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246 Cards in this Set
- Front
- Back
what is the only reason we have diversity in living things? |
mutations |
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mutation rates are ____ in all genomes contributing slowly to inherited diversity |
low |
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mutations are random and usually deleterious which impaires what? |
the function of the gene or gene product |
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mutations generate inherited genetic diversity that fuels ______ |
evolutionary change |
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mutation rate |
defined and studied differently in sexually reproducing diploids. Is measured as the number of times mutation alters a particular gene per replication cycle or generation |
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_____ Mutations are easier to detect than ______ mutations since a single copy will manifest in the phenotype |
Dominant mutations are easier to detect than recessive mutations since a single copy will manifest in the phenotype
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What factors affect mutation rates? |
genome size and life cycle |
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More complex mutations have a higher mutation rate? T/f |
True this coordinates with the genome size |
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Do larger or smaller genomes have higher mutation rates? |
larger |
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hotspots of mutation |
genes that have higher mutation rates. Normally the larger the gene is, the more likely a mutation event will happen within that gene. |
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A gene has a mutation rate of 44.5 B gene has a mutation rate of 3.3 which gene is bigger? |
agouti because it has a higher mutation rate. less mutations - the smaller the gene is
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Whole genome sequencing |
allows us to assess all mutation rates throughout the genome. previously we could only detect mutations that changed the phenotype |
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There is a genetic component to how efficiently we compare things... this could be why some families are more prone to disease |
just info |
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point mutations (aka localized mutations) |
occur at a specific identifiable position in a gene. Tend to affect one gene in the dna |
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mutation definition |
any change to the dna sequence these mutations have varied consequences depending on the type of sequence change and location of the affected part of the gene |
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coding mutations |
affect coding sequence of the protein and amino acids put into the protein |
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regulatory mutations |
changes things that will appropriately express the gene. The promoter, ribosome binding sites, etc.. (if these were mutated it would affect how well the gene was expressed) |
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Base pair substitution mutations |
the replacement of one nucleotide base pair by another |
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transition mutations |
one purine replaces another purine (a-g) or one pyrimidine replaces another pyrimidine (t-c) |
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tranversion mutations |
a pyrimidine is replaced by a purine or vice versa |
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silent mutation |
a base pair change that does not alter the resulting amino acid due to the redundancy of the genetic code |
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missense mutation |
a base-pair change that results in an amino acid change in the protein |
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nonsense mutation |
a base pair change that creates a stop codon in place of a codon specifying an amino acid |
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what are the 4 coding mutations |
silent missense nonsense frameshift |
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frameshift mutations |
insertion or deletion of one or more base pairs leads to the addition or deletion of mRNA nucleotides altering the reading frame of the message. (the wrong amino acid sequence is produced starting at the point of mutation; premature stop codons may also be produced) |
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Regulatory mutations |
alter the amount (but not the amino acid sequence) of protein product produced these affect promoters, introns, and regions coding for 5' UTR and 3' UTR segments |
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3 types of regulatory mutations |
promoter splicing cryptic splice site |
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Promoter mutation |
mutations that alter consensus sequence of nucleotides of promoters. These interfere with efficient transcription initiation. Some promoter mutations cause mild to moderate reductions in transcriptiojn levels whereas others may abolish transcription |
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promoter mutations affect_____ |
when and where the gene is expressed. The structure isn't changed, just when it will be present |
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splicing mutation |
mutations that alter the specific sequences at either end of the intron (the ones that begin and end splicing) these result in splicing errors and the production of mutant proteins due to the retention of intron sequences within mRNA. you can get reduced gene product (hypomorph). If there is no gene product produced it is an A-morph or null |
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cryptic splice sites |
if a new splice site is created in the wrong position that confuses the splicing and they'll splice the intron at the wrong spot. some base pair substitution mutations produce new splice sites that replace or compete with authentic splice sites during mRNA processing (sometimes results in a frameshift mutation) |
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forward mutation |
converts a wild type allele to a mutant allele (breaks the gene) |
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reverse mutation or reversion |
converts mutant alleles to wild type or near the wild type allele (repair the gene) |
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True reversion |
wild type dna sequence is restored by a second mutation within the same codon. |
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Intragenic reversion |
occurs through mutation elsewhere in the same gene ex: with a mutant wild type we remove 2 nucleotides causing a frameshift, we can correct this by insertint two pars of mutations into a different area to restore it to the correct reading frame |
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second site reversions |
repairs the phenotype caused by the first mutation. occurs by mutation in a different gene that compensates for the original mutation, restoring the organism to wild type. aka supressor mutations (bc they suppress the mutant phenotype) |
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spontaneous mutations |
arise in cells without exposure to agents capable of inducing mutations. usually results in errors in dna replication. DNA POL III doesn't repair ALL of the mistakes it makes |
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Where is it that DNA replication errors commonly occur? |
in repeating sequences. they are hot spots of mutation |
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Strand splippage |
As dna polymerase III is copying the dna with repeat sequences there can be strand slippage where it stutters on this repeat sequence and the repeat sequence can create a hairpin and we pull polymerase back a little bit by doing that so it now copies that repetitive sequence again. This will make the strand much longer where instead of 6 copies of the original repeat there is now 11 on the example on the slide. |
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Trinucleotide repeat disorders |
causes hereditary diseases in humans and other organisms. wild type alleles normally have a variable number of dna trinucleotide repeats. Increase in the number of repeats beyond a certain threshold causes disorders. Ex: Most well known example is huntington disease = cag repeat sequence inside the gene. Normally we only have 10-34 of these cag repeats and that repeat can grow to be 40-200. once we cross the 40 threshold the protein doesn’t function properly anymore. This results in the uncontrolled movement associated with huntingtons disease |
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Tautomers |
spontaneous nucleotide base changes. Every once in a while a nucleotide makes an inappropriate spontaneous chemical structural shift and it is capable of making an inappropriate base pair. The shift is short lived and it will shift back at some point and behave properly. Will not affect anything unless it has a tautomer during dna replication |
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what can lead to base pair mismatches and incorporation of incorrect bases during replication? |
tautomeric shifts |
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what are the most common cause of mutation? |
tautomeric shifts |
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Depurination |
the loss of a purine from a nucleotide by breaking the covalent bond linking the nucleotide base to the sugar. this happens to purines. When we remove the nucleotide base it has a normal phosphate and normal ribose sugar, but is missing its nitrogenous base. The phosphate and sugar are there taking up space, but there is no base. |
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when a depurination event occurs and it is replicated it leaves a _______ site |
apurinic - which is empty with just a phosphate and sugar. The dna tends to just put adenine in the site across from it causing a mutation |
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Deamination
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the loss of an amino group from a nucleotide.
