH - recognising methylated histones Chromo - protein-binding domain that typically recognises methylated histones (BAH and Chromo domain found to have a high Kd to H3K9 methylation from Isothermal Titration Calorimetry (ITC) analysis Isothermal Titration Calorimetry (ITC) DNA Methyltransferase - addition of methyl onto DNA

Answer 8

SUVH2/9 lack a post-SET domain, this leads to non-functional histone methyltransferases

Answer 9

- Re Maintenance of non-CG RdDM by - interaction with Defective in RNA-directed DNA Methylation 1 (DRD1) a chromatin remodeller - Recruitment of RNA Pol V - evidenced by Co-IP assays

Answer 10

SRA domain to bind methylated DNA Lacks functional histone methyltransferases = STRICT reader function - Interacts with transcriptional activators such as DNAJ1/2 - Required for expression of genes nearby to TEs

Answer 11

SUVH4(KYP) - read and write, forming DNA methylation and H3K9 methylation feedback loop SUVH2/9 - read and recruit DRD1, maintain non-CG methylation in RdDM = feedback loop SUVH1/3 - read and recruit transcriptional activators, preventing TE related silencing = homeostasis

Answer 12

Methyl CpG Binding Domain (MBD) MBD5/6 - binds methylated CG and recruits SILENZIO to silence genes = regulates downstream signalling MBD7 - Recruits demethylators such as ROS1 to prevent hypermethylation, binds to methylated CG = homeostatic regulation

Answer 13

Repression = AGDP1 - binds H3K9me2 Activation = BRAT1 binds H4ac, recruiting ROS1 resulting in demethylation and promotion of transcription Dual reader - EBS, has BAH (repression binds H3K27me3) and PHD (activator binds H3K4me3)

Answer 14

De-methylation of transposable elements (TEs), especially of CHH methylation - Evidenced by GUS reporter analysis that fused GUS to LTR transposon, and upon addition of flg22 there was an increase in GUS expression

Answer 15

Reduction in the suppression of genes leading to pathogen resistance met1 and ddc (triple) mutants (unable to methylate) had lower leaf bacteria concentration = Increase in pathogen related 1 (PR1) even in the absence of the pathogen ACTIVE demethylation by enzymes such as ROS1 is required for activation of R genes

Answer 16

ROS1 is important for demethylation and activation of R genes - ros1 mutants were hypermethylated in the W box of the promoter of an R gene RLP43 + It was found that 72% of transcription factors are inhibited by DNA methylation

Answer 17

Pericentromeric decompaction - Pericentromeres, which are rich in TEs that flank the centromeres are found to interact more with the rest of the genome Evidenced by Hi-C results

Answer 18

DNA hypermethylation in CG, CHG and CHH regions Priming - faster, greater response Transgenerational stress memory assay found that plants that had ancestors exposed to salt stress were better able to respond to such stress

Answer 19

Treatment leads to enhanced disease resistance by the plant - Doesn't induce the transcription of PR1 but primes it for activation, so that levels increase more rapidly

Answer 20

1. Increased chromatin accessibility - hypomethylation in WRKY areas 2. Removal of DNA methylation - removal of H3K27me3 increased drought/ salt tolerance 3. Removal of repressive histones 4. Addition of activation associated histones - activation to promote WRK29 expressio 5. Poised Pol II - having Pol II already bound before TSS 6. Trans-acting mechanism - presence of 21/22nt sRNAs can bind AGO1 to direct RNA Pol II to increase transcription (Parker et al., 2022) NOTE: priming doesn't have to involve methylation changes, and methylation changes doesn't have to prime the plant

Answer 21

FORGETTER1 (FGT1) Loss of function mutants exhibit reduced stress memory capacity

Answer 22

1. Changes in expression patterns of genes 2. Movement of transposons - Presence of H2A.Z in ERGs attracts TE insertion that can generate (epi)genetic diversification by HR and DNA methylation or histone modification changes (Parker et al., 2022) 3. Changes to regulatory pathways

Answer 23

Transcriptional changes can precede changes to DNA methylation e.g. transcription can change 7 days post induction, while methylation changes can occur 21days later Rate of reversal is also faster for transcriptional changes

Answer 24

- Cost of priming is high (e.g. increase in protein amount) - Priming might suppress other functions, for example general responses to light might be hindered by always being primed to high light

Answer 25

1. Epigenetic changes - Visualised using whole genome bisulfite sequencing (WGBS) = increase in CMT2 led to greater DNA methylation - Expression of memory transcripts were found to be regulated by DNA methylation changes in rice (Li et al., 2019) 2. Production of miRNAs - can degrade mRNAs e.g. in drought response in switchgrass 3. Changes to phytohormone production - increase in JA and ABA was found to be associated with drought in rice

Answer 26

Chain termination PCR and addition of tagged bases that are subsequently read out - 500-1500bp read length - 99.99% accuracy

Answer 27

Illumina (sequencing by synthesis) 1. Fragments formed and adapters bind to the fragments 2. Adapters then bind to the flow cell 3. Strands are amplified to generate clusters 4. Sequence Sanger style

Answer 28

1. Amplification bias 2. Requires sufficient numbers for amplification and detection 3. Read length limits (hard to map repetitive regions)

Answer 29

Pacbio - incorporation of a fluorescent base which emits light that is then detected. Oxford Nanopore - strand is passed through a nanopore membrane which undergoes changes in voltage. - specific changes in voltage are associated to different bases and with different modifications

Answer 30

- PacBio had a lower accuracy pre Hi-Fi. HiFi allows for circularisation and multiple reads of the same strand - Oxford Nanopore struggles with accuracy in regions that are homopolymers

Answer 31

1. De novo assembly - reconstruction of whole genomes, including centromeres and other areas with high number of repetitive elements 2. Structural variation analysis - allows for analysis of changes such as deletions and insertions which may have been hard to piece together when chromosome is fragmented 3. Long read sequencing of RNA - allows for analysis of splice variants, doesn't require conversion to cDNA which could generate biases

Answer 32

Forms the link between DNA and spindle fibre which is important for proper segregation of chromosomes - Failure to link properly can lead to aneuploidy and cell death

