Genetics exam 2 study guide
Genetics exam 2 study guide BIOL 3060
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This 19 page Study Guide was uploaded by Myla Pereira on Sunday February 7, 2016. The Study Guide belongs to BIOL 3060 at Southern Utah University taught by Laurie Mauger in Winter 2016. Since its upload, it has received 20 views. For similar materials see Genetics in Science at Southern Utah University.
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Date Created: 02/07/16
Genetics Exam 2 Study Guide Major topics and description: Lecture 7: Regulation of Gene Expression in Eukaryotes (Ch. 12) At any given time in a cells life history, only a fraction of the RNA’s and proteins encoding in its genome are expressed. Regulation can take place at mRNA level (through alterations in splicing or the stability of the mRNA) and after translation by modifications of proteins. Post-transcriptional gene regulation o RNA acts post transcriptionally to repress gene expression, called gene silencing. miRNA, ncRNA, and siRNA participate Transcriptional gene regulation o Molecular signals from inside/outside cell lead to binding of regulatory proteins to specific DNA sites outside of protein- encoding regions o Binding of these proteins modulates rate of transcription o These proteins may directly or indirectly assist RNA polymerase in binding to its transcription initiation site (promoter) or they may repress transcription by preventing the binding of RNA polymerase. Both bacteria and eukaryotes use sequence-specific DNA-binding proteins to modulate level of transcription. o Eukaryotic genomes are bigger so regulation is more complex Requires more regulatory proteins and more types of interactions with adjacent regulatory regions in DNA DNA packaged into nucleosomes, forming chromatin where bacterial DNA lack nucleosomes In eukaryotes, chromatin structure is dynamic and essential in gene regulation o Ground state of bacterial gene is on so RNA poly can bind directly to a promoter when no other regulatory proteins are around to bind to DNA. Transcription initiation is prevented or reduced if binding RNA poly is blocked, usually though binding of repressor regulatory protein. Activator regulatory proteins increase binding of RNA poly to promoters where a little help is needed. o Ground state of eukaryote genes is off. Transcriptional machinery (RNA poly II and GTF’s) can’t bind to promoter in absence of other regulatory proteins. Binding of transcriptional apparatus not possible because nucleosomes positioned to block the promoter. So chromatin structure usually has to be changed to activate eukaryotic transcription. Changes depend on binding of sequence- specific DNA-binding regulatory proteins. Structure of chromatin around activated or repressed genes can be quite stable and inherited by daughter cells. (epigenetic regulation is the inheritance of chromatin states). Transcriptional regulation in eukaryotes o Larger number of genes o Patterns of gene expression complex o Timing of gene expression varies o Amount of transcript produced varies o Eukaryotic gene expression must be able to Ensure that expression of most genes in the genome is off at any one time while activating a subset of genes Generate thousands of patterns of gene expression o Regulatory proteins Promoter-proximal elements They are cis-acting regulatory DNA sequences that interact with proteins in RNA poly II and GTF’s to initiate transcription (located near promoter of a gene). Bound by transcription factors that affect expression of many genes Enhancers Regulatory proteins that consist of transcription factors that bind to cis-acting regulatory DNA sequences May be far from gene promoters. Bound by transcription factors that control the regulation of smaller subsets of genes. Will act in only one or a few cell types o Control caused by how specific transcription factors control access of general transcription factors and RNA poly II o Transcription requires binding of general transcription factors to additional promoter-proximal elements. Mutations on these sites could have dramatic effect on transcription Level of transcription usually reduced o To modulate transcription, regulatory proteins possess one or more functional domains. Functional domains of regulatory proteins A domain that recognizes a DNA regulatory sequence (the protein’s DNA binding site) Domain that interacts with one or more proteins of the transcriptional apparatus (RNA poly or a protein associated with RNA poly) Domain that interacts with proteins bound to nearby regulatory sequences on DNA such that they can act cooperatively to regulate transcription Domain that influences chromatin condensation either directly or indirectly Domain that acts as a sensor of physiological conditions within the cell Gene regulation