Epi T 3

What is "escape" from X inactivation?

"Escape" from X inactivation means that a gene or region of genes has not been inactivated due to the dosage compensation mechanism. This can mean that the expression of these genes in females is different than it is in males.

Describe the steps necessary to determine if epigenetic changes seen in a cancer are due to correlation or due to causation.

1. Check if 5mC and expression state correlate and can be reversed 2. Check if 5mC correlates with tumor state• 3. Document how 5mC changes during tumor progression 4. Check if phenotype can be reversed by active copy of gene 5. Establish function of gene 6. Document gene function in vivo

What are three functions of cytosine methylation in mammals?

1. Regulation of gene expression (promoter methylation) 2. Silencing of transposable elements 3. X chromosome inactivation

What are PREs?

A PRE is a polycomb response element. It is a region of DNA that allows the polycomb response complexes to bind and silence a stretch of DNA. PRC2 recognizes this region and causes H3K27me. This methylation is then recognized by PRC1, which then binds to the methylated PRE leading to DNA silencing and formation of condensed DNA.

Describe a typical imprinted gene cluster in mammals.

A typical imprinted gene cluster has differentially DNA-methylated regions (DMRs). These regions are methylated differently depending on the sex of the parent that they came from, and they tell the genome to express certain genes as imprinting regulating elements (ICE). The expression of an imprinted gene in the genome of one parent does not ensure that all other genes in the cluster are expressed in that parental genome. One gene in the cluster might be expressed from one parental genome, but the rest of the genes could be expressed from the other parent. There are also long non-coding RNAs present that can be expressed in both of the parents.

Compare and contrast active versus passive DNA demethylation and give an example of where we see them in nature.

Active DNA demethylation involves the activation of proteins that demethylate DNA by removing the methyl group from cytosine, while passive DNA demethylation involves the inactivation of maintenance methyltransferases for a particular DNA locus. As DNA is replicated, maintenance methyltransferases such as Dnmt1 help maintain CpG methylation patterns in the daughter genome. However, in passive DNA demethylation, maintenance DNA methyltransferases are not activated for that locus. As the genome replicates, only the original strands of DNA will continue to be methylated at that locus, with the relative amount of 5mC present in that region decreasing as more copies of the genome are made. This effectively leads to demethylation at that locus because the methylation is not being maintained over replication. Active DNA demethylation is seen in the paternal genome of mice, while passive DNA demethylation is seen in the maternal genome of mice. OR Active DNA demethylation usually involves an eraser, such as the TET enzyme, that actively comes in to demethylate methylated DNA. Passive DNA demethylation involves not having a writer, so that as methylated DNA undergoes DNA replication, it becomes hemimethylated. Without Dnmt1 for maintenance methylation, the hemimethylated DNA undergoes more DNA replication. The hemimethylation eventually dilutes away until the DNA is fully demethylated.

How do epigenetic mechanisms contribute to cancer?

Altered DNA methylation can contribute to cancer. Losing methylation (hypomethylation) can lead to genome instability. For example, losing methylation in transposable elements will no longer suppress them. Additional methylation (hypermethylation) to promoter region can silence genes such as tumor suppressor genes which will contribute to cancer. DNA methylation sites can be deaminated causing a mutation. DNA methylation sites are also a stronger target for UV radiation which increases UV-induced mutations In addition, some carcinogens have preferential impact on areas of the genome that are methylated

What is Beckwith-Wiedemann syndrome? Make sure to explain its molecular cause and describe patient phenotypes.

Beckwith-Wiedemann syndrome can be caused by errors in imprinting. It leads to higher risk of childhood cancer as well as abnormally large abdominal organs. The cause of Beckwith-Wiedemann syndrome is either the loss of methylation at the IC2 region on the maternal genome or the gain of methylation at the IC1 region on the maternal genome. Normally, the IC1 region is methylated on the paternal genome, while the IC2 region is methylated on the maternal genome. The gain of methylation at IC1 leads to expression of Igf2 from the maternal genome, and no expression of H19. The loss of methylation at IC2 leads to silencing of two genes in the region.

What is CRISPR/Cas9-mediated homologous recombination?

CRISPR/Cas9 is a protein-nucleic acid complex from the bacterial immune response that can cut out DNA. This DNA can then either be repaired, or it can be replaced by homologous recombination.

Compare and contrast cis- and trans-genetic effects that can lead to epigenetic defects.

Cis-genetic effects involve mutations in the regulatory region of the gene, while trans-genetic effects involve mutations in the actual regulatory protein. For example, in Cis-genetic effects, the addition of a repeat sequence in the CpG island of a gene promoter can lead to differential expression and epigenetic regulation of that gene and also disorder. In trans-genetic effects, there can be a mutation in a protein such as a histone acetyltransferase that causes it to lose its function. Without the function, the targets cannot be epigenetically modified, potentially causing disorder.

Explain the concept of parent-of-origin effect using the Agouti mouse model.

Depending on what parent the epiallele comes from, the epiallele will be expressed differently in the offspring. If an Agouti viable yellow mother is crossed with a Pseudoagouti father, most of the offspring will appear yellow, while some will be mottled (brown and yellow spots). However, if the father is Agouti viable yellow while the mother is Pseudoagouti, then the offspring will be a mixture of pseudoagouti, mottled, and yellow. This clearly shows that there is some bias based on parental sex on what epiallele will appear in the offspring. OR Mice offspring had different phenotypes depending on whether the mother or father had Avy, indication that parent of origin had an effect. If the mother had Avy, there was a shift towards yellow phenotype in offspring. If the father had Avy, there was a shift towards pseudoagouti phenotype in offspring.

What is dosage compensation?

Dosage compensation occurs when the genome of the two sexes differ. For example, in humans, males have a sex chromosome genotype of XY, while females have a sex chromosome genotype of XX. Since most of the genes on the X chromosome are essential and aberrant levels of expression of these genes can cause problems, the amount of genes being expressed must be equalized between the sexes. In humans, this occurs by the inactivation of one of the X chromosomes in the female, which causes relatively the same amount of X chromosome expression in the female as occurs in the male.

What aspect of their DNA methylation landscape distinguishes mammalian embryonic stem cells from most other cells?

Embryonic stem cells have non-CpG context DNA methylation. Most other cells only have DNA methylation within a CpG context.

How would you determine if a new phenotype you see in your mouse strain is maintained by epigenetic inheritance?