Answer 33

1. Analysis of resistance genes - useful for following R genes during introgression (movement of genetic material during hybridisation or crossing) e.g. TM-2 that provides resistance against TMV 2. Tracking structural variation - Short reads provides bias towards detecting SNPs compared to other changes e.g. SIKLUH gene thought to increase fruit size due to a SNP in the promoter, but turns there was an additional copy of the gene 3. Haplotype phasing -identifying which alleles are inherited together to look at inheritance patterns. whole sequencing of many individuals fast and cheap can be used to compare and phase haplotypes

Answer 34

1. Bisulfite sequencing - conversion of non-methylated Cs to U and sequence to find differences in 2. Nanopore - different voltage change when a methylated base is passed through compared to a non-methylated base 3. PacBio - different time lag associated with the incorporation of the tagged base 4. CHIP-seq - bind antibody that is complementary to specific histones around a DNA fragment, then fragment and isolate before sequencing

Answer 35

1. Change priorities towards healthy foods 2. Sustainably intensify farming practices 3. Global governance of land and oceans 4. Reduce food waste

Answer 36

During the formation of gametes (sperm and egg) two alleles for the trait will separate Independent genes will independently assort these alleles (Independent assortment)

Answer 37

The likelihood of two closely located genes to be inherited together Degree of linkage is calculated by recombination frequency, which is the percentage of offspring in which recombination occurs

Answer 38

Location of genes and marks along a chromosome based on recombination frequencies Measured in centiMorgans - Important when looking at breeding at it shows chance of recombination

Answer 39

May show two genes being closely located, however, if near the centromere were crossover is minimal then highly unlikely that these genes will cross

Answer 40

Traits determined by multiple genes - Makes it difficult to classify roles of certain genes as they may cause a gradient of changes - In polyploid genomes (such as wheat) there may also be redundancy, so knock out of a gene may not show anything (CRISPR has been useful to overcome this problem, allowing multiple copies of one gene to be knocked out in one go)

Answer 41

1. Influenced by the environment - measurements may vary depending on temperature, soil pH e.g. Hydrangea flower colour 2. Regulation by multiple genes - polygenic inheritance 3. Trait may be continuous e.g. height, unlike flower colour

Answer 42

1. Acquire individuals with genetic variation 2. Locate genetic markers that are specific to the gene 3. Determine heritability of the phenotype and accuracy of measurement 4. Statistically test - See whether there is significant association between the phenotype and the genetic marker for the gene

Answer 43

Quantitative Trait Locus analysis 1. have genetic markers that help distinguish between two parents 2. Use these markers to characterise the genome of the F2 population as either heterozygous/ homoA or homoB = genetic map . Uses the marker as the assay 3. Phenotype e.g. for length 4. Generate a table with the genotype at the different marker points 5. Generate a table with the average value of the phenotype from a certain genotype e.g. homoA (average between no of plants that are homoA at marker 1) - used to find genes that are important for the phenotype by calculating the difference between homoB and homoA (NOTE: if heterozygous is bigger this could suggest hybrid vigour)

Answer 44

Genome Wide Association Studies (GWAS) Analysis genome for SNPs that are statistically associated to certain traits = Manhattan plot higher dots mean greater association Positives - Doesn't require specific lines being crossed (useful for human analysis) Negatives - requires many marker genes to annotate the whole genome - for rarer alleles, recombination frequency below 5% doesn't provide enough statistical power to generate conclusions as to whether the gene could be relating to trait 1. Measure the trait 2. Genotype the population 3. Statistically analyse the correlation between the trait and the phenotype

Answer 45

Multi-parent Advanced Generation Intercross (MAGIC) Hybrid between GWAS and QTL - Less diversity than GWAS but uses known markers from the known set number of parents max around 12 Intercrosses are carried out to increase genetic diversity within the population

Answer 46

Depends on the population size and type Also depends on the availability of markers - More recombination means greater resolution as it allows for distinction between different markers - Also can help show any linkage traits within the genome

Answer 47

Proposed that areas with the greatest diversity in the wild are where humans first cultivated them Against - Hard to know where the crops originated - Crops have moved around a lot and may have undergone major evolutionary changes in a certain area, rendering secondary places also very important. For example the movement of maize from low-lands to high lands in Central America

Answer 48

Teosinte domesticated to what is now known as maize - Increase in ear and protective leaf size - Sweetening of kernels also - shorter size due to dwarfing gene - aim for determinate development so that plants mature at similar times for easier harvesting

Answer 49

Polyploid formed from the cross over event between a tetraploid and diploid to form a hexaploid - there has been greater genetic exchange with the tetraploid wild relative, while the D (diploid) genome is largely 'locked' showing little diversity (suggested that D genome captures less than 10% of the diversity found in wild types) ++ Opposingly for Durum wheat, this was formed solely from the tetraploid

Answer 50

Bottlenecking - events that limit genetic variation in a population e.g. only 45% of the variation found in wild type wheat is found in the domesticated AB genome, and only 10% in the D genome

Answer 51

1. de novo domestication of wheat, would allow for introduction of variety from the D genome e.g from A.tauschii 2. Genetic engineering and CRISPR techniques, e.g to knock out negative regulators

Answer 52

Crossing of two plants with desirable phenotypes to get these in one plant

Answer 53

Norman Borlaug Following the development of the Harber-Bosch process that produced nitrate for fertilisers - Application of fertilisers led to extreme growth and an increase in height which caused lodging of plants Dwarfing of the plant (Semidwarf1) meant that upon application of fertilisers less plants lodged

Answer 54

The use of genetic markers within the chromosome to select for certain traits + faster way to quantify, easier also for R genes that require infection and pathogen to find real effect + Found the gene Flood Resistance 13A which was resistant to submergence and flooding, this was easily selected with IR64 that produces high yield

Answer 55

1. Traditional breeding 2. Marker assisted selection 3. Mutation breeding 4. Genetic modification 5. Gene editing

Answer 56

Irradiate seeds to mutagenise and analyse phenotypic effects Commonly done to generate fruits such as grapes and oranges with no seeds Adv - Generates mutations that may not exist - Rapid development of changes, compared to traditional strategies Disadv - Unpredictable - Regulatory hurdles, whether they are classed as genetically modified

Answer 57

Insertion of new genes that leads to acquisition of a new trait Transgenics - transfer of genetic material from one species to another e.g. sweet potato has bacterial DNA in it e.g. transfer of pathway that stores oils from algae as omega3 in fish to plants. reducing fishmeal for fish farming Cisgenics - transfer of material between different varieties but same species e.g. transfer of R genes from a wild potato from Peru into Maris species lead to greater resistance in the absence of pesticides