via chromatin o Modifying local chromatin structure around gene regulatory sequences influences gene transcription o Eukaryotic chromosomes packaged into chromatin (composed of DNA and proteins that are mostly histones) o Basic unit of chromatin is the nucleosome Nucleosome has 150 bp of DNA wrapped 1.7 times around a core of histone proteins Nucleosome core has 8 histones (H2A, H2B, H3, and H4) organized as 2 dimers of H2A and H2B and a tetramer of H3 and H4 Linker histone H1 surrounds core, which can compact the nucleosomes into higher-order structures that further condense the DNA o Because DNA packaged into chromatin, much of DNA not readily accessible to regulatory proteins and transcriptional apparatus. (explains why genes are off unless activated) So modification of chromatin structure needed 3 mechanisms to alter structure chromatin remodeling: moving nucleosomes along DNA histone modification: in the nucleosome core replacing common histones in a nucleosome with histone variants chromatin remodeling move histone octamer along DNA. Nucleosome position changes. Moving of nucleosome away from promoter activates transcription. Moving towards promoter represses gene o Repressed when promoter and flanking sequences wound up in a nucleosome Modification of histones Histones are almost identical in all eukaryotic organisms Histone tails are amino-terminal ends making electrostatic contacts with the phosphate backbone of surrounding DNA Post translational modifications (PTMs): reactions modifying histone tails by attachment of acetyl and methyl groups. Take place after histone protein has been translated and even after histone has been incorporated into a nucleosome. o Covalent modification of histone tails contributes to a histone code (reminiscent of the genetic code) For histone code, info is stored in patterns of histone modification rather than sequence of nucleotides Histone acetylation, deacetylation, and gene expression o Histone modification o Acetyl group added by histone acetyltransferase (HAT) and removes by histone deacetylase (HDAC) o Hyperacetylated: histones associated with the nucleosomes of active genes are rich in acetyl groups o Hypoacetylated: inactive genes are underacetylated o Transcription is activated by acetylating nearby histones Neutralizes (+) charger of lysine residues Reduces interaction of histone tails with (-) charged DNA backbone Results in more open chromatin Binding of regulatory proteins Deletion of acetyl group Compressor-facilitates gene repression but is not a DNA binding repressor Histone methylation can activate or repress gene expression Post translational modification of arginine and lysine residues in histone tails can result in altered chromatin and gene expression. Histone methyltransferase (HMTase) adds one, two, or three methyl groups to a specific amino acid in the tail of histone H3. o Unlike acetylation, addition of methyl groups can either activate or repress gene expression. o Methylation of specific lysine residues (doesn’t affect charge) creates binding sites for other proteins that either activate or repress gene expression depending on the residues modified. Inheritance of histone modifications and chromatin structure Epigenetic inheritance: inheritance of chromatin states from one cell generation to the next o In DNA replication, both DNA sequence and chromatin are passed to next cell generation o Chromatin structure can change in the course of the cell cycle and during successive generations of cell division Nucleosomes in parental strands have to disassemble then reassemble in daughter molecules. During this process, the old histones from existing nucleosomes are randomly distributed (epigenetic marks) to daughter molecules and new histones are delivered to the replisome. This randomly distributed old histones serve as templates to guide the modification of new histones. Histone variants Consensus histones most common histones added during replication Histone variants can replace consensus histones already assembled into nucleosomes Provides quick way to change chromatin DNA methylation: another heritable mark that influences chromatin structure Addition of methyl groups to DNA residues after replication, therefore it is not a histone modification. Mammals o Methyl group added to cytosine in CG dinucleotide o Symmetric methylation o CpG island: unmethylated CG dinucleotides found in clusters near gene promoters o C methylation associated with inactive regions of the genome DNA methylation marks can be inherited Hemimethylated: semiconservative replication generates daughter helices that are methylated on only the parental strand. This is DNA molecules methylated on only one strand o Methyl groups are added to unmethylated strands by DNA methyltransferases DNA methylation more stable than