First, I would see if the phenotype is inherited at all. I would breed the mouse for a few generations and observe if the phenotype remains in the offspring. Next, I would compare the DNA sequence of the mice where the phenotype is present with mice of the same lab strain where the phenotype is not present to observe if the phenotype is due to a mutation in DNA sequence. If the phenotype is inherited by offspring and does not seem to be from a DNA sequence mutation, then it is possible that the phenotype is maintained by epigenetic inheritance. OR a. The first thing to determine is whether the phenotype is heritable across generations. I would breed the mouse till the F3 generation and observe if the phenotype remains in each generation to ensure its heritable and not due to any other cause. I would compare the DNA sequence for each generation to determine whether the phenotype is observed in each generation without a change in DNA sequence. I would also cross foster the F3 generation to ensure that the phenotype isn't due to maternal effects. b. In addition, I would compare the DNA sequence of mice where the phenotype is present with mice of the same lab strain where the phenotype isn't present to determine whether the phenotype is due to a mutation. c. If the phenotype is observed in the F3 generation, and if the phenotype isn't due to a mutation, it's probable that the new phenotype is maintained by epigenetic inheritance.

What are homeotic transformations? Explain with an example.

Homeotic transformations involve the substitution of one part of a flies body with another. These can be caused by mutations in the polycomb genes. One homeotic transformation is Antennapedia, where legs develop where antennae are supposed to be.

Your grandfather comes to you with an exciting piece of news. He has just read that eating bacon is going to make him live longer, and he is ready to embark on a new bacon-rich diet. Given your knowledge of human biology, you are skeptical of this piece of news, and you track down the primary article that this piece of news is based on. How will you determine if this finding is reliable?

I would investigate the credibility of the primary article. I would check to see if it's from a journal with reasonable quality control, such as being peer-reviewed. I could further look into the author to see if the author has written any other credible papers. I could also look through the data and methods myself to see if data truly represents the result and if the methods are replicable. OR 1. check the author's background to see if they are qualified and experts in this field of research 2. check the reputation of the journal that the article was published in; the article could be published in a "fake" online journal; a popular magazine could have "clickbaited" this information or drew false conclusions from the primary research 3. check the funding of the research; the funding could come from a third party company that could affect the objectivity of the data 4. read the article and look for reliability and validity flaws such as low sample size

Describe a genetic disorder that impacts DNA methylation. Make sure to identify the gene causing the disorder and explain its role.

ICF syndrome usually involves a mutation in a methyltransferase enzyme such as Dnmt3B. For whatever reason, the mutation in Dnmt3B causes chromosomal instability in lymphocytes, leading to lackluster immune response in a recessive manner. This leads to many individuals born with ICF to succumb to infectious diseases before they become adults.

Explain how the Polycomb groups of genes was discovered in Drosophila.

Mutations in the polycomb genes led to homeotic transformations where the development of various different limbs were affected. Study of these homeotic transformations led researchers to identify specific mutations that these phenotypes were associated with, leading to discovery of the polycomb genes.

What is the "n-1" rule?

N is the number of X chromosomes present in the cell. N-1 denotes how many Barr bodies will be formed due to X inactivation. There will only be one active X chromosome per cell, so 1 is subtracted from the number of X chromosomes.

What is the "two-hit" model for cancer?

Normal cells have two undamaged chromosomes, one from the mother and the other from the father. People with hereditary susceptibility to cancer inherit a damaged gene on one chromosome, so their first "hit" or mutation occurs at conception. Even if it isn't hereditary, if a cell receives damage to the same gene on the second chromosome, that cell can produce cancer (second hit).

Compare and contrast the two polycomb silencing complexes.

PRC2 is present in every eukaryotic model system, while only certain parts of PRC1 are present in every eukaryotic model system. PRC2 causes H3K27 methylation, while PRC1 binds to H3K27 methylation to cause some response. Thus, PRC2 acts as a methylation writer and PRC1 acts as a methylation reader.

Explain how PRC1 and PRC2 function using the "writer-reader-eraser" model for epigenetic marks.

PRC2/PRC1 binds to a polycomb response element and PRC2 causes methylation of H3K27, acting as a histone methylation writer. PRC2 can then dissociate from the PRE, while PRC1 can read the newly formed H3K27me and bind. PRC1 stays bound to that PRE, leading to silencing of the gene .

What is the relationship between polycomb and trithorax?

Polycomb genes and trithorax genes bind to the same region. Polycomb genes methylate the histones in this region, leading to gene silencing, while trithorax genes acetylate the histones in this region, leading to gene activation.

What is the difference between random X inactivation and imprinted X inactivation?

Random X inactivation means that there is no parental preference for whichever X chromosome is inactivated. This means that regardless of which parent the X chromosome came from, the inactivation of X chromosomes happens at random. Imprinted X inactivation means that there is a parental preference for whichever X chromosome is inactivated. For instance, the paternal chromosome might be preferentially inactivated during X inactivation and the maternal chromosome might have a much higher chance of not being inactivated.

What is Rett syndrome?

Rett syndrome is caused by a mutation in MeCP2, which is an X-linked gene. A child with Rett syndrome has neurons that develop normally for a time period between 6 months and 18 months. However, at some point between those two times, neurological regression occurs. If the child was able to walk before this period, it may be unable to walk after the regression begins.

What is a human disorder affecting chromatin in trans? Make sure to explain what is meant by "in trans".

Rubinstein-Taybi syndrome is a human disorder that affects chromatin in trans. "In trans" means that chromosome modifying factors have been lost. In the case of Rubinstein-Taybi syndrome, a mutation in a histone acetyltransferase gene leads to the loss of the histone acetyltransferase function. Since there is no HAT, genes associated with that HAT cannot be regulated, so disorder occurs.

Describe the experiment by Bowman et al that illustrates how the polycomb system regulates Hox gene expression in the Drosophila embryo.

They used Gal4 and Gal80 reporter gene constructs to turn on reporter gene mCherry, in a specific segment. mCherry is only activated where Gal4 is present but Gal80 is not. Because mCherry can be turned on in specific sites, they can isolate cells in that specific activated segment. They took the embryos and the cells from those segments and looked at K27 methylation at hox genes (diff segments have diff active hox genes), and predicted a difference in methylation and silencing. Results: PS4 is silent and enriched in methylation, PS5 has lower K27 methylation and active UBX, PS6 has even lower methylation and more expression overall, PS7 has virtually no methylation and only the back part of the cluster was methylated. This shows how the polycomb system is involved in segment and developmental regulation in these genes.

List four types of genes that belong to the trithorax group?

Trithorax genes can be ATP-dependent remodelers, Transcription factors, mediator subunits, and growth factors.