For example, when cytosine is deaminated, an oxygen atom usually takes its place, converting the cytosine into uracil DNA mismatch repair removes the uracil from the DNA and replaces it with cytosine,restoring the wild-type sequence |
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when methylated cytosine is deaminated a thymine base is produced which is now base paired with guanine. What mechanism will repair this and what happens if repair doesn't happen? |
mismatch repair will repair the system and either restore the wild type g-c or replace it with a mutant A-T. If repair doesn't occur replcation will produce two sister chromatids. One with the mutant A-T pair and one with the Wild Type G-C pair |
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Induced mutations |
produced by interactions between DNA and physical, chemical, or biological agents that generate mutations. |
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mutagens |
agents that cause dna damage leading to mutations |
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List the six types of chemical mutagens note: the first five are ways we chemically modify bases inappropriately with the wrong nucleotide. the 6th is a little different |
nucleotide base analogs deaminating agents aklylating agents oxidizing agents hydroxylating agents intercalating agents in the quiz he will give you an example of something and you determine which agent it is. see slides for examples to study |
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Base analog |
a chemical compound with a structure similar to a DNA nucleotide base. The dna is unable to tell them apart from a normal nucleotide base. If they’re present they will get misused. DNA polymerase can’t distinguish these chemicals from the normal nucleotide bases. |
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deaminating agents |
removes amino groups from nucleotide bases Some nucleotides have amino groups in their nitrogenous base, this removes that and changes the structure of nitrogenous base and effect how it base pairs with the other nucleotides. After two rounds of replication we will have a double helix with a point mutation |
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Alkylating agents |
added bulky side groups (called bulky adducts) such as methyl and ethyl groups to nitrogenous bases. bulky adducts can interfere with dna base pairing and may distort the dna double helix |
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Hydroxylating agents |
add hydroxyl groups to a recipient compound. Ex: hydroxylamine adds a hydroxyl group to cytosine creating hydroxylaminocytosine. this can mispair with adenine creating a C-G to T-A transition mutation |
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Oxidative reactions |
oxidation is a chemical process of electron transfer by addition of an O2 atom or removal of an H2 atom. Oxidizing agents such as bleach and hydrogen peroxide cause oxidative reactions that can lead to mutations |
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Intercalating agents |
Some molecules are able to fit between DNA base pairs (intercalation) These agents distort the DNA duplex Distortionleads to DNA nicking that is not efficiently repaired, resulting in frameshift mutations due to added or lost nucleotides. Proflavin (for example) can shimmy in between base pairs, it jams itself in and affects the overall structure of the double helix. It causes a slight bulge. This distortion leads to dna nicking (breaking the phosphodiesterbackbone of the dna double helix) and those DNA nickings are inefficiently repaired. This leads to some added or lost nucleotides. This leads to a frameshift mutation. |
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Radiation induced dna damage |
includes photoproducts, pyrimidine dimers and thymine dimers |
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what is the most common radiation |
UV |
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Photoproducts |
are aberrant structures with additional bonds with additional nucleotides caused by UV radiation. They are the results of light exposure in our DNA |
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Pyrimidine dimers |
produced by the formation of one or two additional covalent bonds between adjacent pyrimidine nucleotides |
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thymine dimers |
one common photoproduct which is formed btwn the 5 and 6 carbons of adjacent thymines. also formed between the 6-4 carbons. Just know there are two ways to get thymine dimers, but its the same process for both. Uv light makes these things called pyrimidine dimers which tend to occur just between these thyamine things so they are thyamine dimers |
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Explain how UV light causes thymine dimers |
The uv waves carry some energy and when it hits our dna it hits some electrons in the nucleotide molecules and excites them. Excited electrons on the two neighboring thyamines create a new covalent bonds between the nitrogenous bases (thyamine dimers) see notes for drawing (4-20) |
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what happens if pyrimidine dimers aren't repaired |
they can cause disruption in replication because the complementary adenines of the new DNA strand cannot form hydrogen bonds with the thymines (bc the thymines are stuck together) Replication stalls at this point and then reinitiates at an adjacent rna primer site leaving a gap |
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When gaps are left by stalled DNA synthesis they are filled with___________ which is carried out by specializing DNA polymerases that can replicate accross gaps. These dna polymerases lack proofreading activity so they are error prone |
translesion dna synthesis |
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x-rays and gamma rays are other types of radiation. what kind of damage do these cause and what are the repercussions |
they are high energy and break existing covalent bonds. these cause damage. The most serious damage caused is double stranded or single stranded breaks in DNA. these breaks block dna replciation |
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UV light energy level |
low
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The Ames test
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§The Ames test mimics what happens when animals are exposed to chemicals, and tests the chemicals and their breakdown products for mutagenic potential Bacteria are exposed to experimental compounds in the presence of mammalian liver enzymes In animals, ingested chemicals are transported to the liver, where they are broken down by enzymes Biologist use the ames test to determine if a chemical can cause mutation |
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example of Ames test procedure |
The bacteria (his- bacteria) carries mutations in them. They interfered with the organisms ability to make the amino acid histidine Can’t make its own histidine, can only grow if we give it histidine. Cannot grow on the his- media. If they expose this to a mutagen that causes mutations we will get lucky and get a reversion mutation so the bacteria grows on the histidine-media where it could not grow before |
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If trying to evaluate an Ames test the chemical is mutagenic if...... |
there is More bacteria growing on his- plate = mutagen If it is a mutagen = his – strain (for example) will grow on a his- plate (will grow without us providing it histidine) The more mutagenic something is the more and more colonies you should get. |
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Which type of mutation converts a nucleotide to an alternative structure with the same composition but a slightly different placement of rare, less stable hydrogen bonds that cause base-pair mismatch? Depurination Deamination tautomeric shift transition transversion |
tautomeric shift |