histone modification so often associated with regions of genome that are maintained in an inactive state for the entire lifetime of an organism. Activation of genes in a chromatin environment o Transcription levels are made finely adjustable in a chromatin environment by clustering binding sites into enhancers o Several different transcription factors or several molecules of the same transcription factor may bind to adjacent sites. o The binding of multiple regulatory proteins to the multiple binding sites in an enhancer can catalyze the formation of an enhancesome Large protein complex Catalyzed by binding of multiple regulatory proteins Synergistically activates transcription When the binding of transcription factors to sites that are the correct distance apart lead to an amplified, or supperadditive, effect on activation transcription. When an effect is greater than additive, it is said to be synergistic. B-interferon gene is an example of an enhancesome Encodes antiviral protein interferon Normally off Activated to high levels of transcription on viral infection Enhanceosome key to gene activation o Enhancer-blocking insulators Enhancers could interfere with the regulation of nearby genes, so to prevent promiscuous activation, regulatory elements called enhancer-blocking insulators have evolved When positioned between an enhancer and a promoter, it prevents the enhancer from activating transcription at that promoter. Don’t have an effect on the activation of other promoters that are not separated from their enhancers by the insulator. DNA is organized into loops containing active genes Insulators act by moving a promoter into a new loop, where it is shielded from the enhancer. Long term inactivation of genes in a chromatin environment o Gene silencing is a position effect that depends on the neighborhood in which genetic info is located. Two types of chromatin o Heterochromatin is highly condensed Constitutive heterochromatin Inaccessible to regulatory proteins o Euchromatin makes up the majority of the genome Loosely condensed during most of the cell cycle Where most genes found o Condensation changes during the cell cycle Position-effect variegation in drosophilia reveals genomic neighborhoods o White eye mutation induced with x rays Some progeny had eyes with patches of red and white Inversion can occur Heterochromatin can spread into neighboring euchromatin and silence genes Chromatin state inherited in next generation of cells o PEV is the ability of heterochromatin to spread into eurchromatin and silence genes Common phenomenon Variegation suppressors (Su(var)) Reduce spread of heterochromatin Variegation enhancers (E(var)) Increases spread of heterochromatin Heterochromatin might spread into adjoining regions and inactivation genes in some cells but not others in the absence of barriers. Barrier insulators prevent spreading of heterochromatin by creating local environment not favorable to its formation. o Ex: a barrier insulator could bind HATs and, in doing so, make sure that adjacent histones are hyperacetylated. Gender-specific silencing o Genomic imprinting Certain autosomal genes have unusual inheritance patterns. o Maternal imprinting Copy of the gene derive from the mother is inactive. o Paternal imprinting Paternal copy of gene is inactive o Identical gene can be active or inactive depending on who it was inherited from o Methyl groups added to regulatory regions of imprinted genes in one sex only Methylation blocks binding proteins needed for transcription DNA of genes that are shut down for an entire lifetime are usually highly methylated o X-chromosome inactivation Umber of transcripts produced proportional to number of copies of a gene in a cell Mammalian x chromosome has 1000 genes Dosage compensation: makes amount of gene products in females equivalent to males One x chromosome randomly inactivated in each female cell Barr body (Xi) inactivated chromosome Gene silenced on Xi o Epigenetic marks associated with heterochromatin o Most gene on Xi remain inactive in all descendants of that cell Lecture 8: Mutation, Repair, and Recombination Part 1 (Ch. 16) Fundamental properties of mutations o Whether a mutation happens is unrelated to any adaptive advantage it may confer o Mutations don’t arise because and organism needs it o Selective techniques select mutants already present in a population Point mutations o Alteration of a single base pair of DNA or of a small number of adjacent base pairs o Two types of point mutations Base substitutions: one base pair is replaced by another Transition: the replacement of a base by the other base of the same category. Purine replaced by purine (from A to G or from G to A) or a pyrimidine is replaces by a pyrimidine (from C to T or T to C) Transversion: replacement of a base of one category by a base of