Discuss three ways how epigenetics can influence human disease.

cis, trans, imprinting********* OR a. Environment affects your genotype which affects epigenome which can be seen in phenotype. b. Small non coding RNAs, heritable, can cause disease phenotype c. Imprinting disorders d. Infertility - epigenetic patterns aren't completely erased to make new epigenome needed for sperm of oocyte

You are about to start your own epigenetics lab. You need to choose your model organism and defend your choice to your department chair. Pick a model organism, and explain your choice. Make sure to discuss at least 4 different characteristics of your organism and explain how it is superior to other model systems used for epigenetics studies.

refer to number 1 If I were starting my own epigenetics lab, we would be focused on human diseases. For that purpose, I would choose to work with mice. The mouse genome is much closer to the human genome, and they also have biological systems that are similar to humans. Mice are also usually smaller than rats, so working with mice over rats is preferred because the animal facility would take up less space. Working with monkeys causes the biosafety level of the research facility to be increased in order to deal with the risks of cross-species infection, which costs money. Thus, working with mice as a mammal system is preferred for most epigenetic studies.

What types of epigenetic changes are often seen in cancer?

DNA methylation and other histone modifications contribute to cancer formation such as hypomethylation, hypermethylation, 5mc deamination, increased UV-induced mutations, and carcinogen induced mutations. In addition, DNA methylation patterns change in the beginning of cancer progression, causing mutations in chromosomes and chromosome fragments which can lead to a tumor. OR DNA methylation/inactivation of tumor suppressor gene promoters.

How do the polycomb and trithorax systems contribute to cellular memory?

Polycomb genes recognize PREs and cause genes to be silenced. PRC2 binds to a PRE and causes H3K27 methylation in that region. This is then recognized by PRC1, which binds to H3K27me and causes the DNA to become more condensed. PRC1 can then remain attached to this methylation as long as the methylation is maintained. The trithorax genes recognize TREs, which are the same sequences as PREs. However, whereas polycomb genes add silencing methylation to the regions, trithorax genes acetylate the TRE regions, potentially activating the genes.

What characteristics of mice make them useful model organisms?

1. small and easy to take care of 2. 95% genetic similarity to humans 3. relatively short lifespan and generation time 4. sequenced genome Mice are small and easy to take care of. This is important because it allows less space or resources to be used in caring for animals than would be required if an animal such as a monkey was used in research. 95% of the mouse genome is the same as the human genome, allowing for more feasible applications to human epigenetics to be tested compared to a model that is much less related such as S. pombe. Mice have relatively short lifespans, with some mice only living around 2 years. They also have relatively short generation times, which allow heritable mammalian epigenetics to be studied faster than a model such as monkeys, which can have much longer generation times. The mouse genome has been completely sequenced, allowing for precise studies of genetics.

Describe the distribution of 5mC in the mammalian genome.

5mC is usually only studied in CpG contexts in mammals since the only known mechanism for maintaining methylation in mammalian cells occurs at CpG sequence contexts. However, 5mC can be found in non-CpG contexts, usually in cells that do not tend to divide. 70~80% of CpGs in mammals are methylated. OR 5mC in the mammalian genome is found mainly in CpG contexts.

What is a CpG island?

A CpG island is a stretch of DNA that is around 1 kb long and consists of high amounts of CG residues. They are usually found in the promoter region of genes and appear to have a regulatory effect on the expression of genes. Namely, CpG methylation in CpG islands seems to be inversely related to the expression of that particular gene. However, most CpG islands are unmethylated. OR CpG islands are a chunk of DNA in the promoter region that have a lot of CG dinucleotides in a short span. They're usually about 100 bp and often unmethylated (implying that the gene is active).

Explain how paramutation fits the definition of epigenetics we use in this class.

A paramutation occurs when a plant with a specific paramutable, active allele is crossed with a plant that has the paramutagenic, inactive allele. The introduction of the paramutagenic region causes the paramutable locus to become paramutagenic. In practice, this can occur when maize that is purple is crossed with maize that is lighter. The paramutagenic allele from the lighter maize causes a region inherited from the purple maize to become silenced. Thus, the progeny of the purple maize X lighter maize will never produce any purple maize. The actual genomic sequence of the maize is not changed, but the expression of the gene is changed and this change of expression can be inherited. OR A paramutation is when one allele induces a heritable change in the other allele, often in the pattern of DNA methylation or histone modifications. Since this change is a mitotically and/or meiotically heritable change that affects gene function and involves no change in DNA sequence, this aligns with the definition of epigenetics we use in this class. OR Paramutation is the process that homologous DNA sequences communicate in trans to establish meiotically heritable expression states. This occurs at the maize b1 locus in maize. Trans-interaction between the B' silent allele and the B-1 active allele changes the active allele into a silent allele. The actual genomic sequence of the maize is not changed, but the expression of the gene is changed and this change of expression can be inherited.

How do ATP-dependent remodelers function?

ATP-dependent remodelers hydrolyze ATP and can lead to chromatin remodeling via four different mechanisms. The nucleosome can slide away from the region that the remodeler was bound to, causing the chromatin to be less condensed and possibly opening the region to transcription. A histone can be exchanged for a modified version, which can either allow for more transcription or less transcription to occur. A nucleosome can be removed, leading to more free DNA, potentially activating a region of DNA. Lastly, the nucleosome structure can be modified, allowing the DNA to be more loosely bound to the histone octamer. This can allow this DNA to be transcribed and lead to activation of a region of DNA. nucleosome remodelers can slide nucleosomes, evict nucleosomes or exchange nucleosomes using ATP

Compare and contrast the following two methods to assay DNA methylation: BS-seq and Nanopore sequencing.

BS-sequencing involves treatment of the DNA with bisulfite, causing all unmethylated cytosines to become uracil, while all 5mC remain cytosine. PCR is required to sequence the sample of DNA. Sequencing of this DNA allows identification of 5mC. Nanopore sequencing is able to determine if 5mC is present without any chemical modification of the DNA or PCR. Nanopore sequencing involves the production of an electric field and determination of DNA sequence based on minute changes to the field caused by different bases. However, nanopore sequencing is much less stringent than conventional sequencing methods with an error rate much higher than other methods. Thus, more fine-tuning of the nanopore method is required before it becomes preferable to Bisulfite sequencing. OR BS-Seq: When you treat DNA with bisulfite, it turns unmethylated cytosine into uracil while methylated cytosine remains cytosine. After this treatment, you can determine in what percentage of your cells cytosine by comparing the ratio of Ts (unmethylated) to Cs (methylated). It cannot determine between 5mc and 5hmc. Nanopore sequencing: Feeds DNA through tiny pores. An electrical current runs across the pore and that current differs depending on what nucleotide it is. There is a tiny difference in the current of methylated and unmethylated cytosine that nanopore sequencing can read. It can determine between 5mc and 5hmc. Nanopore is also considerably faster than BS-Seq.

