Suggested study guide for exam 2.
Lecture 8. Nondisjunction and Chromosomal Rearrangements. (Chapter 6 & 7).
Know five terms used to describe the chromosomal state of a cell or organism? ∙ Aneuploidy – Addition or deletion of whole chromosomes from the expected balanced number
∙ Euploidy – Cell contains 1 or more complete sets of chromosomes (N, 2N, 3N, 4N…)
∙ Polyploidy – Cell has more than 2 sets of chromosomes (triploid, tetraploid) ∙ Monosomy – a diploid cell/organism missing a single chromosome (45, XY) ∙ Trisomy – a diploid cell/organism with an extra chromosome (47, XX)
What is nondisjunction? How can you determine whether nondisjunction occurred in meiosis I or meiosis II?
Don't forget about the age old question of What is co-immunoprecipitation?
∙ NonDisjunction – failure of chromosomes to segregate properly during cell division ∙ To see if it happened in meiosis 1 or 2 you have to look at zygote formation o Meiosis I = results in
2 trisomic zygotes from disomic gamete
2 monosomic zygotes from nullisomic gamete
o Meiosis II results in
1 trisomic zygote from a disomic gamete
1 monosomic zygote from a nullisomic gamete
2 euploid zygotes from euploid gametes
How might mitotic nondisjunction result in a mosaic individual? ∙ Sister chromatids may also undergo nondisjunction or loss during embryonic mitosis to produce mosaic individuals for aneuploid cells
∙ The clinical impact of somatic aneuploidy depends on when the aneuploid arose during development (how many cells involved and where they are located) ∙ If mosaicism arose from nondisjunction of sex chromatids early in embryonic mitosis, these individuals are termed “gynandromorphs” because they have both male and female parts (EX. RED/YELLOW BIRD)
What are five techniques used to obtain fetal cells and/or DNA for prenatal testing? Which has the lowest risk to fetus? To the mother? We also discuss several other topics like What is the humanitarian reform?
o Sampling amniotic fluid using hollow needle inserted into uterus If you want to learn more check out What is the corpus callosum?
Don't forget about the age old question of What is discrete random variable?
o Fetus 1416 weeks
o Oldest form and most prominent
2. Chorionic villus sampling
o A tiny tissue sample is taken from the villi of the chorion, that forms tissue part of the placenta
o Issues with knowing neurological development since test is performed so early in development
3. Fetal Cell Sorting
o Takes blood from the mother and tests the fetal cells in the mother’s blood for any development issues We also discuss several other topics like What is the least developed sense at birth?
4. Percutaneous umbilical cord blood sample (PUBS) RISKY o Examines blood from fetal umbilical cord to detect fetal abnormalities
Define uniparental disomy? How might it occur (2 ways)?
1. Uniparental Disomy – A condition where BOTH copies of a chromosome come from ONE parent
2. It might occur in one of 2 ways:
1. Very early embryo in cells that are initially trisomic but loose one copy to restore disomic state
2. When the gametes of both parents are aneuploid for the same chromosome 3. Results in
∙ Expression of autosomal recessives (cystic fibrosis) in individuals who should be heterozygotes
Why are autosomal aneuploidies more severe than sex chromosome aneuploidies? ∙ Autosomal aneuploidies are more severe than sex chromosomes because if >2% of the chromosome is affected it ends in fetal death Don't forget about the age old question of Do we have case morphology in english?
∙ Sex chromosome aneuploidies are not as severe as autosomal aneuploidies because of X inactivation and because the Y chromosome doesn’t have enough functional genes to make a huge difference
∙ Sex chromosome aneuploidies are less severe (and more common) than autosomal aneuploidies due to inactivation of supernumerary X's and the near absence of functional genes on the Y
How does haploid autosomal length (HAL) relate to severity of chromosomal imbalance?
∙ Haploid autosomal length imbalance < 1% will most likely result in a live birth ∙ Haploid autosomal length imbalance > 2% will most likely result in abortion ∙ 1% of haploid autosomal length is about 20cM or 20 million base pairs of DNA
What chromosome is involved in Down’s Syndrome. Is this observation consistent with the threshold of HAL?