Answer 58

Kompetitive Allele Specific PCR - for detection of SNPs Uses 3 types of primers Common primer - binds to region of DNA that is identical in both alleles Allele-specific primer - primers that are specific to different alleles (pairing to the different SNPs) PCR product generated is distinct to which allele-specific primer is used, with fluorescence being specific to either A, B or both alleles

Answer 59

Uses oligonucleotide baits that selectively enrich certain areas of the genome - Allows for focused analysis and reduces cost, especially for large genomes such as wheat e.g. Exon capture to look at coding regions

Answer 60

Mutation Resistant Gene Enrichment Sequencing 1. Mutate seeds that are resistant to the pathogen 2. Grow them and look for susceptibility 3. Use NLR baits to capture R genes and then sequence 4. Use multiple independent lines to confirm that mutations in that specific R gene is the cause for susceptibility (as there are many R genes and many mutations can occur, accuracy of pinpointing which gene is responsible is hard without multiple lines)

Answer 61

Generates the ability to introduce multiple R genes at once to prevent the pathogen from evolving resistance - introduction of one or two, the pathogen may be able to overcome resistance

Answer 62

Previously small fruit, indeterminate structure - Target CLAVATA to have more locules in the fruit Mutated repressors of traits, in order to enhance certain traits

Answer 63

Tetraploid rice (O.alta) had many favourable traits such as drought and waterlogging resistance - Mutations were made in the tetraploid genome to enhance the yield and growth characteristics of the tetraploid rice to have the beneficial traits that Oryza sativa has (this was done by introducing a casette targeting many genes in one go, much easier than trying to transfer multiple genes that give rise to a quantitative trait from O.alta in O.sativa) Challenges - successful transformation of the polyploid - ensuring heritability (of these genes in the rice example is not known)

Answer 64

1. Genetic robustness - increase in genetic diversity with additional chromosomes carrying various alleles 2. Hidden variation - redundancy against genetic loss, preventing loss of crucial functions - only clear phenotypes may be seen in higher order mutants which are hard to generate if gene is not fully known. This is commonly overcome by looking at sequence similarity of genes in diploids e.g. mutation to increase yield may be 5% in wheat while 20% in rice, however, knockout of the genes in all chromsomes can lead to 20% increase in yield in wheat 3. Large chromosomes - makes it hard and expensive to sequence, in the case of wheat the genome is also highly repetitive - takes longer to backcross and select, however, new technologies have helped overcome this 4. Gene dosage effect - increase in fruit size and yield due to greater gene dosage

Answer 65

C4 cereal that is commonly grown in Africa, gluten-free alternative - Produces very small seeds that are sewn by dispersal = harder for mechanised farming - Prone to lodging which leads to yield loss = SD1 gene targeted to reduce height, however, the straw from the plants are also useful for building and liveihoods

Answer 66

Addition of growth regulators GRF4-GIF1 - promote the proliferation of the callus which can be transformed in AgBac media BabyBoom, Wuschel and GIF1 - are important for retaining undifferentiated state to promote proliferation

Answer 67

- Improvements in quantitative traits are commonly associated with changes in the expression of genes in terms of strength and location of expression e.g. fas in tomato, 294kb inversion causing change in the expression profile that increases locule number. This leads to quantitative variation with many different locule numbers NOTE: hard to see if it is due to expression as these analyses commonly involve grinding tissue and which includes many cells that can mask the changes in expression

Answer 68

Mutation in cis-regulatory element of VRT2 found changes to spatial patterning, visualised by in situ hybridisation with labelled probes - Found ectopic expression (expression where not commonly found) VRT2- plays a role in floral development and vernalisation

Answer 69

1. Retrotransposon (Class I) - encodes a reverse transcription that generates a dsDNA strand that can be integrated - move via an RNA intermediate 2. Class II - encodes a transposase that allows for cut and paste of the element - move as DNA 3. Helitron - doesn't have border repeats, replicates via rolling circle

Answer 70

1. Inhibition of gene function (transfer within gene) 2. Change expression of gene (transfer into promoter) 3. Promote re-arrangement of chromosomes 4. Attract heterochromatin resulting in silencing e.g. by H3K9me2

Answer 71

Genome size doesn't reflect complexity -Many plant genomes are large due to highly repetitive regions

Answer 72

1. Cytological analysis - DAPI binds to AT rich regions of DNA - Heterochromatic areas stain more brightly due to high intensity packing of DNA 2. HiC - analysis of DNA-DNA interactions across the chromosomes - achieved by crosslinking closely located DNA using formaldehyde 3. CHIP-seq - look for the areas in which certain proteins such as histones bind to 4. ATAC-seq (assay for transposase-accessible Chromatin) - transposase that fragments DNA that is easily accessible to add primer and tag -

Answer 73

1. HP1 (methylation reader attached to GFP), photobleaching of the area saw rapid recovery suggesting the area is active 2. Gene switching between different epigenetic states can occur e.g. FLC on/off 3. transcription of non-coding RNAs that are responsible for maintaining heterochromatic state suggests the area is not silenced e.g RdDM 4. regulation of the epigenetic status of TEs is variable, with more TEs opening up in stress conditions e.g. pericentromeric decompaction in heat stress

Answer 74

Changes in the expression of a gene due to changes in its position along a chromosome - For example transfer nearby a heterochromatic region can lead to its silencing e.g. movement of the white gene causing changes in Drosophila eye colour

Answer 75

Mutation in IBM1 a gene encoding a H3K9me2 demethylase - This resulted in an increase in methylation which silenced the gene Bonsai - This caused the A.thaliana phenotype to be small

Answer 76

21-24nt in length - Responsible for maintaining methylation and modifications within a specific region of TEs

Answer 77

1. siRNA enrichment in heterochromatic areas, where DNA methylation is dense 2. Viroid mediated RdDM - viroids are RNA genomes that lack coding abilities - Tobacco transformed (inserted) with DNA copy of Potato Spindle Tuber viroid to produce lots of viroid - Infection with viroid RNA led to increase in DNA methylation Evidenced by a southern blot where the size of DNA fragment was larger, because the enzyme was unable to cut (due to more methylation) 3. Promoter hairpin-driven silencing - Trigger and target transgene, with target conferring resistance against kanamycin - solely have target leads to kanamycin resistance INSERT inverted promoter sequence of target leading to complementarity and dsRNA forming - Southern blot found changes in the DNA fragment size, suppression of methylators returned fragment sizes to unmethylated control - Greater presence of 21 and 24nt