Explain why utilizing bisulfite sequencing for assaying DNA methylation results in a higher resolution than MeDIP analysis.

Bisulfite sequencing allows for direct sequencing of DNA methylation, whereas MeDIP only allows you to know the general region in which the DNA methylation is located. MeDIP is a modified version of chromatin immunoprecipitation where an antibody for methylation is used to pull down regions that have methylated cytosines in them. This only allows a researcher to identify regions that have DNA methylation in them, and not individual methylated bases. Bisulfite sequencing involves the modification of all nonmethylated cytosine residues into uracil, while 5mC remains intact as cytosine. This allows a researcher to sequence the DNA and identify specific cytosines that have been methylated. OR Bisulfite sequencing allows you to visualize where methylation is along the analyzed strand. MeDIP can only give a peak at the approximate location where methylation is enriched.

Describe the DNA methylation dynamics of the maternal and paternal genome in the pre-blastoderm mouse embryo.

Both the maternal and paternal genome in their respective germ cells are methylated until zygote formation. After zygote formation, the amount of 5mC in the paternal genome decreases rapidly, while the amount of 5hmC increases rapidly. This indicates that the Tet enzymes are being used to actively demethylate the paternal genome. The maternal genome maintains its 5mC levels until the paternal genome has reached a peak of 5hmC levels. At this point, cell division occurs, and the maternal genome is demethylated via diffusion of methylation, while the paternal genome appears to passively lose 5hmC. This slow decrease in methylation levels continues until the blastocyst forms, at which point the methylation increases due to cellular differentiation.

Explain briefly how the ChIP-seq method works.

CHIP-seq involves a combination of chromatin immunoprecipitation, in which DNA and proteins that are interacting with that DNA are crosslinked. The DNA is then cut to allow for separation and an antibody against the protein is used to pull down the protein/DNA complexes. After that, the DNA attached to the protein can be sequenced to see what DNA sequences the protein was interacting with in those samples.

How does the cytosine methylation state change during the early stages of mouse development?

Cytosine methylation from the paternal genome decreases faster than cytosine methylation from the maternal genome after fertilization. Demethylation of the parental genome from gametes must occur because the gamete DNA is methylated in a manner that allows it to function as a differentiated gamete instead of a fibroblast. The decrease of methylation in the paternal genome correlates with an increase in hydroxymethylation of the paternal genome, indicating that demethylation is occurring via the TET enzymes. However, demethylation of the maternal genome is not paired with an inverse increase in hypomethylation of the genome, indicating that the demethylation of the maternal genome proceeds via a TET-independent mechanism. This can occur via passive demethylation, where the maintenance of 5mC is shut off for the maternal genome. As the zygote divides into more cells, the methylation on the maternal genome is lost because it is not being maintained.

What is the problem with the statement "cytosine methylation is associated with gene silencing"?

Cytosine methylation is only associated with gene silencing when the methylation occurs in gene promoters. Cytosine methylation in gene bodies is actually associated with gene activation. OR While cytosine methylation is generally associated with gene silencing, but it does not always have a repressive function (such as in the case of gene body methylation).

Describe the DNA methylation dynamics during early mouse development.

DNA methylation decreases immediately following fertilization of the new mouse. Once the zygote begins dividing, the cells begin to differentiate into more specialized cells, causing DNA methylation to increase. At a certain point, the methylation pattern of the mouse somatic cells increases as they differentiate more, while the embryo's germline cells begin to demethylate as they become the highly undifferentiated primary germ cell. Upon sexual maturity, the primary germ cells will divide and be differentiated into egg and sperm cells, meaning that the DNA methylation patterns increase.

How do DNA methylation levels change throughout mouse development?

DNA methylation decreases immediately following fertilization of the new mouse. Once the zygote begins dividing, the cells begin to differentiate into more specialized cells, causing DNA methylation to increase. At a certain point, the methylation pattern of the mouse somatic cells increases as they differentiate more, while the embryo's germline cells begin to demethylate as they become the highly undifferentiated primary germ cell. Upon sexual maturity, the primary germ cells will divide and be differentiated into egg and sperm cells, meaning that the DNA methylation patterns increase. OR Early in mammalian and mouse development, maternal and paternal germ cells begin with high methylation that needs to be erased. The timing of this erasure is different between paternal and maternal differs. Paternal DNA 5mC levels decrease first due to active TET enzymes working on the cytosines. Maternal DNA 5mC levels decrease later through passive dilution of 5mC after divisions.

Why is the Drosophila dosage compensation complex also called the MSL complex?

Deletions in any one of the genes involved with the creation of the male specific lethal complex in male flies will cause the fly to die. Thus, these genes are lethal if mutated specifically for male flies. This is because the MSLC is responsible for upregulating the genes on the male X chromosome to match the female fly's expression levels. Some of the genes on the X chromosome are essential to fly viability, and thus expression of the genes on the X chromosome need to be the same across the sexes. If the male X chromosome is not upregulated, it can lead to death.

Describe what happens to chromatin structure during spermatogenesis in rhesus monkey.

During spermatogenesis in rhesus monkey, various different regions of DNA associate with one another, forming a TAD. Throughout the differentiation of sperm, these TADs begin to become much more loosely established, reaching a peak of TAD loss at the PAC stage. However, these TADs are reestablished by the time that differentiation has completed and mature sperm have formed, meaning that there is some mechanism by which the TADs stay associated with each other and reform during spermatogenesis.

Describe how to generate a gynogenetic embryo via nuclear transfer.

First, an egg must be fertilized with sperm. However, before the sperm nucleus fuses with the egg nucleus, the sperm nucleus must be removed, and a nucleus from another egg must be added. This then causes the fusion of two maternal nuclei, which produces a gynogenetic embryo, which will be fatal. OR A gynogenetic embryo has two copies of the maternal nucleus, which can be done by removing the paternal nucleus and sticking a second oocyte into the embryo.

For an experiment you want to do, you need to have Arabidopsis plants with different levels of DNA methylation. Describe how you might manipulate your wildtype Arabidopsis strain to obtain plants with three different levels of DNA methylation.

First, you could keep the wild type as one of the levels of DNA methylation. This would be as if you did not modify anything that can regulate methylation and can act as a control. The second strain could be hypermethylated by overexpressing the gene coding for the de novo DNA methyltransferase. This could be done with insertion of a transgene or by identifying a positive transcription factor for the DNA methyltransferase and increasing the levels of that transcription factor. The third strain could be produced by increasing the expression of DNA glycosylases, which are a type of DNA demethylase that is present in plants.

Describe the two models of trans-generational epigenetic inheritance.