∙ The chromosome involved in Down’s Syndrome is 21. This observation is not consistent with the threshold of HAL but the chromosome is so small that a trisomy on this chromosome doesn’t result in abortion because there isn’t enough genetic information available to cause a significant amount of damage
What process in gametogenesis might explain the association of Down’s Syndrome with advanced maternal age?
What are nine classes of chromosomal rearrangements?
1. Deletion: A SINGLE break in a chromosome can lead to a deletion 2. Tandem Duplication:
4. Pericentric Inversion:
5. Paracentric Inversion:
6. Reciprocal Translocation:
7. Robertsonian Translocation:
How does an inversion suppress genetic recombination?
How does a translocation result in reduced fertility?
Lecture 9. Mechanisms of sex determination & sexual development (chapters 6 & 7)
Explain the XO and XY models of sex determination and the taxa in which each was discovered. Which model exists in mammals? In birds?
∙ Sex Determination in Butterflies and Moths (XO)
o Determination of sex is by ratio of gene expression on X chromosome XO Autosome and 1 X chromosome = MALE
XX Autosome and 2 X chromosome = FEMALE
∙ Sex Determination in beetles, birds and humans (XY)
o Based on presence of dimorphic sexual chromosomes
XX = FEMALE (homogametic sex)
XY = MALE (heterogametic sex)
ZW = FEMALE (heterogametic sex)
ZZ = MALE (homogametic sex)
THIS IS OPPOSITE FROM MAMMALES!!
o Fruit Fly (Drosophila)
Has an XY chromosome set but sex is determined by X:Autosome ratios (XOlike)
∙ When autosome exceeds X chromosomes you get a species that is male (ratio ≤ 0.5)
o X/2A – MALE (infertile because NO Y)
o XY/2A – MALE
o XY/3A – SUPER MALE
o MALES NEED Y CHROMOSOME TO BE FERTILE
∙ When X chromosomes exceed autosomes, you get a species that is female (ratio ≥ 1.0)
o 3X/2A – SUPER FEMALE
o 3X/3A – FEMALE
o 2X/2A – FEMALE
∙ If ratios fall between 0.5 and 1 the species is intersex (ratio 0.5< X < 1.0)
o 3X/4A – INTERSEX
o 2X/3A – INTERSEX
What is “dosage compensation” and why is it biologically important? ∙ Dosage Compensation – mechanism that regulates the gene expression of sex linked genes (mechanism to normalize gene expression between the sexes o Equalizes expression of genes on the X chromosome between male and female
o This is important because it balances gene expression in both sexes which is required for normal development and function in animals
o And X chromosomes (>2000 genes) are much larger than Y chromosomes (<200 genes)
∙ Mechanisms of how this is accomplished is different in different organisms ∙ There is 2 times the level of expression of genes in X females and it must be reduced to a 1time level so the XY males have an equivalent gene expression as the females
Know how to predict phenotypic ratios in calico cats segregating at the black/orange locus on the X chromosome.
What is a Barr body? What is ectodermal dysplasia?
∙ Barr Body – inactive X chromosome (shown as a densely staining mess in a somatic nuclei of mammalian females, attached to the edge)
o Normal in females because 1 X needs to be inactivated
∙ Ectodermal Dysplasia – a group of conditions in which there is abnormal development of skin, hair, nails, teeth or sweat glands
When does X inactivation occur during human development? How does this explain the occasional observation of skewed Xchromosome inactivation in females? ∙ X inactivation occurs at the 816 cell stage of embryogenesis (this is before the embryo is implanted at 24 days post conception)
∙ It is possible to observe skewed X chromosome inactivation in females by chance meaning that all of 1 X is inactivated when it is supposed to be 50/50 ∙ Nothing specific causes this but decreased chance precursor pull will increase the chances of it happening (when there are many cells)
∙ IT is simply by chance or it can be due to growth rate changes
o Allows for expression of Xlinked recessive alleles in females
o Normal distribution expected to be 50/50 inactivation of X chromosome o All of the cells that come from the original X inactivated chromosome will be x inactivated
What is the genetic cause of “sexreversal (sxr)”?
∙ This is caused by unequal crossing over between X and Y chromosomes
What are four conditions in humans caused by nondisjunction of sex chromosomes?