Answer 78

Virus detected as foreign and the plant generates small RNAs to reduce the transcript load of the virus - In plant that is expressing GFP under a 35S promoter, addition of TRV with part of the GFP sequence will lead to the plant suppressing the expression of GFP due to VIGS - Similarly when the virus contains part of the 35S promoter that is driving the GFP - Transcriptional run on analysis found that in the TRV 35S plant transcription is suppressed, while in the TRV:GFP modification is post-transcriptional Heritability - promoter suppression that is transcriptional is heritable due to methylation changes, while TRV:GFP suppression is not inherited

Answer 79

Analysis of the FWA gene in A.thaliana that regulates flowering - Commonly the FWA promoter region is methylated and leads to normal flowering = introduction of unmethylated FWA in a drm2 mutant leads to lack of de novo methylation and late flowering - however, introduction of FWA in wildtype, the gene is methylated and flowering is normal

Answer 80

1. SHH1 and CLASSY helps recruit PolIV to heterochromatic mark - C terminal domain of PolIV is important for SHH1 recruitment which is a histone reader 2. Pol IV along with RDR2 generates dsRNA 3. DCL3 generates 24nt length siRNAs 4. 24nts bind to AGO4/6/9, which using RNApolV scaffolf recruits DRM2 5. DRM2 deposits DNA methylation

Answer 81

Archaebacterial lack long tails *long tails in eukaryotes are important for post-translation modifications - However, function the same by binding and wrapping DNA to form a nucleosome like structure

Answer 82

Constitutive - always present e.g. at transposons Facultative - depends on cell type and conditions of the cell as to whether present or not e.g. vernalisation

Answer 83

1. Circadian clock CONSTANS 2. Temperature - vernalisation 3. Light = APETALA genes need to be switched on

Answer 84

MADS transcription factor that represses floral integrator genes Forms a heterodimer with Short Vegetative Phase (SVP) also a MADS-box TF FLC is promoted by FRIGIDA (FRI)

Answer 85

Northern blot showing that FLC levels remained low after 20days no cold which followed 3 or more week of vernalisation VRN2 + LHP1 - responsible for vernalisation memory, as mutants had levels of FLC increasing again after only 10-15days = CHIP-Seq found an increase in heterochromatic marks on the promoter of the FLC gene

Answer 86

Mutagenesis of Arabidopsis plants that didn't respond to vernalisation (so after cold didn't flower) Found VRN1/2/5 and VIN3

Answer 87

Retain memory of vernalisation Homologue of Polycomb Group that is responsible for gene silencing - An orthologue was found to be responsible for H3K27methylation : Although the mRNA levels of VRN2 was not found to differ in cold conditions

Answer 88

Plays a role in vernalisation and its transcript levels increase in cold conditions, as FLC decreases - has a PHD (chromatin reading) and FIN III domain (protein interaction) = Recruits Polycomb Repressive Complex 2 (PRC2) for H3K27me3 (repression) Suggested to play a role in establishment of vernalisation prior to VRN1/2 complex that regulates FLC Experiment - Mutation looks like a vernalisation mutant, where in FLC:LUC assays, FLC is still expressed and flowering is suppressed

Answer 89

Quantitative vernalisation achieved by accumulation of switches between bistable epigenetic states (Csorba et al.,. 2014) 1. Before cold VRN2 is bound but not active 2. After vernalisation VRN2 is activated by VIN3 and VRN5 - Early in vernalisation there is an increase in COOLAIR transcript (antisense of FLC to support suppression) 3. This leads to stable repression of FLC by H3K27methylation of the promoter region - With upregulation of antisense FLC transcript to antagonise sense also

Answer 90

mutant analysis ELF6 - histone demethylase required for the removal of H3K27 methylation - Measurement of FLC transcript n mutant found no change in the in vernalisation FLC suppression response, but FLC levels were lower during resetting Overexpression leads to global demethylation - suggest that there is no resetting, plants that don't experience vernalisation Finnegan et al., 2021 suggests that there is no clear resetting and the processes of activating FLC expression in young embryo of progeny of non-vernalised and vernalised plants are the same. In both cases, there are major chromatin changes during reprogramming for gametogenesis. Null and weak mutants of elf6 had no effect on resetting, suggested to play a minor role.

Answer 91

Occurs post-embryonically Meaning changes can occur during lifetime and seeds from different branches on a tree may be genetically different e.g. Shorea tree found a linear increase in the number of mutations with branch distance

Answer 92

1. Single stranded break 2. Mismatch 3. Damaged base 4. Intra-strand cross-link 5. Inter-strand cross-link

Answer 93

1. Base excision repair 2. Nucleotide Excision repair 3. Homologous recombination 4. Non-homologous end joining 5. Mismatch repair - Show some redundancy as seen in experiments that have knocked out pathways and radiated plants and they still look healthy

Answer 94

C:G to A:T transition due to the deamination of cytosine to uracil which can be fixed to cytosine HOWEVER - 5-methyl cytosine is deaminated to thymine, so harder to judge which is correct

Answer 95

Essential genes - mutations are less likely Mutations are more likely to occur in areas depleted of nucleosomes e.g. upstream of the TSS - It is possible that epigenetic marks help recruit machinery involved in DNA repair, so there are less mutations that remain

Answer 96

Mutation and changes in the epigenome e.g. methylation to non-methylation - These rates are faster than DNA mutation rates - Could be important for adaptability of organisms to new environments by changing gene expression

Answer 97

Fixes double stranded breaks - Homology independent, so RAD51 independent Error prone due to the lack of template strand Experiment: Require ligase IV, which is responsible for rejoining broken strands Alternative NHEJ - uses microhomology in the overhanging strands that can result in sections of the DNA being lost

Answer 98

CRISPR-Cas9 introduces a break in the DNA strand - Relies on NHEJ to introduce a mutation due to error prone repair

Answer 99

Homology dependent, uses a template strand 1. Strand invasion, transferring genetic material from one strand to another (if they are two chromatids then unlikely that a difference will be seen, however, if this is during meiosis then differences in SNPs may be seen) EVIDENCE - GUS analysis with breaks and complementary strands in different positions along the chromosome were able to repair to generate a functional GUS protein