For epigenetic changes to be inherited over generations, either some epigenetic changes need to be maintained through the initial epigenetic reset of the embryo or some other factor such as RNA must be maintained from the gametes in order to direct other factors to specific regions to cause epigenetic change. Without one of these factors in place it is difficult to explain any trans-generational epigenetic inheritance with our current understanding of epigenetics. OR One model of trans-generational epigenetic inheritance states that there is incomplete erasure of maternal epigenetic marks. While the paternal genome is demethylated, maternal marks slip through and gets into the embryo so that the epigenetic information gets passed on. Another model of trans-generational epigenetic inheritance states that it's done through paternal germline-transmitted RNA. While the maternal genome is demethylated, RNA from the sperm reestablishes the epigenetic mark.

What is a human disorder affecting chromatin in cis? Make sure to explain what is meant by "in cis".

Fragile X syndrome is a human disorder that affects chromatin in cis. "In cis" means that there is a mutation in the regulatory region (i.e. where chromatin or DNA modifiers would bind) of the gene, in this case, the promoter of the FRM1 gene. This mutation causes differential regulation of the gene in question, which can cause disorder.

What is gene body methylation?

Gene body methylation is methylation that occurs within the transcribed region of a gene, either within an exon or an intron. Gene body methylation is associated with increased expression of a gene, while methylation in a gene promoter is associated with decreased expression of a gene. OR Methylation of gene promoters is associated with gene silencing. Methylation of the gene body itself is remarkable in that it's positively correlated with expression.

What is genomic imprinting, and how does it fit the definition of epigenetics we use in this class?

Genomic imprinting involves the differential expression/regulation of genes depending on the sex of the parent that it was inherited from. In most cases, the gene sequence is the same between parents, but different epigenetic modifications present on the DNA cause the expression patterns to be different between parental alleles. There is no change in the gene sequence, but a heritable change in traits can be observed between parental alleles. OR Imprinting is an epigenetic phenomenon in which certain genes are expressed in a parent-of-origin-specific matter in a diploid cell. This fits our definition of epigenetics in that it is a mitotically and meiotically heritable change that affects gene function and involves no change in DNA sequence.

What is an obstacle to using mice to study epigenetic inheritance within germline cells?

Germline cells have to be obtained in order to study epigenetic inheritance. However, it is difficult to obtain these cells, as surgery must be performed in order to get eggs from the mother. After that, a few generations must pass after any initial treatment in order to observe if the changes were caused by flaws in the experimental design (i.e. surgery) or by the actual treatment being studied. If the mice are not observed for a few generations, it is difficult for researchers to conclude if any inheritance will actually be maintained. OR A pregnant mouse contains the F1 embryo, and the F2 germ line cells. Exposing a pregnant mouse to any chemicals will be transferred to the offspring and the F2 germline cells. Thus, the F3 generation must also be examined to observe whether the change is heritable and independent of any treatment. The F3 generation mouse would need to be cross fostered to confirm that the F3 isn't impacted by the mother. In order to study epigenetic inheritance in mammals, there are many obstacles due to the biological anatomy of them.

It was noted in the 1970s that elephants, despite being much bigger than humans and living similar lifespans, rarely suffered from cancer. This observation remained unexplained until the elephant genome was sequenced and it was discovered that elephants have approximately 20 copies of the important tumor suppressor gene p53. Explain why this increased copy number might lead to a lower cancer risk.

Having more copies of a tumor suppressor gene lowers cancer risk because those extra copies act as "backup" copies should one of those genes mutate and become non-functional. Humans have just one copy of p53. Should that one copy mutate and lose functionality, then that human's risk of developing cancer rises dramatically. Whereas if one copy of p53 failed in elephants, they still have 19 copies of p53 that are functional and suppressing tumor development.

What are Hox genes?

Hox is a family of genes that regulates the location of limb formation in embryo development. They have transcription factors that contain a homeobox domain that allows them to turn on genes associated with whichever part of the embryo that they are specific for.

Describe how epigenetic mechanisms can contribute to the two hits in the "two-hit" model for cancer.

If a tumor suppressor needs to be shut down to get from a normal cell to a cancerous cell, DNA methylation can serve as the first or the second hit. This is because DNA methylation can shut down the gene without mutating it.

Explain how imprinting at the mouse Igf2/H19 locus works. You may use a diagram, but make sure to clearly label your diagram and explain it.

Igf2 is located upstream of H19, which is located upstream of an enhancer region. In the paternal genome, the H19 promoter is methylated, shutting down the expression of H19 and allowing the enhancer to cause Igf2 expression. However, in the maternal genome, the H19 promoter is not methylated. This allows the insulator protein CTCF to bind, which stops the enhancing effect from travelling to Igf2. Thus, H19 is expressed from the maternal genome, while Igf2 is expressed from the paternal genome.

How are imprints regulated during development?

Imprints are first placed into the sperm and egg before fertilization. The imprints are maintained throughout the growth of the embryo, even during the epigenetic reset seen in developing embryos. The imprints are erased in primary germ cells, which then get imprinted differentially depending on the sex of the embryo.

Explain how X inactivation in female mammals leads to mosaicism as evident in calico cats.

In cats, coat color is an X-linked trait. Inactivation of different X chromosomes at different parts of the body can lead to different coat colors being present on the same organism. While the genome of the cells are the same, the X chromosome that is inactivated can vary, meaning that differences in the epigenetics between the cells leads to a mosaic.

Compare and contrast how dosage compensation is achieved by two different species.

In humans, dosage compensation is achieved by inactivation of an X chromosome in females. Male Humans have an XY genotype, so female humans inactivate one of their X chromosomes to silence it. In Drosophila, males only have one X chromosome, while females have two. The X chromosome in male drosophila is upregulated so that gene expression occurs at a rate similar to female drosophila.

What are four functions of PRC2 complexes?

In mammals, the PRC2 complexes regulate the Hox gene clusters, control cell proliferation, contribute to X chromosome inactivation, and contribute to gene imprinting. 1. regulate Hox gene clusters 2. control cell proliferation 3. contribute to X chromosome inactivation 4. contribute to gene imprinting

Describe one piece of early evidence that hinted at the existence of imprinting.

It was discovered that in certain arthropods, the genome of one of the parents was completely thrown out based on the sex of that parent. Since the actual genes present in the eliminated genome was essentially the same as the retained genome, there must have been something other than gene sequence that told the offspring which genome to throw out. OR In arthropods, there was paternal-specific genome elimination, in which the entire paternal genome was thrown away.

Describe how the X chromosome is regulated in C. elegans. Make sure to discuss both germline and soma in both sexes.

