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Lecture

Lecture 21

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Department
Human Biology
Course
HMB265H1
Professor
Chris Bovaird
Semester
Fall

Description
Lecture 21 -epigenetics above the genome -genotype=phenotype these twins are identical but if you look closely, there are still subtle differences between them even with monozygotic twins, differences can become more apparent as they age especially in terms of health vy -but, these mice are genetically identical All have genotype =A /a differences due to epigenetic changes but a fat mice and yellow coat and these have wild grey coat colour but same DNA/same genes allele combinations -epigenetics ”heritable modifications in gene function not due to changes in the base sequence of DNA” such that one allele is expressed and other isn’t -molecular mechanisms: 1) DNA methylation at CpG (dinucleotides) islands-repression of gene expression  -2) covalent modification of histones- e.g. acetylation, methylation, ubiquitination, etc these modifications can lead to repression or activation of transcription -3) non-covalent modification of histones-chromatin remodeling, histone variants due with chromatin remodeling complexes -4) non-coding RNAs- transcriptional silencing -epigenetic changes cause coat colour differences as well as other differencesthes changes can be influcneced by the environment -agouti gene is a pleiotropic gene one of these traits is coat colour, viability and obesity thus some mice being obese or skinny fat mice are prone to diabetes and cancer these changes can be influenced by the environment vy -regulation of gene expression in A /a mice in agouti viable yellow mice, in mice that have this viable yellow allele, there is a retrotransposon inserted upstream of the wild type promoter this retrotransposon inserted in it a cryptic promoter this promoter activates expression of the agouti gene ectopically turned on all the time constant expression of this A allele resulting in increases in size, yellow coat colour, diabetes and risk for cancer in these mice with yellow coat colour, this retrotransposon is unmethylated in mice that have this viable yellow allele but are brown and skinny their retrotransposon is methylated, and methylation of this retrotransposon results in gene silencing so the agouti gene is under its normal developmental controls normal promoter of this agouti allele is turned on normal developmental expression -mice yellow in colour=bigger in size etc have unmethylated retrotransposon and influences on this agouti gene and misregulates it so it is always turned on but methylated DNA and the agouti wild type gene is under its normal developmental control leading to wild type phenotype -mother’s diet during pregnancy can cause epigenetic changes in offspring methylation of this retrotransposon results in silencing of it take viable yellow mothers a diet with methyl donors, there is a switch almost from yellow to the brown/grey agouti coat colour not 100% but a lot of these yellow mice give rise to offspring that are wild type a change in phenotype a result of feeding these foods that are high in methyl donors -if we also have these genes present in humans, a mother’s diet during pregnancy for overall phenotype of the offspring these epigenetic changes can also be made in humans based on diet -epigenetics is required for normal development DNA and genome as stored information the epigenome is the software etc and genome hardware this software tells the computer when to be turned on and expressed and how much governs or controls the computer this is the epigenome controls the expression of genes so it regulates development activation and silencing mechanisms that control gene expression this epigenome is this organized information all born with stem cells and all identical to each other and contain the same genes so exact same DNA through epigenetics, all born with 25,000 genes in these stem cells and epigenetic mechanisms they all give rise to the final patterning of your cells and each cell contains its own epigenetic pattern -X –chromosome inactivation random X-chromosome inactivation in females early during (or individuals with more than one X chromosome) development (after 100 cell stage)inactivation is hereditary through cell division i.e. clonal (except reactivated in germ cells) reason-dosage compensation (2 copies of X chromosome, twice as many X linked genes as male counterpart so twice the amount of X linked gene expression so equalize gene expression from X chromosome from males and females, inactivate most of the X chromosome) this is hereditary, inactivated in one cell this inactivation remains the same in all progeny cells reactivated in germ cells when an organism is forming after 100 cell stage, in a female with 2 X chromosomes, 1 X chromosome inactivated (random), so one cell you might have paternal X chromosome and in another maternal progeny would have the same X chromosome inactivated but when germ cells form, in final mature gametes, this X chromosome is activated again since we know the gametes are haploid they require an active X chromosome -X-inactivation produces coat colour in calico cat females are mosaics, clusters of cells and some cells with one X chromosome and other cells with other X chromosome inactivated all these different clones and progenitors pathces of orange, black and white fur this is always seen in normal females, normal males are all white, orange or all black zygote, one X chromosome from mother and another from father, then random X-chromosome inactivation all the progenitor cells have same X chromosome inactivated these inactivated chromosomes are called Barr bodies since they are darkly stained and they are condensed heterochromatin structures some of the cells that have 1 X chromosome inactivated and coat colour genes are on X chromosome and are x linked if one X chromosome is inactivated, so that means the other one will be seen then all the progenitors for example will have an orange phenotype the other x chromosome give you a black phenotype white colour is another gene on the X chromosome that is responsible for coat colour and it is epistatic to orange and black, when this gene remains active it gives you the white coat colour and masks effects of orange and black allelesmales just produce the phenotype of the X –linked allele they inherited from their mother -mechanism of X-inactivation coating of chromosome with XIST RNA hypoacetylation of a Lys in two histones methylation of inactivated X region on the X chromosome called the XIC (x inactivation center) and this region contains a gene called XIST this gene is transcribed into this XIST RNA this mRNA isn’t translated to proteins it remains in nucleus travels in a bidirectional manner and coats and condenses most of the X chromosome since some genes still remain active creates heterochromatin or silenced chromatin XISt mRNA also binds or recruit chromatin remodeling complexes and these complexes also participate in the silencing of the X chromosome all these mechanisms contribute to transcriptional silencing (hypoacetylation, methylation) only occurs in 1 X chromosome this XIC is present in both X chromosomes in females but not sure why one is inactivated and not the other -parental imprinting (example of epigenetics)genes rendered inactive depending upon parental source (made inactive depending on whether inherited from mother or father)e.g. only paternal mouse Igf2 gene expressed (maternally imprinted) e.g. only maternal mouse H19 gene expressed (paternally imprinted) first two imprinted genes discovered in mice, deletion in Igf2 gene resulted in a skinny mice only if this was inherited from the paternal chromosome, if deletion inherited from mother, mice remained normal, only paternal Igf2 gene (insulin like growth factor 2) is expressed H19 (noncoding RNA
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