Answer 100

Recognises a specific 18bp sequence and cleaves to generate an overhang - Base pair sequence that is unlikely to occur in most genomes, so can introduce a transgene with this sequence and introduce a DSB cut at a specific location by adding the enzyme

Answer 101

Coats single stranded DNA that invades the strand (similar to DMC1, but RAD51 is also for somatic recombination) - Promotes finding of homology in another chromosome knockout of rad51 results in impairment of repair mechanisms that require another chromosome

Answer 102

Both part of HR Single-strand annealing (SSA) - no donor - can lead to large deletions of the genome 1. 5' resection of the strand 2. Binding of the strand in another place due to homology e.g. due to repetitive sequences, generating flaps are cleaved 3. Gaps again are filled with polymerase Synthesis-Dependent strand annealing (SDSA) 1. Resection of DNA, followed by strand invasion into 'donor' 2. Generates a Displacement (D) loop in the donor strand 3. Starts synthesising the new strand using a donor as a template until homology is found with the broken strand 4. Broken strands find homology and bind, gaps are filled by polymerase - repair without excising large regions of DNA, preserving more genetic information

Answer 103

ATM/ATR kinase signalling - Break sites leads to activation and phosphorylation of targets = DNA repair mechanism recruitment or apoptosis - In animals cells a target is p53 1. In plants a target is H2A.X that is phosphorylated upon DSB formation 2. Recruits repair mechanisms to the site

Answer 104

Reason - Mitosis DNA recombination uses the strand for repair - Meiosis DNA recombination uses the strand to generate diversity by crossing over DSB formation - Mitosis DSB formation is random - Meiosis SPO11 can promote DSB to promote cross-over Factors and complex involved - Mitosis RAD51 - Meiosis RAD51, DMC1 and synaptonemal complex Types of DNA recombination - Mitosis uses HR and NHEJ (no RAD51, no homology required) - Meiosis uses HR and not NHEJ (no crossing over)

Answer 105

1. Chromosome expansions 2. Telomeres to prevent degradation 3. Histone code and RNAi to suppress deleterious genes from expressing 4. Meiosis (crossing over)

Answer 106

Pair of chromosomes one from maternal and one from paternal origin Contain the same genes in the same order, however have allelic variation - Form and align during prophase I in meiosis

Answer 107

Prophase I - synapsis occurs when homologous chromosome pairs align and the synaptonemal complex forms. Recombination occurs which can lead to crossing over First division occurs where the homologues separate Second division occurs where the sister chromatids separate

Answer 108

Cohesin rings hold chromatids together until anaphase Following anaphase when the cell divides the chromatids are split Formation of progeny identical to parent

Answer 109

Similarities - In both cases the genome is duplicated at the start - Cohesin rings are present Differences - Meiosis, crossover occurs in DNA recombination - Orientation of the centromere differs - Meiosis has two rounds of division

Answer 110

Sexual recombination allows for integration and sharing of genes and mutations In asexual populations, for an individual to have all the changes present would require sequential adaptation and mutation in the same lineage = Sexual mutation better for evolution of new traits which is essential for survival ++ recombination allows for Hill-Robertsone effect splitting deleterious and beneficial linkage drag + Muller's Ratchet - absence of recombination leads to build up of deleterious

Answer 111

SHUGOSHIN1 (SGO1) is a phosphatase that removes Pi added by CDPK - This protects the cohesin from being degraded and keeps the sister chromatids together At the second round of division SGO1 is no longer present so the two chromatids can be separated

Answer 112

Complex forms between two homologous chromosome pairs Promotes breakage, crossing over and repair of the chromosome ZYP1- axial protein that ensures recombination between homologous and not non-homologous pairs Recombination nodule - with MSH4/ MLH1 SPO11 to introduce DSB

Answer 113

Chromosomes are arranged in loops extending from the middle axis Following DSB the broken strand is moved into the axis for recombination - The broken strand is coated by DMC1 that likely interacts with proteins that drive this movement - DMC1 only present in meiosis!!!!

Answer 114

Class I - interfering 85% of most recombination Class II - non-interfering

Answer 115

Few crossovers occur at the centromeres - likely due to heterochromatic marks More crossovers occur at the ends, evidenced by comparison between physical and genetic map and fluorescent staining of recombination nodules that are found to be located near the ends

Answer 116

Many DSB are made however few actually cross over Requirement of at least 1 crossover to ensure proper segregation Interference - reduces the chances of multiple crossovers in an area e.g. fusion of chromosomes in C.elegans which elongated the physical length of the chromosome still only had 1 crossover e.g. non-linear relationship between size of the genome and number of cross-overs

Answer 117

Topoisomerase that regulates DNA unwinding - Generates a break in the phosphate backbone to reduce torsion 1. Becomes covalently bound to DNA via tyrosine residue 2. Cleaved off and retains a DNA tag 3. This can be purified and sequenced to find where SPO11 binds = Found that SPO11 binds inter-genic regions e.g. promoters where there is a lack of nucleosome

Answer 118

- Knock out of RdDM genes leads to increased recombination due to a lack of methylation (except for knockout of CG methylators ddm1 and met1) - Location of SPO11 binding is in nucleosome free areas with minimal epigenetic marks - Recombination rate is much lower in heterochromatic regions such as the centromere

Answer 119

Distribution of the E3 ligase HEI10 Coarsening model - suggests that accumulation of HEI10 leads to aggregate HEI10 and further accumulation targeting recombination to that area and away from others Experiment: transformation of HEI10 gene to increase copy number was found to increase the amount of recombination

Answer 120

Female - single cell undergoes meiosis in the ovule's megaspore mother cell Male - occurs in anthers, in the 4 locules so much easier to see

Answer 121

Required for proper segregation 1. Enable breeding - conventional breeding to select for an agronomic trait 2. Limit breeding - unevenly distributed along a chromosome, can prevent crossover of 25% of genes in cross-over cold spots

Answer 122

When genes bring along other genes commonly closely located

Answer 123

1. Increase in genetic diversity 2. 25% of protein coding genes are located in cross-over cold spots

Answer 124

1. Reducing cross-over suppressors - while many breaks occur, only few cross-overs form - e.g. fancm, mutagenised zip4 plants and found fancm mutants to be fertile. FANCM is responsible for unwinding D loops and preventing holiday junction formation 2. Promoting cross-over promoters - e.g. ZIP4 promotes cross-over, mutants show no cross-overs and plants are infertile