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Fragile X is an example of a trinucleotide repeat disorder caused by a mutation that increases the length of the long arm of the X chromosome and ultimately decreases mRNA stability. Which type of mutation causes Fragile X? Promoter Polyadenylation splice site triplet-repeat expansion frameshift |
triplet repeat expansion |
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________________ are nucleotide structures that have the same composition and general arrangement but a slight difference in hydrogen bonding. |
tautomers |
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What kind of DNA damage is caused by UV (low energy) radiation? pyrimidine dimers DNA double strand breaks DNA single strand breaks trinucleotide repeat expansion silent mutations |
pyrimidine dimers |
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What kind of DNA damage is caused by ionizing (high energy) radiation? pyrimidine dimers DNA double strand breaks trinucleotide repeat expansion silent mutations |
DNA double strand breaks |
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Which type of damage is likely to be the most lethal to cells? DNA double strand breaks Thymine dimers |
DNA double strand breaks |
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You have conducted an Ames test on a given compound. Which of the following would be classified as a positive result on the Ames test? his- strain grows on an his- plate. his- strain grows on an his+ plate. his+ strain grows on an his- plate. his+ strain grows on an his+ plate. his+ strain grows on either an his- or an his+ plate. |
his- strain grows on an his- plate. |
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Which type of damage is likely to be the most lethal to cells? |
DNA double strand breaks |
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Which of these statements best describes gene expression in a damaged cell? In a damaged cell, p53 is low and BAX transcription is inactive, so Bcl-2 represses apoptosis. In a damaged cell, p53 is high and BAX transcription is active, so Bcl-2 is repressed and apoptosis is induced. In a damaged cell, p53 is low and BAX transcription is active, so Bcl-2 represses apoptosis. In a damaged cell, p53 is low and BAX transcription is active, so Bcl-2 is repressed and apoptosis is induced. In a damaged cell, p53 is high and BAX transcription is inactive, so Bcl-2 is activated and apoptosis is induced. |
In a damaged cell, p53 is high and BAX transcription is active, so Bcl-2 is repressed and apoptosis is induced. |
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Which method of double strand break repair results in fewer mutation events? NHEJ SDSA Neither is error prone |
SDSA |
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When during the cell cycle is SDSA double strand break repair possible? G0 G1 S G2 M |
G2 |
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When during the cell cycle is NHEJ double strand break repair possible? G0 G1 S G2 M |
All of the above |
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Why do you think aneuploidy (having an extra, or missing, copy of a chromosome) is inviable for most cases in humans? Answer |
gene dosage |
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Trisomic individuals tend to have decreased fertility. This is due to errors in which process? DNA Repair Mitosis Meiosis DNA Replication Transcription Translation |
Meiosis |
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How many of the four gametes produced during mitosis will be abnormal if nondisjunction occurs during meiosis II? 0 1 2 3 4 |
2 |
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How many of the four gametes produced during mitosis will be abnormal if nondisjunction occurs during meiosis I? 0 1 2 3 4 |
4 |
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Gametes produced following a non-disjunction event are: Haploid (n) Diploid (2n) Haploid and missing a chromosome (n-1) Diploid and missing a chromosome (2n-1) Haploid with an extra chromosomee (n+1) Diploid with an extra chromosome (n+1) |
Haploid and missing a chromosome (n-1) |
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A pair of homologous chromosomes cary the genes A1 and B1 or A2 and B2. If a pair of Holliday junctions between A and B, one is resolved by a NS cut and the other by an EW cut, the resulting chromosomes ________. are nonrecombinant and carry the combinations A1 and B2 or A2 and B1 are nonrecombinant and carry the combinations A1 and B1 or A2 and B2 are A1 and A2 or B1 and B2 are recombinant and carry the combinations A1 and B2 or A2 and B1 are recombinant and carry the combinations A1 and B1 or A2 and B2 |
are recombinant and carry the combinations A1 and B2 or A2 and B1 |
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What type of aneuploidy is responsible for Turner syndrome in humans? trisomy 13 trisomy 18 trisomy 21 monosomy XO monosomy YO |
monosomy YO |
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When nondisjunction occurs early in embryogenesis rather than gametogenesis, what would you expect in the resulting karyotype? Monosomy Trisomy Mosaicism uniparental disomy random X-inactivation |
Mosaicism |
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Prader-Willi and Angelman syndromes are caused by which type of chromosomal mutation, where both copies of chromosome 15 originate from a single parental homolog? Monosomy Trisomy Mosaicism uniparental disomy random X-inactivation |
uniparental disomy |
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A region of a chromosome spanning the centromere is broken and reattached in the reverse direction. This is an example of which type of chromosomal defect? paracentric inversion pericentric inversion pericentric translocation dicentric inversion dicentric translocation |
pericentric inversion |
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A chromosome contains the following gene order: A B C D • E F G H Which of the following rearrangements represents a pericentric inversion? A B C • D E F G H A F G D • E B C H A C B D • E F G H A B F E • D C G H A B C D • E F B C |
A B F E • D C G H |
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A chromosome contains the following gene order: A B C D • E F G H Which of the following rearrangements represents a paracentric inversion? A B C • D E F G H A F G D • E B C H A C B D • E F G H A B F E • D C G H A B C D • E F B C |
A C B D • E F G H |
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Be sure to review LC April 18th |
info |
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The most direct way to repair DNA lesions is to identify and then reverse the DNA damage. What is one direct mechanism that does this? |
the proofreading activity of DNA Pol III (3'-5' Exonuclease activity). does not correct all of the errors though, especially not chemical or radiation mutagens |
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mismatched nucleotides that escape DNA polymerase proofreading or are tautomeric may be detected and repaired by _______ |
mismatch repair |
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In mismatch repair, how do repair enzymes distinguish between the original correct nucleotide and the new mismatched nucleotide? |
the original strand is methylated |
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explain the process of mismatch repair |
the protein of mismatch repair recognizes and binds to the dna, a phosphodiester bond is broken on the 5' side of the mismatch on the unmethylated daughter strand. The exonuclease digests the nucleotide from the "nick" through the mismatched nucleotide. DNA polymerase fills the gap in the daughter strand, DNA ligase completes the repair. Dam methylase methylates the adenine of the GATC sequence on the daughter strand. |
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Nucleotide Base Excision repair |
DNA glycosylases recogniz and remove bases creating an apurinic/apyrmidic site (AP site(hasno nucleotide base)). Once there is an AP site the AP endonuclease removes the APsite leaving a gap where we need to put in the appropriate nucleotideDnapolymerase fills the gap, but it can’t build the last phopshpidester bondso we need DNA ligase. Ligase fills the last phosphodiester bond and we've repaired the nucleotid. Very limited repair 1-3 nucleotides |