∙ Occur during anaphase of meiosis 1 or 2
∙ Turner’s Syndrome
o 45 chromosomes, XO
o Short stature/webbed neck
o Normal IQ and life expectancy
∙ Klinefelter’s Syndrome
o 47 chromosomes, XXY
o Slight mental retardation
o Normal life expectance, long limbs and large hands/feet
∙ Triplo X
o 47 chromosomes, XXX
o Taller than normal
o Menstrual irregularities
o Average intelligence
o 2 Barr Bodies
∙ Jacob’s Syndrome
o 47 chromosomes, XYY
o No symptoms except for taller heights
o Aggressive personality traits
How does one distinguish between true hermaphrodite and pseudohermaphrodite? ∙ Pseudohermaphrodites
o Individuals where the gonads do not correspond to the chromosomal sex
o Have either ovaries or testes, but external genitalia are opposite fo gonad sex or are ambiguous (undifferentiated)
o Secondary sex characteristic do not match the primary ones
o Four disorders predominate
TFM (testicular feminization)
5 alpha reductase deficiency
Congenital adrenal hyperplasia
Congenital adrenal hypoplasia
∙ True Hermaphrodite
o Refers to individuals with ambiguous gonadal sex
Gonads comprise of BOTH ovaries and testicular tissue
∙ Are almost always sterile
What are the phenotypic consequences of the genetic defects in Tfm, 5 alpha reductase deficiency, congenital adrenal hyperplasia, and congenital adrenal hypoplasia?
Lecture 10. Replication and Chemical structure of DNA/RNA (chapter 8 in text)
Know the five classical experiments to identify DNA as the genetic material. 1. 1928 – Griffith demonstrated the transforming principle could be transferred between bacterial strains
o Griffiths experiment demonstrated that bacteria were suitable for studying genes
2. 1940 – Avery, MacLeod and McCarty showed that the transforming principle was DNA
o Avery and people… experiment showed that DNA, NOT PROTEIN, was the genetic material
o Took bacterial colonies of salmonella (1 rough and 1 smooth) and gave it to mice
The rough strain (nonvirulent) mouse lived
The smooth strain (virulent) mouse died from pneumonia
Smooth strain (heat killed) mouse lived (not virulent)
Mixed rough (nonvirulent) and smooth (heat killed) strain
mouse died from pneumonia
Then treated the smooth virulent bacteria recovered from last test and treated with
∙ Protease killed mouse
∙ DNase mouse lived
3. 1950 – Hershey and Chase showed that nucleic acid was the genetic material in viruses too
o35S was used to label viral proteins
o32P was used to label nucleic acids
o When viruses infected E. coli and replicated only the 32P was present in the bacterial cells
o The labeled viruses were placed in E. coli
o After 10 minutes they were blended to stop replication process
o For 35S there was no growth in bacteria but there was growth for the control
o For 35P there was growth meaning that the virus was injected into the E. coli and able to reproduce
4. 1952 – Chargaff’s Rule: purines = pyrimidines
o DNA is composed of purines (A/G) and pyrimidine (C/T) in a 1:1 ratio and concentrations of A=T and C=G
o G:C bonds are more stable than A:T bonds due to presence of 3 hydrogen bonds rather than 2
5. 1953 – Watson and Crick announced their model of DNA
o Modeled DNA and provided chemical explanation why double stranded DNA obey Chargaff’s Rule
o The double helix model suggested a mechanism for replication and the nature of the genetic code
What is Chargaff’s Rule?
∙ Chargaff’s rule states that DNA is comprised of purines (A/G) and pyrimidines (C/T) in a 1:1 ratio
∙ Also that A=T and C=G
What are the chemical differences between DNA and RNA?
∙ DNA composed of phosphate, deoxyribose sugar and 4 different nitrogen bases A/T and G/C and is DOUBLE STRANDED
∙ RNA composed of ribose sugar, and 4 nitrogen bases A/U and G/C and is SINGLE STRANDED
∙ The sugars are also different
o DNA has a deoxyribose sugar where it has an H on the 2’ carbon o RNA has a ribose sugar where it has an OH on the 2’ carbon
What does “semiconservative replication” mean?