Answer 125

Class I - promote by ZIP4 (85%) Class II - repressed by FANCM (15%) (fancm mutants had higher rates of class II non-interfering cross-over)

Answer 126

1. cmt3 mutants that had reduced CG methylation can open up the centromere and lead to an increasing in crossover (cM) 2. suvh5/6 mutants that had reduced H3K9e2 also had increase in cM 3. Differences between melon and cucumber arise from TE expansion which can suppress crossing over

Answer 127

Increase in the number and movement of transposable elements - This can lead to upregulation of deleterious genes - Additionally can cause genomic instability

Answer 128

Targeted recombination using Cas9 or Cas12 - Homologous recombination occurs when there is cross-overto finsih

Answer 129

Cross-over -> cut and swap event, reciprocal exchange of genetic information. Genes are physically swapped -- Detected by changes in flanking markers -- Non-crossover -> copy paste, where genes around the DSB will be replaced using a copy of the homologous chromosome

Answer 130

1. Targetting of cross-over suppressors or activators still doesn't unlock heterochromatic regions 2. Targeted recombination could used to overcome this and to separate linkage drag 3. Although there can be more breaks, the actual amount of cross-over that occurs isn't as high 4. Unlocking by removing epigenetic marks in crops is not feasible and can lead to genomic instability

Answer 131

- Doesn't work for all loci - Low transmission to the next generation - Can lead to whole chromosome loss

Answer 132

Tomato - had two mutant alleles, one for yellow (mutation in the gene), and one that generated bicolor phenotype (mutation in the promoter) - Generated an F1 hybrid that was bicolor then induced DSB in the region between the two mutation sites by cas9 - in NHEJ more mutation would lead to yellow fruit, HR would lead to spotted or red, depending on whether repair was early in fruit development, or if the gene was fully repaired e.g. by crossing over and generating un-mutated fully functioning PSY1 gene Maize - Generated a targeted cut site for recombination and cross over to occur then backcrossed to one of the homozygous parent. This easily showed if the crossed parent had undergone reciprocal exchange (crossover) - Saw a 10 fold increase in occurrence of recombination around the DSB cut site, and these changes were heritable

Answer 133

COOLAIR (Csorba et al., 2014) PRE- VIN3 and other histone remodelling 1. Antisense long RNA transcript helps reduce amount of FLC - mutant had a slower decrease in FLC transcript 2. Destabilises H3K36me3 that acts as an activator for FLC - mutant had no reduction in this activator histone mark 3. Alternative polyadenylation and splicing leads to different roles. Proximal polyadenylation leads to enhanced H3K4me3 demethylase activity, while distal polyadenylation leads to increased FLC transcription 4. FLD recruitment as a histone demethylase = Found to be independent from H3K27me3 so co-ordinated but not an obligatory connection/ feedback loop present. Mutant in coolair still had normal increase, but ratio of promotive/suppressive marks differed COLDAIR (Kim et al., 2017) 1. Long noncoding RNA sense transcript that interacts with PRC2 (complex containing VIN3 and VRN2) 2. Maintains this COOLAIR-initiated repression of FLC Central region binds to FLC and modular motifs binds to PRC2. Mutation in modular motif leads to lack of PRC2 interaction and insensitivity to vernalisation

Answer 134

1. Inversions doesn't allow for cross-overs due to the differences in orientation 2. Drive high recombination rates to the sub-telomeric regions to compensate. Also supporting the idea that cross-overs are required for proper segregation - Used to lock important stretch of genes to prevent loss - Can also result in genetic bottlenecking as recombination can't occur - Presence of inversions identified due to next gen sequencing techniques

Answer 135

Targeted CRISPR can invert a whole section of the genome - Evidenced by increase in recombination frequency

Answer 136

1. Colchicine - binds to tubulin and prevents spindle fibre formation 2. In vivo - chromosome elimination

Answer 137

Histone variant (CENH3) in the centromere is responsible for binding microtubules In chromosome elimination - asynchronous cell division - CENH3 fails to load onto one of the centromeres and so chromosome is lost and this generates a micronuclei COOPERATIVE BINDING - once CENH3 is loaded onto the centromere, more will follow. Favouring WT over the Haploid Inducer

Answer 138

- Haploid induction easier - Intraspecies induction CENH3 tail modification for GFP - When crossed with WT, the chromosome segregation was weaker so produced: Uniparental haploid (desired), Hybrid diploid (normal segregation) and aborted seeds

Answer 139

1. Haploid inducer will have less functional CENH3 than the WT 2. When crossed, CENH3 from WT will bind to kinetochores 3. This leads to segregation and haploids with WT chromosomes being formed 4. HI chromsomes are lost as they couldn't bind to kinetochore due to competition from WT

Answer 140

Generate parents, from desirable lines, that when crossed will regenerate these lines 1. Prevent further crossing over - Achieved by silencing DMC1 or MSH5 2. Inducing haploid 3. Doubling the haploid

Answer 141

Inducing clonal reproduction by MiMe and parthogenesis (fertilisation without meiosis) 1. Inhibiting crossovers by inhibiting SPO11 from generating DSB 2. REC8 - cohesin ring that hold sister chromatids together is removed, to allow for sisters to separate 3. OSD1 - omits second meiotic division to generate diploid and not haploids

Answer 142

By haploid induction

Answer 143

1. Baby Boom 1 (BBM1) expression - not normally expressed in the egg cell, overexpession results in parthenogenesis (embryo without fertilisation) 2. PAR expression - not commonly present, but results in embryogenesis without fertilisation

Answer 144

DNA doubling without the cell dividing

Answer 145

- Formation of clonal seeds e.g. from hybrids - Fixing of specific traits - Reduce spread of disease in plants that are propagated vegetatively, most plant pathogens can't spread through seed

Answer 146

Combination of two plants already showing hybrid vigour to generate another plant with greater phenotype Crop breeding - Generation of tetraploid tomato that was larger in size, but cost of not being able to produce seed (Wang et al., 2024) - Greater resistance to stressors commonly seen in polyploid plants - Increase in genetic diversity - Clonal transfer of beneficial diploid genes into polyploid plants NEGATIVE may lead to genome instability

Answer 147

- Colonise the host - Modulate host defence - Promote nutrient uptake - Alter plant morphology and secretion