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What is the difference between nucleotide base excision repair and nucleotide excision repair? |
nucleotide base excision: fixes nucleotides damaged by chemicals nucleotide excision repair: fixes UV damage |
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Nucleotide excision repair recognizes and removes ________ _______ |
bulky adducts such as UV light damage that distort DNA |
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Nucleotide excision repair process |
enzymes recognize and bind to the damaged region. A segment of nucleotides is removed from the damaged strand. DNA polymerase fills in the gap and DNA ligase seals the sugar phophate backbong ( much like mismatch repair process) |
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What is the most common mutagen for most organisms? |
UV radiation |
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What two mechanisms repair UV induced DNA Damage and which is specific to bacteria? Which is specific to human cells? |
photoreactive repair- bacteria UV Repair- used in organisms such as human cells |
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Photoreactive repair |
Pyrimidine dimers are directly repaired in bacteria, plants, and some animals, but not humans. Photolyase binds to DNA, collects a little energy and breaks the bonds between the pyrimidine dimers. the enzyme photolyase esentially uses light energy to repair the damage caused by light (UV) it is a solar powered enzyme. |
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UV Repair |
a form of nucleotide excision repair. CErtain proteins recognize DNA damage and other proteins excise a short segment of DNA containing the photoproduct. New DNA is synthesized to replace the removed nucleotides. Ligase seals the sugar phosphate backbone Fixes thymine dimers |
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DNA Damage Signaling Systems |
biochemical mechanisms recognize the presnece of DNA damage and initiate a repair response. |
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Ingeneral our cells respond to dna damage. If we have an excessive amount of dna damage it would be innappropriate for the cell to proceed through the cell cycle so the cell intiaties |
apoptosis |
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The p53 repair pathway controls cell responses to mutation by..... |
pausing the cell cycle at the G1 to S transition to allow time for repair (while p53has this stopped it starts a timer, if the cell can’t fix itself within acertain amount of time p53 recognizes that it is not repairing well and it willtrigger cell death.) or Direct the cell to undergo programmed death |
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Explain the p53 Repair Pathway pausing the cell cycle. |
In a normal cell p53 levels are low because there is no DNA damage and it can progress into s phaseanytime it likesIf ATM senses cell damage it turns on p53 1.Dna damage stimulates ATM 2.. ATM activates p53 3. p53 stops the cell in G1 1.P53 turns on p21 2.P21 inhibits cdk/cyclin complex formation3.P53 activates BAX (if left on toolong it will slowly inhibit BCL2) 1.BCL2 keeps cell from killing itself 2.Without BCL2 the cell dies |
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Which proteins listed in the p53 Repair pathway can be mutated in certain forms of cancer? |
All of them: Atm P53P21BAXBCL2 |
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Mutationsin genes that participate in _______________cause an organism to be highlysensitive to chemical mutagens and to radiation They also greatly increase the organism’s susceptibility to cancers caused by exposure to mutagens |
DNA damage repair |
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Translesion DNA Synthesis Repair |
This SOS repair system is a last-ditch effort of a heavily damaged bacterial system to replicate its DNA and divide. It is accomplished by activation of translesion DNA polymerases,or bypass polymerases. These error-prone polymerases have no proofreading ability, and can replicate across lesions that would stall DNA polymerase III |
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Bacteria will try to ____ the cell that is too damaged to replicate humans would rather ______ the damaged cell |
Repair the cell at all costs b/c it is better to live with it than kill cells kill, because it isn't worth living with |
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What damage occurs from exposure to X-rays and certain types of oxygen radicals? |
double strand DNA break |
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What two mechanisms carry out double-strand break repair? |
1.Non-Homologous End Joining (NHEJ) 2.Synthesis-Dependent Strand Annealing(SDSA) |
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Nonhomologous End Joining |
A double-strand break that occurs during G1prevents completion of DNA replication . Non-homologous end joining(NHEJ)allows cells to regain the ability to complete DNA replication |
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NHEJ always results in mutations 100% of the time t/f |
True, Every time you do NHEJ you lose sequence ~ 10 nucleotides and you create mutations every time |
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4 Steps of Non homologous end joining |
1.X-rays or Oxidative damage produce double-strand breaks in DNA. 2.Double-strand breaks are recognized by a protein complex that attaches to the broken ends of the DNA. 3.The complex trims back the free ends of the breaks (nucleotides are lost!!!) 4.The blunt ends produced by resection are ligated by DNA ligase I. |
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Synthesis Dependent Strand Annealing |
only works during G2 the sister chromatid is used to repair the break. error-free |
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steps of synthesis dependent strand annealing |
1.One chromatid has a double-stranded break. 2.The broken strands are trimmed and Rad51 coats the undamaged chromatid. 3.Rad51facilitates the invasion of the intact sister chromatid by the resected end of the broken strand.The strand invasion process displaces one strand of the DNA duplex on the sister chromatid, forming a displacement(D)loop. 4.Replication within the D loop synthesize s new DNA from the intact template strand. 5.Sister chromatids re-form by dissociation and annealing of the new strand to the other side of the break, resulting in replacement of the excised DNA with a duplex identical to the sister chromatid. |
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How do we get crossing over during meiosis? |
SDSA causes it |
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homologous recombination (crossing over) takes place during.... (1 answer for in bacteria, another for eukaryotes) |
In bacteria, and similarly in archaea, it occurs during conjugation and as a consequence of double-strand break repair In eukaryotes, homologous recombination takes place in prophase I of meiosis(crossing over) |
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______________________is the exchange of genetic material between homologous DNA molecules |
Homologous recombination |
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homologous recombination in eukaryotes is initiated by_____ |
controlled DNA double strand breaks |
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what errors would occur without homologous recombination |
nondisjunction |
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The Double-Strand Break Model of Meiotic Recombination |