Male C. elegans have XO genotype, while hermaphroditic C. elegans have XX genotype. In Males, the somatic cells have no change to X chromosome activation, while in hermaphrodites, each X chromosome is inactivated around 50% to produce expression levels similar to the males. In all germline nuclei, the X chromosomes are inactivated. However, in the hermaphrodite, the X chromosomes are partially reactivated during oogenesis.

What is the "parental conflict" hypothesis?

Most imprinted genes control growth of the embryo. Genes imprinted by the mother usually stall growth, while genes imprinted by the father usually promote growth. Each parent has different interests for their offspring. The father does not carry the child, and so he has interest in ensuring that his child that the mother carries is the strongest embryo that gets the most nutrients. This ensures that the father in question's embryo is not beaten out by other father's embryos for nutrients. The mother carries the child, so there is no doubt that it is her child. The mother genome tries to stunt growth so that she can produce as many children as possible by allocating nutrients equally. OR "Parental Conflict Hypothesis" explains why maternal and paternal nuclei have different effects on the embryo. This is due to the parents having different interests for their offspring. Evolutionarily, paternal interest wants his embryo to get the most nutrients possible, so genes promote nutrient acquisition, so they want placenta growth around their embryo specifically. On the other hand, maternal interest wants to spread nutrients around equally to individuals and limits how much each embryo can suck up.

Explain how Muller's observations about the white gene led him to postulate the existence of a dosage compensation system in Drosophila.

Muller was studying the white gene, which is found on the X chromosome. Males only have one X chromosome, while females have two. Muller postulated that there was some regulatory mechanism that helped compensate for this difference between the number of X chromosomes in male and female flies. Since the gene for eye color is on the X chromosome, one would expect male flies to have lighter eyes than female flies if gene expression is solely based on number of genes present. He observed that deletion of the white gene on one of the chromosomes in female flies caused light red eyes to occur, while the addition of another white gene in males caused dark red eyes to occur. This meant that there was some kind of mechanism that caused differential expression of the X chromosome between male and female flies.

You have isolated three mutant alleles of your favorite gene, fav, that all lead to the decompaction of the inactive X chromosome in female mice. While the phenotype of the inactive X chromosome is identical for female mice with the three mutant alleles, you discover while none of the mutant alleles produce FAV protein, allele 3 produces an mRNA of the same size as is seen in the wildtype. Explain how it is possible that alleles 1 and 2 produce no mRNA and no protein, while allele 3 produces an mRNA similar to wildtype but no protein.

On alleles 1 and 2 there are mutations on both copies but on allele 3 there is a mutation only one copy. Thus, you can still make mRNA on allele 3 but due to post transcriptional modifications regulated by the second copy, a recognizable protein isn't formed, so a signal does not show up in the Western Blot.

Explain how PEV fits the definition of epigenetics we use in this class.

PEV in Drosphila can be observed in strains that are exposed to X-rays. An inversion occurs so that the white gene is placed closer to heterochromatin. This heterochromatin then can spread to the white gene and cause it to become inactive. The genomic sequence changes due to the inversion, but most of the time no actual genes are cut in half when the inversion occurs. Thus, the silencing of the white gene is epigenetic in nature since it is not a mutation of the white gene sequence and involves the formation of heterochromatin. OR PEV stands for position-effect variegation in which the phenotype of a gene is affected by its positioning relative to heterochromatin. Since this change is a mitotically and/or meiotically heritable change that affects gene function and involves no change in DNA sequence, this aligns with the definition of epigenetics we use in this class.

Compare and contrast the way primary and secondary deviations can lead to non-random X inactivation.

Primary deviations involve the inactivation of a specific parent's X chromosome as a preference while secondary deviations involve the preference for cells in which a specific parent's X chromosome is inactivated. Primary deviations occur at the initial choice of X inactivation, while secondary deviations occur after the X chromosome is inactivated.

Explain briefly why loss of SETDB1 might lead to decompaction of chromatin.

SETDB1 is a H3K9 methyltransferase protein. The presence of H3K9me can lead to chromatin remodeling via binding of heterochromatin forming proteins similar to Swi6. Loss of this H3K9me can cause defects in heterochromatin formation and lead to less condensed chromatin.

Why are human sex chromosome disorders more easily tolerated than other chromosomal disorders?

Sex chromosome disorders in humans are more easily tolerated as long as there are functioning copies of the X chromosomes. At least one functioning X chromosome is required for a zygote to be viable. If there are multiple X chromosomes, all except one of the X chromosomes will be deactivated to lower expression levels closer to the male. Thus, something like three or four X chromosomes is tolerated by inactivating two or three X chromosomes respectively. The other chromosomes do not have a silencing mechanism such as this, so having multiple copies of autosomes can lead to much higher expression of genes and can be fatal.

What is a TAD?

TADs (Topologically Associated Domain) are multiple regions of DNA that appear to interact with each other. There is no quantitative definition of a TAD with the method that is used to define a TAD being based on the researcher's own interpretation of a Chromosome Conformation Capture heatmap.

Describe the Agouti viable yellow (A^vy) allele, its molecular structure, and the phenotypes that can arise.

The A^vy allele is an epi-allele that needs to be turned on and off during development so that mice can get a brown hair color. In its promoter region is an IAP, which is a transposable element that causes expression. If IAP is methylated and Agouti is off, you get a brown mouse. If IAP is unmethylated and Agouti is always on, you get a fatter, yellow mice. A mottled yellow and brown mouse is also possible if IAP is methylated in different cells. OR The Agouti gene needs to turn on during mouse development for the coat to turn brown. It is on for a very short amount of time to allow for a short stripe of yellow to appear in the coat. The agouti viable yellow epiallele makes the mouse yellow. It has an IAP element, a repetitive Transposable element, upstream of the agouti gene. The IAP element has a promoter that can cause expression of Agouti. If the IAP element promoter is not regulated, Agouti will always be turned on, causing the mouse to appear yellow. There can be a mottled patchwork of yellow and brown in which some cells have methylated/inactivated the IAP or not methylated/not inactivated the IAP. There can also be a phenotype with the IAP methylated in all cells, causing the mouse to appear brown.

What is the Barr body?

The Barr body is the inactivated X chromosome in female mammals. It consists of a highly condensed X chromosome which is transcriptionally inactive.

Compare and contrast the two chromatin silencing systems used by many organisms.

The HP1a system involves H3K9 methylation and is present in areas that are generally always silenced, such as the centromere or telomeres. The polycomb gene system involves H3K27 methylation and it works throughout the genome. The polycomb system is usually associated with the developmental regulation of genes, such as the Hox genes.

What is the Hi-C method? Describe the basics of this technique in 2-3 sentences.