Answer 148

Cell surface receptors: RLKs or RLPs - Detect apoplastic effectors - Commonly require a co-receptor

Answer 149

1. Detected by FLS2 - Uses BAK1 as a co-receptor 2. Pi and activates BIK1 3. Kinase that activates immunity

Answer 150

Nucleotide binding Leucine-rich Repeats N-terminal domain = TIR/CC NB-ARC domain = modular activation (ATP=on) LRR = pathogen recognition site - Intracellular receptor in ETI

Answer 151

Direct Indirect - decoy or guardee binding that results in conformational change that is detected by the NLR - integrated domain e.g. HMA domain in rice NLRs => Greater robustness one NLR can detect more effectors

Answer 152

ZAR1 binds and detects Forms a pentameric structure known as a resistosome that is inserted into the plasma membrane and acts as a channel promoting ion flow CC-domain is responsible for triggering cell death

Answer 153

Don't move and form a channel, instead function as enzymes - Binding (not ATP) leads to activation

Answer 154

Sensor and Helper NLRs work together - Sensors don't have MADA motif that is responsible for triggering cell death - Binds and associates with Helper NLRs - Generates a degree of redundancy where some sensor NLRs can work through multiple helper NLRs which helps generate a robust immune system

Answer 155

1. Intracellular receptors can increase the expression of BIK1 and MAPKs 2. Pre-activating ETI an lead to greater ROS production when flg22 is detected (just activating ETI by AvrRPS4 didn't trigger ROS increase) 3. PTI potentiates ETI-induced cell death

Answer 156

ZED1 a pseudokinase - modification is sensed by ZAR1

Answer 157

Organised in clusters Commonly in sub-telomeric regions Generated by tandem duplications High recombination rates

Answer 158

Selection of R genes for amplification, carried out by generating probes that bind to conserved domains such as the NB-ARC domain - Sample then purified and don't have to sequence the whole genome

Answer 159

1. Lifespan differences - e.g. trees require more 2. Pathogenic load - more load = more stress = more variation required 3. Plant size - greater target for pathogens 4. Pathogen types - airborne, aquatic etc. 5. Plant lifestyle - parasitic plants with minimal roots have lost NB-ARC genes - crops grown in high density such as rice and maize have greater repertoire of resistance genes

Answer 160

1. Gene loss - e.g. truncated FLS2 gene in Melon that wasn't able to confer resistance against flg22 2. Polymorphism - mutations in R genes that results in susceptibility due to lack of resistance

Answer 161

1. Tandem duplications resulting in - sub-functionalisation = slightly different functions (2functions) - neo-functionalisation = completely new function - specialisation = more specific function 2. New gene fusions 3. Integration of new domains - helper and sensor commonly genetically linked

Answer 162

Unequal cross-over resulting in first half of NLR1 and second half of another NLR joining in Chinese spring variety - Not tested whether the new NLR actually works - Requires full genome sequencing to see that there is a large indel (missing stretch)

Answer 163

Sensor = balancing evolution, expansion to detect more effectors Helper = purifying selection, reduced variation and greater redundancy

Answer 164

Slightly different functions - NRC3 in benthi has subfunctionalised and diverged from potato and tomato - In Solanum NRC3a works with Rpi-Avrblb2 to provide resistance against Avrblb2 from P.infestans - Divergence likely due to higher pathogen load on potato and tomato than benthi

Answer 165

Rx and Gpa2 are near identical NLRs with 88% sequence similarity Rx- confers resistance to virus Gpa2 confers resistance to nematode

Answer 166

1. Transcriptional level - Epigenetic modification regulating accessibility - WRKY TFs regulating transcription 2. Post-transcriptional level - miRNA silencing 3. Post-translational level - Chaperones regulating stability - Ubiquitination 4. Inter NLR regulation - e.g. RGA5 and RGA4

Answer 167

EFR cell surface receptor that detects elf18 - In Brassicaceae moved to tomato and generated greater resistance

Answer 168

- Function and effectiveness in different host species - Broad or specific resistance - Number of R genes that need to added (need to know pathogen load/ diversity) - Interaction between different immunity genes

Answer 169

Transferance of certain genes from one species to another isn't able to generate resistance e.g. transfer of Bs2 gene from Solanaceae that confers resistance against Xanthomonas by detecting AvrBs2 to non-solanaceae - due to it being part of the NRC network

Answer 170

Previous - Random mutagenesis found that single amino acid change in R3a increases target range of Avr3a molecules Newer - Targeted (structure-guided) mutagenesis - Single amino acid mutation meant that Sr50 was able to detect new variation of AvrSr50 QCMJC

Answer 171

- single amino acid change in the NRC2/3 prevented it from being targeted by the SS15 effector ! Important to ensure that mutation doesn't alter how the protein oligomerises! - Resistance was still present before as the sensor could act via NRC4

Answer 172

1. Immunise llama with effectors and antigens from the plant pathogen 2. The mammalian immune system will generate nanodies (like antibodies but smaller) 3. Isolate these nanobodies 4. Integrate these nanobodies in place of the ID domain in NLRs = formation of pikobody 5. Confers plant immunity to the pathogen

Answer 173

Decoys - Modification of a decoy (e.g. PBS1) so that is can be targeted by more effectors - Ensure its recognition site by NLRs (RPS5) isn't modified and can still be detected Integrated domains - HMA from rice introduced to other NLRs to confer resistance against more effectors

Answer 174

1. Promoter engineering - changing expression of genes 2. Plant vaccines 3. Engineering decoys 4. Resurrecting NLRs 5. Receptor engineering 6. Transferring R cassettes between plants

Answer 175

1. Innovation - new ideas 2. Manufacturing requiring standardisation - e.g. certain enzymes, processes etc. are used

Answer 176

1. Type IIS restriction enzyme that cut outside recognition site 2. Generates overhangs that can be matched with DNA ligase 3. Cleaved and the recognition site disappears so can't be cut again

Answer 177

1. Simple on/off 2. Feedback Positive - generate oscillatory dynamics Negative - suppresses induction, damping effect

Answer 178

Three repressors that repress each other in a loop - generates a cycling of readouts

Answer 179

Antimalarial drug - Study of the processes and then pathway synthetically generated

Answer 180

Cloning technique that allows you to assemble DNA fragments that have overlapping ends, without using restriction enzymes or ligation 1. Exonuclease creates overhangs 2. DNA pol fills gaps in annealed areas 3. DNA ligase seals the nick