∙ Semiconservative replication – means that when DNA is replicated: o The first generation – has 2 strands with 1 being the parent and the other being the new synthesized strand
o Second generation – replication of the first generation has 4 total strands replicated (2 being completely new synthesized strands, 2 being 1 strand from parent and 1 newly synthesized strand)
Name the functions of each component of a replisome?
∙ Helicase – unwind the double stranded DNA
∙ Polymerases – synthesize new strands of DNA (can only replicate 5’ to 3’ direction)
o DNA polymerase gamma – Mitochondrial DNA replication
o DNA polymerase Alpha – Chromosome replication (initiation by RNA primase, Okazaki fragment priming)
o DNA polymerase delta – Double strand break repair chromosome replication (elongation of 3’ to 5’ strand [lagging strand], and proof reading DNA repair)
o DNA polymerase epsilon – Chromosome replication (elongation from 5’ to 3’ strand [leading strand], and proof reading DNA repair)
o Telomerase – Telomere maintenance (binds to the distal end of DNA molecule and adds repeat RNA primer for extension to protect against chromosome shortening)
∙ Exonuclease – excise mismatched bases
Explain the need for leading/lagging strand synthesis in terms of DNA polymerase activity.
∙ Replication in prokaryotes is catalyzed by Pol I through III
∙ Replication in eukaryotes is catalyzed by Alpha, Delta and Epsilon o Polymerase can only replicate in 5’ to 3’ direction
o Alpha (primase and lagging strand)
o Delta (lagging strand)
o Epsilon (leading strand)
∙ Since the polymerase can only copy 5’ to 3’ this is why leading/lagging synthesis is needed
What is a telomere? What is telomerase and when is it expressed? ∙ Telomeres control expressions of some genes
o Early in life the telomere is longer and loops to come in contact with with certain genes
o But as the cells divide and get older, the telomere length gets shorter and is not able to interact with that same gene that results in age related changes in gene expression
∙ Telomere length declines with dividing cells as we age (a function of the number of times a cell completes the cell cycle)
o So depending on how fast these cells divide determine the length of the telomere at a certain age
o The more frequently a cell divides the more rapidly the telomere will get smaller
∙ Disorders that may involve telomerase
o HutchinsonGilford Syndrome – LMNA (laminin A, a scaffolding protein, when absent fo defective, causes increase cell death and progeria fatal by age 12
o Werner’s Syndrome – (WNR helicase is important in DNA repair and telomere maintenance) develops in teens and young adults and causes death by age 50 due to atherosclerosis or cancers
Why don’t bacterial chromosomes have telomeres? ∙ Because they are circular instead they have a single ori
Lecture 11. Transcription. (chapter 9 in text).
Know the organizational features of a typical eukaryotic gene.
GENOMIC DNA (gDNA) – IN NUCLEUS
∙ CAAT Box
o Located about 75 base pairs from the initiation codon (found upstream from start codon)
o Is the 1st promotor sequence and basically says hey get ready dude, we are about to start transcription (1st alarm for an 8am)
∙ TATAA Box
o Located about 30 base pairs away from the initiation codon (found upstream from start codon)
o The 2nd promotor sequence saying HEYYYYY you really need to get up now transcription is coming SOON (2nd alarm for 8am)
∙ Cap Site
o This is located on the 5’ end
o It is the transcription start site
∙ Kozak Sequence
o This is your initiation sequence in a eukaryotic gene (REMEMBER IT IS DIFFERENT FOR BACTERIA!!)
∙ Termination Sequence
o TAA, TGA, or TAG
o Stop codon that cuts off the reading for transcription
∙ Polyadenylation Signal AATAAA
o Located downstream from the stop codon
o This is the site for the addition of the poly A tail when transcription occurs
mRNA PRECURSOR (the sequence right after transcription) – IN NUCLEUS ∙ Cap
o Located at the beginning of the 5’ end of the sequence
∙ Contains both exons and introns
o Exons are Expressed
o Introns are CUT OUT
∙ Poly A tail
o Located at the 3’ end of the precursor RNA
MATURE mRNA (FINAL PRODUCT) – IN CYTOPLASM
o Located on the 5’ end, just stays throughout the process
o This is the sequence that matters!!