1.recombination is initiated by the protein Spo11, first discovered in yeast.Spo11is a dimeric protein that generates an asymmetric double-stranded cut in one chromatid 2.The proteins Mrx and Exo1 associate with Spo11, Spo11 degrades,and Mrx/Exo1resect the single strands of the cut chromatid (5’à3’exonuclease 3.Two proteins, XRCC3 and Rad51, assemble on the 3’ ends of the broken DNA strands.4.The proteins Rad51 and Dmc1 (RecA homologs) help facilitate strand invasion. 5.Strand invasion results in: AD loop, A heteroduplex region(double-strandDNA formed from DNA of different homologs that may contain nucleotide mismatches because of allele differences), A Holliday junction two strands that appear to cross over one another. 6.The invading strand is extended, with DNA synthesis guided by the intact template strand creating a second heteroduplex region 7.Strand extension and ligation fills the gap on the upper chromatid. 8.The3' end of the invading strand is ligated to the 5'end of the original DNA strand. A double Holliday junction(DHJ)structure forms. The Holliday junctions can be resolved in two ways, only one of which results in recombination.. |
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What are two ways to cut the Holliday junction and which results in crossing over? |
same sense-cut them both the same way (same sense resolution) Cut them both north south or both east west Opposite sense = one north south, and one north west – swap dna creating recombinant chromosomes see drawing in notes. More common than same sense. Results in crossing over!! |
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An ______ _______ ______ is when one Holliday junction is resolved by a NS cut and the other by an EW cut More common The resulting chromosomes are recombinant, and lead to production of recombinant progeny |
opposite sense resolution |
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_____ _____ _______ involves either two north-south (NS) resolution cuts or two east-west (EW) resolution cuts. flanking markers do not recombine,though hetero duplex regions remain |
same sense resolution |
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Gene conversion |
is a process of directed DNA sequence change in which one allele is altered to another allele present in a heterozygous individual |
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Thanks to ______ _______, in the heteroduplex regions, we have potential mismatches in alleles which can create new genetic combinations in offspring |
crossing over |
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Heteroduplex DNA can include nucleotide mismatches Different ratios can result from a heteroduplex,depending on how the mismatches are repaired |
If you get a heteroduplex region where there is an allele difference (G-A) dna mismatch repair will repair it.Technically both strands are correct so which ever way we repair it (g-c) or(A-T) this might remove one of the alleles |
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The normal number of chromosomes found in the nucleus for a given organism is? |
The euploid number ours is 46, and they are in pairs because they are diploid (2n) |
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If cells contain a number of chromosomes that are not euploid, the chromosome number is called: |
aneuploid for ex: 2n+1 |
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The failure of homologous chromosomes or sister chromatids to separate as they normally do during cell division is called |
Chromosome nondisjuction |
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Iff we get a nondisjunction event a cell will have an _______chromosome or a ________chromosome |
extra missing In somatic cells, it can result in one daughter cell with an extra chromosome (2n+ 1)and the other missing one chromosome (2n- 1) |
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There are few examples of aneuploidy cells in animals. Why? |
bc of their poor survival rate |
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Non disjunction is mainly a problem in meiosis. What do we get with there is a nondisjuction in Meioises 1? Meiosis 2? |
Meiosis 1: two n+1 cells and two n-1 cells. all abnormal Meiosis 2: two n-1 cells and two normal cells see notes for a drawing |
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The number of copies of a gene |
gene dosage (which aneuploidy affects) |
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Plants can generate their trisome, but humans and animals can't. Why? |
because we are very sensitive to gene dosage, but plants tolerate gene dosage changes |
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Which conditions containing trisomies will result in living births? |
Trisomy 13, least frequent and most severe (Patau Syndrome), Trisomy 18 intermediate, results in early death (edward syndrome), Trisomy 21, the most frequent (Down syndrome) there is abnormal development, but not an early death |
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How many autosomal monosomies are observed in humans? |
none
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Sex chromosome trisomies and monosomies happen and are associated with 4 aneuploidys. what are they? |
XXY Klinefelter syndrome, males XYY Jacob Syndrome, males XXX Triple X syndrome (females XO Turner syndrome, females |
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What causes approximately half of all miscarriages? |
trisomies or monosomies in the infant |
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What can be correlated with the majority of Down Syndrome symptoms |
when there are three copies of those genes in the critical region it makes the down syndrome symptoms |
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Asmall number of genes on chromosome 21 are responsible for the cognitivedisabilities and heart abnormalities that are principal symptoms. what are they? |
THe down syndrome critical region also DYRK is known to produce defects in mice and flies DSCAM is associated with formation of the heart and nervous system |
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explain what happens in XO (Turner syndrome) |
In XX we turn off most of the secondary X except for the SHOX protein making it insufficient to direct normal development. the haploinsufficiency of this gene plays a central role in producing the symptoms of the syndrome |
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In trisomics,chromosome segregation during meiosis is disturbed because of failure to properly pair and segregate. 2 patterns of synapsis are possible. what are they |
a trivalent synapsis or bivalent and univalent arrangement. Neither mechanism can segregate 3 chromosomes and will lead to problems with creating normal gametes. It will almost certainlyalways make the n+1 |
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semisterility |
In trisomics,meiosis results in two chromosomes moving to one pole and one chromosome moving to the other Thus half the gametes contain two copies of the chromosome; these will produce trisomic offspring that are unlikely to survive. Only some of the offspring produced are viable |
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Mosaicism |
anytime the chromosomes fail to separate appropriately For example, 25–30% of Turner syndrome cases occur in females who are mosaic, with some 45, XO cells and some 46, XX Some Turner syndrome individuals carry 47, XXX cells too |
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Uniparental disomy |
2 nondisjunction events have to happen at the same time, one in the mom and one in the dad. this results in both copies of the one chromosome coming from the same parent |
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Why can uniparental disomy be bad? |
you’ll be homozygous for every single trait. Any recessive traits given to you from the one parent, you will express because they’re the same parents. So all recessive genes will show up. it is likened to inbreeding |
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trisomy rescue |
more common, We use 1 abnormal gamete and the normal sperm so we make a trisomic embryo. That cell goes through mitosis.There is a mitotic nondisjunction that rescues the cell (gets rid of one of thecopies of the cell) it removes one ofthe extra chromosomes. The normal cells have a growth advantage over the trisomic cell. In the population of growing embryo you will get more and more of the normal cells and they will be outcompeting the trisomic cells. However, with this process wecould end up with uniparental disomy.