Answer 181

- Larger genomes of crops hard to edit - Epigenetic marks and their regulatory role - Compatibility in different species - Repetitive sequences making design harder

Answer 182

EU has less GM crop than USA due to stricter regulations UK having left the EU are revising the precision breeding act and it may be feasible to sell GM crops

Answer 183

Natural bacterial systems that target phage DNA Function - generates a DSB at a target - promotes error prone homologous recombination to generate a mutation in the target gene

Answer 184

WUSCHEL- maintains stem cell identity, so can changes locule number FASCIATED - mutants have larger locules, greater fruit size

Answer 185

To identify genes associated with a quantitative trait (dynamic and governed by multiple genes) - Also help with marker assisted selection in breeding programmes

Answer 186

- Severe growth phenotypes - met1, drm1/2, cmt2/3 = Suggests important for genome stability and transposon regulation

Answer 187

Not much change in phenotype, despite the increase in methylation - Early flowering in quadruple mutant - dme, ros1, dml1/2

Answer 188

1. Changes in DNA methylation around TEs, increasing accessibility of machinery to neighbouring defence genes 2. Overall demethylation to remove transcriptional silencing 3. Changes in histone modification e.g. H2 monoubiquitinationof histones by HBU1/2, mutant shows increased susceptibility to necrotic fungi but greater resistance to hemibiotroph P.syringae in tomatoes - Mutants of rddm pathway are more susceptible to P.syringae - In crops important to be specific as 80% TEs

Answer 189

Chemical inhibition of Pol II to reduce transcript of siRNA 21nt as part of the non-canonical pathway - This leads to less methylation and greater TE movement, leading to (epi)genetic diversification Parker et al., 2022

Answer 190

Complicated and went through intermediate stages - In one accession that moved northwards, its requirement to adapt to the new conditions likely resulted in the loss of some beneficial traits, which was later redomesticated to regain the favourable traits - Studies have found two separate populations of Mesoamerican S.lycopersicum (which can mean that finding the centre of origin for greatest diversity is difficult)

Answer 191

- Domestication of energy rich carb along with a protein rich legume e.g. rice and soya bean - Polyploidy generates an advantage due to greater diversity, but altering dosage of genes could be deleterious - Annual crops are more common, potentially due to the shorter generation time that speeds selection - Ability for self-fertilisation can be beneficial to maintain desirable traits

Answer 192

- Genetic bottlenecking as the allopolyploid is reproductively isolated from its polyploid progenitors - Allopolyploid wheat has been shown to differ a lot from their diploid progenitors which has resulted in them expanding over greater geographical regions - Differing of structure occurs quite rapidly (originated only 10KYA), likely by DNA rearrangements and point mutations - Due to multiple gene copy, mutations are more likely to be tolerated due to the buffering effect - Polyploidisation also alters gene dosage that can have beneficial or deleterious effects - adaptations have resulted in diploid-like expression by elimination and inactivation of genes. = Important to have a plastic genome, selfing (as can't cross with progenitors)

Answer 193

Enhanced adaptability and evolutionary flexibility by: - Greater genetic variation (mutations in somatic (non-repro cells) can be passed on) - Epigenetic diversity (fine tune expression and response to different env conditions) - Clonal variation (variation in plants that reproduce clonally)

Answer 194

1) High rates of recombination - increased by SPO11 topoisomerase enzyme that forms DSB. Also suggested to help form synaptonemal complex 2) Use of homologous chromosome as template - RAD51 and DMC1 important for strand invasion and formation of DHJ, DMC1 is specific to meiosis. - Orientation of spindle fibres and chromosome structure promotes recombination with homologue and not sister chromatid 3) Promoting cross-over outcome - MSH4/5 unsure how but present in meiosis only - Crossover in somatic cells in yeast 5-20%, while in meiosis cross-over rates around 30-50%

Answer 195

The maintenance of cross over levels and aspects such as interference - In spo11 mutants, fewer DSBs where formed. Crossover frequencies were maintained at the expense of non-crossovers in decreasing DSBs PREVENTING too MANY and too FEW crossovers (ALSO 75% of recombination in wheat was found within 10% of the distal end of chromosomes)

Answer 196

1. Reduced gene density - Subtelomeric have fewer genes, less selective pressures to maintain these genes 2. Structure - commonly contain more repetitive regions, which are more prone to misalignment 3.Chromatin structure - less condensed chromatin structure

Answer 197

1) Clonal transfer - transfer traits that are found to be beneficial in diploids to polyploid varieties e.g. strawberry. 2) Increase genetic diversity

Answer 198

Vernalisation has evolved several times and shows convergent evolution - Monocots such as wheat and barley use VRN1/2/3 loop - Eudicots such as Arabidopsis use FRI-FLC loop Vernalisation evolved in grasses that had to adapt to temperature regions, while grasses still in tropical zones lack vernalisation response (Xu and Chong 2018)

Answer 199

1. Find homology to other genes in other plants 2. Analyse protein structure - whether there is a membrane domain 3. Localisation - GFP 4. Co-localisation studies - using Y2H 5. Expression analyses - When the gene is activated in response to different kinds of stress (Mutagenesis unless there are many genes in the family or in crops where redundancy may be present)

Answer 200

DeadCas9 - modification/ addition of an active region to a dead cas9 that has lost nucleolytic activity - modification with either DNA methylator/ demethylase, or histone modifier such as H3K27 methylator EZH2 - targeted and still specific due to CRISPR construct and guideRNA (Brezgin et al., 2019)

Answer 201

Cis - DNA sequences locate within or near the gene they regulate (can effect individual gene) Trans - DNA or protein molecules that regulate transcription different of genes (can affect entire network) e.g. paramutations

Answer 202

1. Inheritance of stress adaptations - as somatic tissues give rise to meiotic cell precursors that can form gametes 2. Silencing of invasive viruses or transposons- due to repeats that are structurally similar to ones in the centromere 3. Mechanism for generating functional homozygosity in polyploids Arteaga-Vazquez + Chandeler 2010

Answer 203

1. Position effect variegation - nearby leads to suppression 2. TEs in regulatory regions - SINE TE in VTE (vitE) promoter, spontaneous changes in TE methylation (rich in CHH), is responsible for epialleles in tomato (Quadrana et al., 2014)

L1: Synthetic Genomes Flashcards

(227 cards)