The Hi-C method is a type of chromatin conformation capture in which formaldehyde is used to covalently attach regions of chromosomes that are close to each other and then attached regions are ligated to each other. These sequences can then be sequenced using Next generation sequencing methods to determine what regions of DNA associate with each other. In Hi-C, biotin is added to the places where the formaldehyde-associated DNA is ligated to each other, allowing for the regions that are close to each other to be pulled down using a biotin antigen.

Briefly explain the basic steps necessary to achieve X inactivation in female mammals.

The Xist gene is transcribed from the X chromosome that is going to be silenced and coats the chromosome. This potentially recruits chromatin modifying proteins which leads to silencing. This initial silencing leads to even more silencing of the X chromosome that the Xist gene is transcribed from, leading to the formation of a Barr body.

Explain the three molecular mechanisms that can lead to an imprinting disorder like Angelman syndrome?

The disorder can have a genetic cause. If a gene is only expressed from the maternal genome, and is imprinted on the paternal genome, then a deletion or mutation of that gene in the maternal genome can lead to problems due to lack of gene product. If a uniparental disomy has occurred for this gene, such as both of the chromosomes containing the gene in question being from the father, then the suppression of the gene due to it being imprinted on the paternal chromosome can lead to disorder. An imprinting defect in the maternal genome so that the gene in question is imprinted on both the maternal and paternal genome can also cause disorder.

What is the relationship between the C. elegans dosage compensation complex and the condensin complex?

The dosage compensation complex (DCC) is active in hermaphroditic XX C. elegans, where it causes partial inactivation of both X chromosomes in somatic cells. Condensin is required for chromosome condensation in mitosis and meiosis. The core proteins of the DCC are homologous to the core proteins of the condensin complex. However, the DCC is made of more proteins than the condensin complex, which are most likely required to ensure that it only targets the X chromosome.

You have done a study assessing cytosine methylation differences between a control group of 10 people and a group of 10 people affected with seasonal allergies. Your study focuses on just one gene, sall, and each cytosine in the gene is represented by a column of circles in the plot to the right. Each row represents data from one individual. Describe the results of the experiment, and address the possible implication for a role of sall in seasonal allergies. Make sure to explain your reasoning.

The effected group has a lot more sall 5mC methylation than control. This could mean that seasonal allergies are caused by differential expression of sall. However, without knowing if in promoter or in gene body/protein expression in affected group.**** OR The affected group show a lot more black circles than the control group, indicating a higher ratio of cytosine methylation in the affected group. This would imply a correlation in people between people having seasonal allergies being methylated at this gene, but this correlation would benefit from a larger sample size, more detailed sampling, and perhaps comparing methylation patterns at other genes. Far more research would be necessary to imply causation.

What is the relationship between Prader-Willi and Angelman syndromes?

The gene that is involved in Angelman syndrome is expressed from the maternal genome, meaning that any kind of imprinting or mutation that causes the maternal gene to not be expressed causes Angelman syndrome. The gene that is involved in Prader-Willi syndrome is expressed from the paternal genome, and any kind of imprinting or mutation that causes the paternal gene to not be expressed causes Prader-Willi syndrome. The gene cluster in question for Prader-Willi and Angelman syndrome are found on human Chromosome 15.

Why were the two RNA components of the Drosophila dosage compensation complex not discovered in genetic screens?

The genetic screens only looked at mutations in a single gene per strain studied. The RNA components are redundant in function, meaning that if one of the RNA components is lost, the other component can cause normal function by itself.

What are the three requirements that any molecular mechanism needs to meet to function as the "imprint"?

The imprints must be present before fertilization, they must be resistant to epigenetic reprogramming in the embryo, and depending on the sex of the offspring, must be reprogrammed in the germline. 1. present before fertilization 2. resistant to reprogramming in the embryo 3. must be reprogrammed in the germline

How do CpG islands control gene expression? Please illustrate with an example.

The methylation of CpG islands may block the ability of a transcription factor for that gene to bind, decreasing the amount of gene expression. Methylation of the H19 promoter prevents enhancer associated elements from binding to it, decreasing expression of H19 and allowing that enhancer to bind to the IGF2 promoter to induce expression. OR a. Methylation of CpG islands may block the ability of a transcription factor for that gene to bind, decreasing the amount of gene expression. b. The Igf2 and H19 model is an example. Expression of the Igf2 gene or the H19 gene depends on whether the imprinting control region (ICR), which is a CpG island, is methylated. c. Methylating ICR results in IGF2 expression, and unmethylated ICR leads to H19 expression.

In the image to the right, you have the partial results of a Hi-C experiment. The intensity of the red color indicates interaction frequency to two locations along the chromosome. Along the X-axis, chromosomal location is plotted. The top half of the graph shows data from wildtype cells, while the bottom half depicts results that you obtained from a mutant lacking a critical chromatin component. Describe what changes you see in the chromatin interactions you see in this genomic region in the mutant. Make sure to use specific regions as examples (give approximate locations).

The wildtype shows more defined TADS from 10,000 to 50,000 than the mutant. The TAD at 10,000 in the wildtype is discarded in the mutant, possibly indicating aberrant gene expression in the mutant. In addition, the mutant depicts more interaction with chromatin regions at a greater genomic distance.

Why is the goal to define the human epigenome so much more difficult than the goal to define the human genome?

There are generally a lot more epigenetic modifications that can occur on a specific gene than alleles that the gene can have. Epigenetic modifications also change depending on what cell type is being studied, as different genes are expressed in different cell types.**** OR The human genome is easy to define because it is well-sequenced and documented, whereas one genome can result in multiple different epigenomes.

Above, you see a screenshot from the UCSC Genome browser, displaying a variety of histone modification enrichment ChIP data from two cell types. Labeled in green are data from the H1 human embryonic stem cell line, and labeled in purple are data from the K562 erythroleukemia cell line. For each histone modification, darker colors indicate stronger enrichment, with black being the strongest, and lighter grey colors indicating weaker enrichment. Using these data, determine if there is a domain regulated by the polycomb system in this region of the genome. Make sure to explain your reasoning, mark the domain on the screenshot or provide coordinates, and explain what is going on in each cell line.

There does appear to be a polycomb domain from around 54,300,000 to 54,575,000 bases. Polycomb regulatory complexes add H3K27 methylation in regions that they bind to. In the H1 human embryonic stem cell line, the amount of H3K27me is very much enriched, with a very dark color along the entirety of the region mentioned earlier. Since H3K27me is appearing in this region at a very high amount, it can be assumed that this is a polycomb domain where PRC2 has added H3K27me and PRC1 can bind to it. In the K562 erythroleukemia cell line, there are some small bits of H3K27me enrichment throughout the region, but there are a few areas with very high H3K27 acetylation between 54,375,000 and 54,425,000. This agrees with the hypothesis that a polycomb domain is also a trithorax domain. The trithorax proteins add acetylation to polycomb domains and activate them, acting as an opposing regulator for the polycomb proteins.