∙ Poly A tail
∙ The functional sequences in gDNA that are missing in mRNA
o CCAAT Box
o TATA Box
So promotor regions
∙ Sequences present in the mRNA but absent in gDNA
o CAP on 5’ end
o Poly A tail on 3’ end
o ONLY THING SIMILAR BETWEEN mRNA and original sequence is th EXONS
(refer to photo on slide 3 of lecture 11)
What are “The Central Dogma” and “The New Central Dogma”? ∙ The Central Dogma
o DNA RNA Protein
DNA dependent RNA polymerase transcribes RNA from DNA template
∙ The New Central Dogma
o DNA ⇔ RNA Functional RNAs
o DNA ⇔ RNA Protein
RNA dependent DNA polymerase (in Retroviruses, aka RNA
viruses) copies RNA into DNA as part of the process to integrate
its RNA gemone into the DNA of the host
How does gene regulation differ between pro and eukaryotic genes? ∙ Prokaryotes
o Operon model of organization and regulation (all genes are organized into functionally related groups of genes = operon)
Multiple genes in same metabolic pathway are under a
common control system
Polycistronic mRNA (meaning 1 mRNA = several different
Genes contain NO intervening sequences (NO INTRONS)
o Each gene has its own regulatory system
Almost all translatable genes contain INTRONS
Very FEW POLYCISTRONIC mRNA in eukaryotes
Explain the roles of functional elements of the Lac Operon. (WHAT DO WE NEED TO KNOW?
∙ Found in Bacteria
o The lac is referring to lactose and the lac operon is only expressed when lactose is present and glucose is absent
o The lac operon is responsible for breaking down lactose into sugars that can be used for metabolism through transcription
o If no lactose is present the lac repressor attaches to the operon and doesn’t allow transcription
o Allolactose is present when lactose is present and it binds to the repressor preventing it from attaching to the operator region on the gene allowing transcription of lactose which is metabolism of lactose into smaller sugars!
o Lac repressor acts as a lactose sensor
Normally it blocks transcription of the operon when glucose is
around, but stops acting as a repressor when lactose is present so
you can metabolize it (sensed through allolactose)
∙ Here is a short summary of this huge mess above!
o Lac operon genes encode enzymes for uptake and metabolism of lactose o The lac operon is ONLY EXPRESSED when
Lactose is available
Glucose is NOT available
o The lac repressor acts as a lactose sensor
o CAP is a glucose sensor (if enough glucose is present then the
transcription of lactose will NOT happen)
∙ Structure of Lac Operon
o Promotor region – is the binding site for RNA polymerase to begin transcription of the gene
o The operator – part where the lac repressor protein will bind
(overlapping the promoter region) resulting in RNA polymerase not being able to bind and start transcription (normally occurs when glucose is around)
o CAP binding site – is the catabolite activator protein this helps RNA polymerase bind to the promoter region and promotes transcription
Name four sequence motifs associated with expression and processing of eukaryotic genes
∙ CAT box (CCAAT)
o Located about 75 base pairs upstream (toward the 5’ end ) of the start codon
o It signals the binding site for the RNA transcription factor (TF)
∙ TATA box (TATAATA)
o Located about 30 base pairs upstream (toward 5’ end ) from the initiation codon
∙ Kozak Sequence (ACC ATGG)
o Tells you the initiation codon for eukaryotic genes!
o Basically, like hey dude this is where you start transcription!
∙ Polyadenylation Signal (AATAAA)
What posttranscriptional modifications are commonly made to precursor mRNA in eukaryotes?