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The presence of three or more chromosomes in the nucleus of an organism (seen in plants) |
polyploidy |
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There are 2 ways to get polyploidy |
autopolyploidy: duplication of chromosome sets within a species allopolyploidy: combining the chromosome sets of different species results from two different species mating and combining chromosomes |
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What are the 3 mechanisms that lead to autopolyploidy? |
multiple fertilizations of one egg by multiple pollen grains meiotic nondisjunction: leading to a diploid rather than haploid gamete mitotic nondisjuction in sex stem cells |
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Autopolyploids result in hybrid offspring that are infertile. why? |
because the chromosome sets are non homologous |
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Chromosomenondisjunction in autopolyploids leads to cells with double the number ofchromosomes so that now each chromosome has a homolog for pairing. Will this result in fertile offspring? |
yes They are infertile if they have an odd number of chromosomes because it makes meiosis virtually impossible |
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F1 hybrid generations of allopolyploids tend to have more rapid growth t/f |
true |
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Hybrid vigor |
more rapid growth, increased fruit and flower production, and improved resistance to disease that occurs in heterozygous allopolyploids produces great in the F1generation, but you can’t replant them bc F2 won’t grow well at all so you have to buy more seeds from the F1 generation bad business practices |
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What contributes to more than half of all flowering plant species |
polyploidy. It has had a made a huge impact on the expansion of species and led to lots of beneficial traits |
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Mutations that result in loss or gain of chromosome segments can produce severeabnormalities due to ________ ________imbalances |
gene dosage |
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When a chromosome breaks, both DNA strands are severed at a location called a |
chromosome break point |
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if a chromosome contains a chromosome break point what could happen to it |
The broken chromosome ends can adhere to other broken ends or the termini of intact chromosomes Or,if the broken chromosome is acentric(lacks a centromere), it may be lost during cell division |
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WHy are acentric chromosomes lost during cell division |
a centromere see drawing (5-2)- there is no place for microtubules to attach during anaphase so they don’t segregate appropriately |
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Partial chromosome deletion |
Loss of a portion of a chromosome Larger chromosome deletions can be detected microscopically and affect more genes |
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terminal deletion |
Detachment of one chromosome arm leads to a terminal deletion; the broken fragment lacks a centromere and is lost during cell division |
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partial deletion heterozygotes |
Organisms with one normal and one terminally deleted chromosome are called partial deletion heterozygotes |
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interstitial deletion |
An interstitial deletion is the loss of an internal portion of a chromosome, and results from two chromosome breaks 1 way is from 2 chromosome breaks, then reattached to each other and it will be accentric and the acentric chromosome will be lost WAGRsyndrome when an interior critical part of the chorosome is deleted |
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an interstitial deletion that results in a partial deletion of the chromosome. This can result in partial duplication on one homolog and partial deletion on the other |
Unequal crossover |
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An organism with one normal and one duplication homolog is a |
partial duplication heterozygote |
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an organism with one normal and one deleted homolog is a |
partial deletion heterozygote |
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when does unequal crossover most commonly ocur |
when repetitive regions of homologs misalign
it doesn't occur often |
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What is used to detect microdeletions and microduplications? |
can use the FISH probe to detectmicrodeletions. Maybe have fish probes for genes ABC and see green band, red band, yellow band, if gene b is deleted we won’t see a spot on the chromosomewhere that probe sticks b/c there is nowhere to stick to. Duplication of gene b will result in two spots of that color on the chromosome. |
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Unpaired loop |
if we prepare chromosomes when two homologous chromosomes are aligned with one another during prophase or metaphase 1 of meiosis. They’re paired up, if one of our chromosomes has a deletion there will be dna that can’t be paired and it loops out from the complex where no pairing is available so the unpaired loop is deletedfrom the chromosome. Unpaired loop= extra dna |
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occurs when a recessive allele is “unmasked” by a deletion that removes the dominant allele on the homolog |
Pseudodominance |
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deletion mapping |
uses psudodominance to map the position of genes in a chromosome |
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if we get the wild type the gene was OR wasn't removed by deletion? if we get the mutant did the gene remove the wild type copy of the gene? |
if we get the wild type the gene was not removed by deletion mutant= the mutant did remove the wild type copy of the gene |
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Sometimes chromosome breakage leads to reattachment of the wrong broken ends
Reattachment in the wrong orientation leads to? reattachment to a nonhomologous chromosome leads to ? |
1. chromosome inversion 2. chromosome translocation |
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chromosome inversion |
breakage and reattach the wrong ends to each other resulting in a chromosome inversion no information is lost, just a change of order. That is fine for a little while, but the real problem happens when we try to do meiosis. |
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Chromosome translocation |
reattachment from one chromosome to other chromosomes = chromosome translocation, no info Is lost, just changed the way its connected together |
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Two types of chromosome inversion |
paracentric pericentric |
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paracentric inversion |
the centromere is outside of the inverted region |
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pericentric inversion |
the centromere is within the inverted region |
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Inversion heterozygotes |
have one normal and one inverted homolog Crossing over that occurs outside the inverted region takes place as normal Crossing over within the inverted region results in duplications and deletions in the recombinant chromosomes |
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(Inversion heterozygotes) Alignment of a normal chromosome with its inverted homologs results in the formation of an_____ ______ |
inversion loop |
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Crossing over that occurs within a paracentricinversion results in a ________chromosome and an ________fragment |
dicentric acentric The dicentric chromosome is pulled toward both poles of the cell and eventually breaks at a random point; both products of the break are missing genetic material |
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Crossing over that occurs within a pericentricinversion results in |
both duplicated and deleted regions in both of the recombinant products Like paracentric inversions, the recombination event yields two normal gametes (from the noncross over chromatids) and two abnormal gametes (from the crossover chromatids) |
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3 Genetic Implications of Recombination in Inversion Heterozygotes |