What is the currently most accepted model for how the Drosophila dosage compensation complex is targeted to the X chromosome?

There is a complex called the Male Specific Lethal complex that is composed of mle, msl1-3, mof, and roX RNA. The roX RNA helps guide the MSLC to the X chromosome in male flies. The MSLC can bind without the roX RNA, but only weakly. The complex first binds to regions that it has high affinity to, after which it moves to the sites with lower affinity, and then finally sites that have the lowest affinity. The complex has acetyltransferase activity, allowing the histones to be acetylated, activating the DNA.

How does DNA methylation regulate expression of H19 and IGF2?

There is an enhancer downstream of both H19 and IGF2 that induces either H19 or IGF2 transcription. When the H19 promoter is not methylated, CTCF transcription factors can bind to the promoter, allowing the enhancer to help induce H19 expression. However, when the H19 promoter is methylated, CTCF is unable to bind to the H19 promoter, and the effects of the enhancer skip the H19 promoter and induce transcription of the IGF2 gene. This is used in paternal/maternal gene imprinting. The paternal H19 promoter is methylated, while the maternal H19 promoter is not methylated. This causes H19 to only be transcribed from the maternal allele and IGF2 to only be transcribed from the paternal allele. OR a. In the paternal allele, the imprinting control region (ICR) is methylated, CTCF is unable to bind to ICR, and the effects of the enhancer skip the H19 promoter and induce transcription of IGF2. Thus, IGF2 is transcribed only from the paternal allele. b. In the maternal allele, the ICR regions is not methylated, allowing CTCF to bind to the insulator regions (ICR) preventing the enhancer from interacting with IGF2 and inducing expression of H19. Thus, H19 is transcribed only from the maternal allele.

You have obtained a new strain of Neurospora called ZYX from a collaborator. Based on the published literature, you suspect that in the ZYX strain, your favorite gene, fav, is unmethylated, which would be different from what you see in your lab's regular wildtype strain. Design an experiment to test how DNA methylation at fav compares between ZYX and your regular wildtype strain.

To compare DNA methylation between fav gene of the ZYX and the wildtype, I could do bisulfite sequencing on both strains. Bisulfite sequencing would allow me to see the pattern of methylation by showing me the location of methylated cytosines to unmethylated DNA. I'll be able the ratio of DNA methylation in both strains and judge how DNA methylation compares at fav between ZYX and the wildtype.

What are UPDs, and how do they illustrate the existence of imprinting?

UPDs are uniparental disomies. They occur when the entire diploid genome of the embryo is contributed to by a single parent. However, most UPDs result in death of the embryo. This implies that there is something that is needed from both parents for proper formation of offspring. This thing needed from both parents is the differential expression of genes based on parental sex. The genes from one parent will be imprinted a single way so that they produce the same amount of product as would when a proper embryo is formed. Thus, if only the mother genome will express a gene, the presence of two maternal genomes should cause double expression of the gene, which can be lethal. If two paternal genomes are used, then it is possible that no expression would occur for that gene, which can be lethal. UPDs are uniparental disomies where an organism has a chromosome or a section of a chromosome in which both copies only come from one parent, showing a parent-of-origin-specific effect.

You are interested in identifying genes that contribute to the formation of the ventral nerve cord in Drosophila embryos. Thus, we have used a collection of mutants and stained the neurons forming the ventral nerve cord (dark stain in the image to the right). The wildtype pattern is seen in A, and in B-D, you see the pattern produced by three different alleles of the gene vnk (vnk1 in B, vnk2 in C, vnk3 in D), which you hypothesize is involved in the formation of the ventral nerve cord. Describe the results of this experiment and determine if vnk is indeed controlling the formation of the ventral nerve cord. Make sure to explain why B-D are not identical.

Vnk does seem to control the formation of the ventral nerve cord. In A, everything appears crisp and the ventral nerve cord appears to bisect the Drosophila embryo. However, once mutations/different alleles begin to occur, the staining pattern is changed. In B, the ventral nerve cord appears to form much larger than in wild type. The bisecting line is much larger as are the segments on the top and bottom of the image. In C, the ventral nerve cord is also increased in size, but it also appears as though the bisecting line does not appear in the right side of the image. However, there is still some dark staining appearing in the region, but it is not in any specific place, meaning that the specificity of the location of expression may have been impaired by this mutation. D appears very similar to the wild-type embryo. The segments are very easily observable, and the bisecting line is similar in size to the wild-type. This could be because the mutation in D is located in a region of the vnk gene that does not lead to loss/change of function. Thus, since different mutations in the vnk gene lead to different types of ventral nerve cord structure, it is safe to say that the vnk gene somehow regulates where the ventral nerve cord forms.

What is bisulfite sequencing, and how does it work?

When you treat DNA with bisulfite, it turns unmethylated cytosine into uracil while methylated cytosine remains cytosine. After this treatment, you can determine in what percentage of your cells cytosine by comparing the ratio of Ts (unmethylated) to Cs (methylated).

Describe how DNA methylation patterns are controlled by writer and eraser class enzymes.

Writer class enzymes use S-adenosylmethionine as a methyl donor and can transfer the methyl group from SAM to cytosine residues at the 5th position in the ring. Eraser class enzymes can remove DNA methylation by oxidizing the methyl group to hydroxymethyl. This 5hmC can then be removed by base excision enzymes and an unmethylated Cytosine can be put in its place.

Describe an experimental approach that would allow you to distinguish the active from the inactive X chromosome in nuclei of female mice.

Xist staining I would perform RNA fluorescent in situ hybridization (FISH) on the X-inactive specific transcript (Xist). First, I would prepare Xist probes by labeling Xist oligonucleotides with fluorescent dye. Then, I would perform immunofluorescence on a slide of cells of interest. The stains would show Xist RNA colocalization cloud and H3K27me3 signal of the inactive X.

Above, you see the results of a 4C experiment. The arrow marks the site for which interaction are mapped below in green. Significant interactions are marked by boxes below the bar graph. Briefly explain how 4C works, and then describe the results obtained in this experiment. Make sure to identify the strongest interacting site for our site of interest and explain your reasoning.

a.4C is used to detect the interaction between an unknown genomic region with a region of interest. DNA protein complexes are crosslinked using formaldehyde. The sample is then fragmented, and the DNA is ligated and digested. b.5086kb has the highest peak indicating the highest amount of interaction between the site of interest and the rest of the genome.