∙ Capping of 5’ End
o Is a methyl G cap that is added to the 5’ end of mRNA during
Regulates export of transcripts from the nucleus
Prevents degradation of transcripts by 5’ endonucleases
Promotes translation by directing correct loading of mRNA onto ribosome
Promotes excision of 5’ proximal intron
∙ Addition of poly A tail to 3’ End
o About 250 A’s are added to the end of the 3’ end of cleaved mRNA transcript
o Promotes export of mature RNA from the nucleus for translation in the cytoplasm and inhibits mRNA degradation
∙ Removal of Introns
o Almost all introns of eukaryotic cells are enzymatically excised by spliceosome complex
o Group I – are selfsplicing
o Group II – are selfsplicing
o Group III – found in archaebacteria and use spliceosome like system of proteins
∙ Editing of RNA (rarely)
o Can change the sense codon to a stop codon giving a nonsense mediated decay
o Can also “improve the specificity of antibodies after clonal selection
Know how to read a ‘sequence logo” representation of a consensus sequence. ∙ Sequence logo is this…
Refer to sequcence logo in slides (rainbow letter diagram looking thing! o Where the larger the letter the more likely it is to appear in that position o This is for Kozak sequence that is ACCAUGG
o So this is saying the A is the most common for the first letter, but it can also be G
o The 2 lower case Cs are iffy, the 1st C can be any base pair, but the second is most likely going to be C or G, with C being more common
o AUG is guaranteed to be the sequence (notice no other letters at the bottom)
What are seven differences between pro and eukaryotic mRNAs?
Differences in Prokaryotic and Eukaryotic mRNA
Methyl Guanine Cap on 5’ end
Start codon identifier
Poly A tail on 3’ end
Start codon amino acid (AUG)
mRNA half life in
∙ ShineDelgarno sequence is the start codon identifier for bacteria, but the sequence AGGAGG identifies the initiation codon like Kozaks BUT the actual start codon is located about 8 base pairs UPSTREAM of the AUG start codon!
Lecture 12. Translation (Chapter 9 in text).
Properties of genetic code.
Types and activities of “functional” RNAs
How do the four types of single nucleotide mutations alter translation? ∙ Synonymous mutation – change to another codon that specifies the same amino acid
∙ Missense mutation – changed from one sense codon to a codon that specifies a different amino acid
∙ Nonsense mutation – change from sense codon to stop codon ∙ Frameshift Mutation – mutation is a small deletion or addition of 1 or 2 bases that changes the reading frame of the mRNA, resulting in downstream premature stop codon
Know the chemical reaction that forms a peptide bond.
∙ The formation of peptide bond is through a condensation reaction (loss of 1 mole of water)
∙ The carboxyl group of one amino acid reacts with the amino group of another and through release of water form the peptide bond
∙ Is catalyzed by enzyme (peptidyl transferase)
∙ And requires ATP energy
Know the processes of translation initiation, elongation, and termination as they relate to the ribosome. – ENGAGED with mRNA
∙ The small subunit is where the mRNA attaches
∙ The larger subunit is where the tRNA attaches
∙ Assembly of the Functional Ribosome in 2 processes
o The small ribosome subunit (30S) engages mRNA via recognition of the initiation sequence (ShineDelgarno for bacteria and Kozak for
o Then the 30S mRNA complex associates with the larger subunit (50S) to make up the 70S ribosome
o Acts on the 30S ribosomal subunit
o The only time when an amino acid occupies the P site on the ribosome before the A site (is the Met amino acid – start codon)
o Growing polypeptide chains
o A site –
o P site – Peptide bond formation on the ribosome is followed by translocation of peptidyl tRNA from A site to P site
So when there is a protein in A site and P site, a peptide bond is formed to make the polypeptide chain
o E site – is where the tRNA will exit the ribosome without its protein attached to it because it is now part of the sequence
o When the ribosome encounters one of 3 stop codons (TAA, TAG, TGA) at the A site
o Polypeptide releasing factor engage the stop codon and causes the release of protein and the disassembly of mRNA and ribosome complex
What are the four levels of protein structure?
∙ Primary Structure – sequence of the amino acids
∙ Secondary Structure – backbone hydrogen bond interactions ∙ Tertiary Structure – folded in certain conformations determined by side chain interactions and SS disulfide interactions of cystine
∙ Quaternary Structure – multiple subunits
What is the “signal hypothesis”?
∙ Provides a mechanism by which secretory proteins move through endoplasmic reticulum are sorted in the Golgi and then stored or released outside the cell ∙
What are “chaperonins”.
∙ Chaperones – are post transcriptional modifications of protein that induces folding (the fold is not predicted by their amino acid sequence)
∙ Protein that assist in the 3D foldin of other proteins without becoming part of those proteins
∙ Remember: if you disrupt a fully folded protein structure it would result in a nonfunctional protein because it can reform by base pairing but not regain the original folded structure
Lecture 13. Epigenetics (chapter 11.8 in text).