1.The probability of crossover within the inversion loop is proportional to the size of the loop 2.Inversion suppresses the production of recombinant chromosomes, identified as crossoversuppression 3.Fertility may be altered if an inversion heterozygote carries a very large inversion |
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_______ ________with one normal copy and one translocatedcopy of each chromosome, may be normal in phenotype if no genes are disrupted by the breakage and reattachment events |
translocation heterozygotes Creates problem during meiosis and might lead to semi sterility |
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3 types of translocations |
unbalanced translocations recriprocal balanced translocations robertsonian transolcations |
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unbalanced translocations
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arise when a piece of one chromosome is translocated to a nonhomolog and there is no reciprocal event |
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Reciprocal balanced translocations |
occur when pieces of two nonhomologs switch places breaka piece off blue chrom put it onto the red then break a piece off of red and put it onto the blue 2 translocations between the pair of chromosomes |
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Robertsonian translocations |
also called chromosome fusions, involve fusion of two nonhomologouschromosomes If we break the chromosome between x and centromere we lose centromere and attach all important genetic information to another chromosome. The centromere is lost. We have an n-1 cell b/c chromosome is lost but we’ve retained all of the genetic information. The person would be healthy b/c they have genetic informiaton |
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Robertsonian- can only happen to a certain kind of chromosome |
telocentric chromosome (when centromere is close to the end of the chromosome) |
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In heterozygotes for reciprocal balanced translocations, none of the four chromosomes has a fully homologous partner An unusual, cross-like structure is formed at metaphase I of meiosis Two patterns of chromosome segreagation are possible. What are they? |
alternante segregation adjacent-1 |
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alternate segregation |
the two normal chromosomes move to one pole of the cell and the two translocated chromosomes to the other. In alternate segregation, the gametes formed contain a full haploid set of genes the only translocation segregation that leads to normal gametes |
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adjacent-1 segregation |
one normal and one translocated chromosome move to each pole of the cell In adjacent segregation, each gamete contains two copies of some genes and the complete absence of others |
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Adjacent 2 segregation |
Adjacent-2 segregation is very rare; homologous centromeres move together to each pole of the cell |
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Adjacent 1– leads to errors, one of the untranslocated chroms with one of the translocated chromosomes (50% of time) Adjacent 2 segregation – very rare, all abnormal gametes Alternate segregation: you get normal chromosome that was intended (50% of time) Translocation: half their gametes will be fine half will be abnormal 50% the gametes won’t produce viable offspring |
info |
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Iftwo pairs of chromosomes fuse by Robertsonian translocation, the number of chromosomes drops to ______ |
2n-2 This mechanism accounts for the difference in chromosome number between humans (2n= 46) and chimpanzees (2n = 48) |
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Carriersof a Robertsonian translocation have one chromosome fusion; the homologs of the fused chromosomes remain separate. what is an example of this |
One type of Robertsonian translocation between chromosome 21 and (usually) chromosome 14 is responsible for familial Downsyndrome A carrier of this translocation is usually normal but may have Down syndrome offspring due to segregation abnormalities |
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Robertsonian Translocation Segregation in Familial Down Syndrome |
One of the chromosomes 21 is telocentric so we can get double strand break that saves almost all of genetic information and we can take that genetic infomraiton and fuse it with 21. one normal 14, one normal 21 and one robertsonian translocation 14+21Run into a problem of how these segregate during meiosis. Microtubules attach to centromeres. 2 will go one direction while one goes the other direction. If we pull the two normal ones one direction and translocation the other, everydaughter cell has the appropriate chroms. If you grab the other two chrom 21s then you’ve duplicated chroms 21 = trisomy translocation event. . Draw these out and see which one will go where. Translocation impacts = how they line up during meiosis and how they segregate during anaphase 1/3 of the time these people will have normal offspring 1/6 of time would be inviable bc theres 2 copies of chrom 21 1/6would be inviable b/c theres no chrom 21 1/6inviable 2 copies of chrom 14 1/6 bc theres no copies of chrom 21 |
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Activator proteins |
bind regulatory sequences to stimulate transcription turn on genes |
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Repressor proteins |
bind other sequences to hinder transcription turn off genes |
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How do we activate or repress genes? |
by regulating its transcription or if we keep the gene to regulate genes at the dna level then we wont make any protein. Can also regulate genes at the level of RNA processing. multiple ways including mrna processing, transcription, translation, and post translation by regulating how long lived a protein is |
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Three sets of regulatory DNA sequences are commonly involved in eukaryotic gene regulation |
Core promoter region (tata box etc) can be used as binding sites for otherthings to turn off and on the gene Proximal– nearby the promoter to help regulates transcription Enhancer and silencer are distant, but can bend over and get close to the promoter |
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cis-acting regulatory sequences |
regulate transcription on the SAME chromosome as the gene they’re controlling 3 regions are all cis acting reg sequences: corepromoter, proximal, and enhancer sequences |
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trans-acting regulatory proteins |
proteins produced by genes on another chromosome the gene product comes and regulates the chromosome where the gene is. |
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At enhancers, aggregations of multiple proteins form large complexes called |
enhanceosomes |
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an example for Control of eukaryotic transmission is the SHH gene present everywhere but inactive until the Brain transcription factors or limb specific transcription factors combine to it. if brain tf is messed up. can limb still function? |
yes |
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______ ______ _____is a highly specialized enhancer that regulates transcription of multiple genespackaged into complexes of closely related genes (ex: globins ) |
locus control region: an example is the globin gene allows us to turn on different types of globin at different times |
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Mutations in the a- and b-globin genes produce a hereditary anemia called |
thalassemia |
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It was found that some thalassemia cases were due to deletions that altered the_____regions, causing abnormal expression of the globin genes |
LCR |
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Enhancer-Sequence Conservation |
Some enhancer sequences are conserved among different species For instance, the b-inteferon gene is conserved among mammals |
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What indicates evolutionary constraint on enhancer diversification |
Enhancer sequence conservation |
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slides 13- 18 on May 4th should be reviewed, can't make cards out of them. Discusses enhancers and silencers in yeast and eukaryotes |
info |
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Insulator sequences |
Their job is that only the gene we want to regulate is affected by the gene enhancer See drawing on 5-4 (on bottom of page) Insulators often form loops that keeps the enhancer from acting with the wrong promoter |
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________modifications alter chromatin structure leading to changes in transcription. |
Epigenetic |
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Is Euchromatin more transcriptionally active than heterochromatin |
yes |
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Five Important Features of Epigenetic Modification |
1.Epigenetic modification patterns alter chromatin structure 2.Theyare transmissible during cell division (will remain heterochromatin if it is originally heterochromatin) 3. Yet They are reversible 4.They are directly associated with gene transcription (hetero – low, eu- high transcription) 5.They do not alter DNA sequence, but can alter gene expression |
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DNA packaging into chromatin is unique to eukaryotes and is regulated in three ways: |
1.Regulate the way dna is bound to histones. (when histone DNA wraps around it to form nucleosome) 2.Chromatin remodelers Change the state of the chromatin (where thehistones are and how they’re behaving) 3.Chromatin Modifiers Change whether acetyl and methyl groups attach to histones which affects if it is heterochromatin or euchromatin |
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Open Promoters |
cause genes to be constitutively expressed; they have a nucleosome-depleted region(NDR) immediately upstream of the transcription start site. Thesepromoters lack a TATA box, but have a poly A/T tract. Thepoly A/T tract contains binding sequences (BS) that bindtranscriptional activators (ACT) and that help position one nucleosome upstream and one downstream (called the +1nucleosome) Thedownstreamnucleosome contains histone H2A.Z thatis easily removed from the transcription start site |
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Covered promoters |
transcription is regulated closed promoters= (wont be transcribed) have a nucleosome covering the core promoter Use to cover TATA box so it isn’t easily accessed by transcription factors. Cant transcribe this gene until we knock the nucleosome out of there revealing the TATA box. Cant get to it unless you dislodge this. |
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___________ __________refers to modifications that reposition nucleosomes so as to open or close promoters and other regulatory sequences |
chromatin remodeling moving histones and nucleosomes around |
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has a relaxed association between DNA and nucleosomes, and allows for access by regulatory proteins (DNaseI Hypersensitive) |
Open chromatin= euchromatin = Dnase 1 hypersensitive- dnase can only cut euchromatin |
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is produced by modifications that cause regulatory sites to be covered by nucleosomes, restricting access by regulatory proteins (DNase I Insensitive) |
Closed chromatin=heterochromatin,= dnase1 insensitive (cant get access to dna bc its closed so it cant cut it up) |
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Chromatin remodelers are protein complexes that move nucleosomes in three principal ways: |
1.Takes a nucleosome and slides it down the dna 2.Cause nucleosomes to reposition (dislodge it from one and associate it with another) 3.Modify the composition of the histone octomer.(if nucleosome contains histone 2az or not) |
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Chromatin modifier |
proteins chemically alter histone proteins in the nucleosomes by adding or removingchemical groups Make chemical modifications that change the strength between the dna and the histone Principal chemical modifications are addition and removal of acetyl and methyl groups |
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“Writer”enzymes ______chemical groups to chromatin; “eraser” enzymes _____ groups from chromatin |
add remove |
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_____ _______activate the chromatin (makes it looser) |
Histone Acetylation |
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what are the writers in histone acetylation |
-Histone acetyltransferases (HAT)= the writers -Take away (or neutralize) the positive charge of the histone so they loosen the attraction between the histone DNA = we get a more euchromatin aka dnase sensitive chromatin. |
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what are the erasers of histone acetylation? |
histonedeacetylases(HDACs)create a stronger attraction between the DNA and histone (by removing acetyl groups I believe) and tightens it back up |
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Histone Methylation |
-When we put a methyl group onto a lysine we increase the strength of the attraction between DNA and histones so it will tighten up the dnamaking it dnase insensitive aka heterochromatin |
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writers of histone methylation |
histone methyl transferases(HMTs) |
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erasers of histone methylation |
Demethylation is carried out by histone demethylases(HDMTs) |
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Differences between somatic cells are ________resulting from different chromatin states, affecting gene transcription |
epigenetic Histones that are knocked off after use and break in half retain either being methylated or acetylated. When they are reassembled we grab one old piece and one new piece so we often retain the chemical modification. If histone was acetylated one of the pieces we grab back is still acetylated. The HATS will fill in any missing acetylation on the other pieces. That’s how we retain these kinds of modifications through generations |
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Epigenetic patterns are often heritable through ________from one generation of cells to the next, so that daughter cells have the same patterns of gene expression as the parent cells Epigenetic differences may also be heritable from one generation of organism to the next through _______—there is some evidence of this |
mitosis meiosis |
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_____ __________ _______play critical roles in gene regulation in eukaryotic cells |
Long noncoding RNAs (lncRNAs) play critical roles in gene regulation in eukaryotic cells |
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Regulatory regions of a eukaryotic gene all contain which of the following sequences, which regulate transcription of genes located on the same chromosome as the sequences? zinc fingers cis-acting regulatory sequences homeodomains trans-acting regulatory sequences leucine zippers |
cis-acting regulatory sequences |
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You have identified a mutation in a gene that also seems to decrease transcription of another gene 2000 bp away from the mutation site. What regulatory sequence, which may be found within another gene, has likely been mutated in this instance? core promoter proximal elements enhancer sequence homeodomain motif upstream activator sequence |
enhancer sequence |
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Typically, methylation of nucleosome N-terminal tails leads to ________. removal of the protein components of the chromatin from the DNA relaxed packaging of the chromatin and increased transcription tighter packaging of the chromatin and reduced transcription increased amounts of euchromatin relative to heterochromatin activation of topoisomerase |
tighter packaging of the chromatin and reduced transcription |
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Typically, acetylation of histone tails leads to ________. removal of the protein components of the chromatin from the DNA relaxed packaging of the chromatin and increased transcription tighter packaging of the chromatin and reduced transcription increased amounts of euchromatin relative to heterochromatin activation of topoisomerase |
relaxed packaging of the chromatin and increased transcription |
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Methylation of histones associated with human promoter sequences would result in __________________________________ transcription. |
reduced |
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A region of chromatin has recently become DNAse I hypersensitive. Which enzyme has been activated to cause this change in chromatin structure? histone methyltransferase histone acetylase histone deacetylase phosphatase kinase |
histone acetylase |