Define “epigenetics” and “epigenome”.
∙ Epigenetics – the process by which inherited changes in gene expression do NOT involve changes in the DNA sequence of the genome
o The idea that every organism has the same DNA sequences but not all genes are expressed in any one cell
o This is the process by which 250 different cell types expresses a subset of genes in the genome needed to perform a certain function
o Helps determine whether genes are turned on or off
∙ Epigenome – The multitude of chemical compounds that tell the genome what to do
o So humans have a bunch of DNA right?
o That encodes for proteins that carry out a variety of functions in a cell… o So how do they know what to do?
The epigenome is the answer!
∙ It is made up of chemical compounds and proteins that can
attach to DNA and direct actions by turning genes on or off
controlling the production fo proteins in particular cells
Who are Lamarck and Lysenko and what is their historical significance in genetics? ∙ JeanBaptiste Lamarck –
o Theory of inheritance of acquired characteristics
o Hypothesized that an organism can pass on characteristics it has acquired during its lifetime to its offspring
∙ Trofim Lysenko
o A Russian geneticist who practiced Lamarckian agriculture
o Convinced Lenin and Stalin that it should become bases for Soviet agriculture (that lead to agricultural collapse)
o He was the person responsible for implementing the theory of acquired inheritance in Russian agriculture in the 192030s
What is neoLamarkianism?
∙ It is the modern use of Lamarckism (1920s)
∙ During this time scientist accepted the theory of evolution but still felt that characteristics acquired during ones lifetime had an effect on inheritance ∙ They studied environmental stresses on how it affected evolution ∙ The movement believed that organisms could go change their own evolution instead of leaving it to random variations
∙ Ties to epigenetic modification studies
o This is because observations that certain environmental conditions can alter the epigenetic marks on the DNA to change the gene expression patterns and that those changes will persist for 23 years (Hunger Winter)
Relate the G0 phase of the cell cycle to epigenetic marking of DNA. ∙
At what stage in development is the human genome marked epigenetically? ∙ During Fetal Development
How do acetylation and methylation of histones and/or DNA relate to gene expression?
∙ Readers, Writers and Erasers in Epigenetics
Methyltransferases – add methyl group (suppress gene
Acetyltransferases – adds acetyl group (increases gene
Deacetylases – remove acetyl groups (suppressing gene
Demethylases – removing methyl groups (increasing gene
Proteins that interpret histone modifications
∙ DNA Methylation (CpG methylation)
o Methyl group is added to the #5 carbon of cytosine located 5’ to guanine o Addition of methyl groups catalyzed by methyltransferase enzyme o Occurs on cytosine bases next to guanine (why it is called CpG – Cytosine phosphate Guanine)
o This suppresses accessibility to the DNA and gene expression o TURNS GENE OFF
∙ DNA Acetylation
o Would add acetyl groups to DNA
o Increasing the accessibility to DNA and gene expression
∙ Methylation of Histones
o Have methyl groups added my methyltransferases that densely packs the DNA where it cannot be accessed for transcription
∙ Acetylation of Histones
o Loosen the histone binding and make DNA accessible to promoter regions that allow gene expression and transcription
∙ GENE ON – when DNA is unmethylated and histones are acetylated ∙ GENE OFF – when DNA and Histones are methylated
What is the significance of “Hongerwinter” to the study of epigenetics? ∙ The Danish “Hunger Winter” of 19441945
∙ Demonstrated the important role of epigenetics in human health ∙ Changes in epigenome caused by environmental stresses are “transgenerational” o Children of pregnant women exposed to famine during Hongervinter were more suseptibel to diabetes, obesity, cardiovascular disease, coeliac disease, schizophrenia ect.
o Children of the women who were pregnant during the famine were also smaller, but when they grew up and had children, their children were also smaller and more susceptible to health problems
∙ This happened because during the development of the firstgeneration child, the embryo was undergoing meiosis so epigenetic modifications of the mother affected the meiotic division of the embryo (females develop their eggs at this stage)
∙ Poor fetal nutrition programs the fetus to early onset adult disease The exam will consist of 40 multiple choice questions.