another, Pyrimidine replaced by a purine (C to A, C to G, T to A or T to G) or purine replaced by pyrimidine (A to C, A to T, G to C, G to T) o Transitions more likely to be silent substitution so it is more common. Also easier to substitute a sing ring structure for another single ring structure. Less likely to result in amino acid substitution Base insertion or deletions o Synonymous mutations Amino acid remains the same Also called silent mutation in exons o Missense mutations Codon for one amino acid is changed into a codon for another amino acid. o Nonsense mutation The codon for one amino acid is changed into a STOP codon (premature) Most produce inactive protein products Closer it is to the 3’ end of ORF, more likely that protein will work o Nonconservative substitution One amino acid may be replaced by a chemically different amino acid and this alteration is more likely to produce sever change in protein structure and function o Frameshift mutations Addition or deletion of a single base pair of DNA changes reading frame for remainder of the translation process Cause entire amino acid sequence translationally downstream of the mutant site. Complete loss of protein structure and function o Consequences in coding and noncoding regions of DNA (whether or not mutation disrupts or creates a binding site) o Harder to predict consequences of point mutations in noncoding regions Spontaneous mutations are naturally occurring mutations and arise in all cells Induced mutations arise through the action of certain agents called mutagens o Mutagens increase mutation rate Germ line mutations occur in reproductive cells and can be passed to offspring Somatic mutations occur in nonreproductive cells and cant be passed to offspring Errors in DNA replication o Transitions can result if one of the bases becomes ionized o Often caught through proofreading o Indel mutations (insertion or deletions) Base inserted or deleted not divisible by 3 Replication slippage Spontaneous lesions o Naturally occurring damage to DNA o Result from Depurination Loss of purine base Interruption of glucosidic bond between the base and deoxyribose and subsequent loss of guanine or adenine residue Apurinic site results: can’t specify a base complementary to the original purine Deamination of cytosine into uracil Converts GC pair to AT pair Oxidative damage Active oxygen species are byproducts of aerobic metabolism Damage DNA and precursors to DNA Implicated in human disease o Trinucleotide repeat disease Common mechanism of genetic disease is expansion of 3 base pair repeat Trinucleotide-repeat Ex: fragile X syndrome Caused by high number of repeats in FMR-1 gene Area of gene that is transcribed but not translated Induced mutations o Mutagenesis Production of mutations in the lab through exposure to mutagens o Incorporation of base analogs Chemical compounds similar to normal nitrogenous bases so occasionally are incorporated Into DNA In place of normal bases Base pairing properties differ so they can cause mutations by causing incorrect nucleotides to be inserted opposite to them in replication o Specific mispairing Not all mutagens are incorporated into DNA Alter a base to ensure a specific mispair is formed Alkylating agents operate by this pathway o Intercalating agents Mimic base pairs and are able to slip themselves in between stacked nitrogen bases Can cause insertion or deletion of single nucleotide pair o Base damage Damage one of more bases so no specific base pairing possible Result is replication block DNA replication will not pass this point, so it stops at damaged base UV light, ionizing radiation cause this Lecture 9: Mutation, Repair, and Recombination Part 2 (Ch. 16) Biological repair mechanisms o DNA is the only molecule that organisms repair instead of replace o Failure in repair mechanisms cause many inherited diseases o Proofreading function of DNA poly Most important repair mechanism DNA poly I and III can excise mismatched bases Direct reversal of damaged DNA o Lesions can be repaired by direct reversal o Mutagenic photodimer Caused by UV light Cyclobutane pyrimidine dimer (CPD) repaired by CPD photolyse Enzyme binds to photodimer and splits it to regenerate original bases Enzyme required light Photoreactivism o Alkytransferases Enzymes that directly reverse lesions Removes alkyl groups added to position of O-6 of guanine by mutagens Base excision repair o Most important mechanism used to remove incorrect or damaged bases after proofreading Carried out by DNA glycosylases Cleave base sugar bonds, causing altered bases to be removed from DNA Generates apurinic or apyrimidic (AP) sites Enzyme called AP endonuclease then nicks damaged strand upstream of AP site Third enzyme called deoxyribophosphodiesterase cleans up backbone by removing a stretch of neighboring sugar-phosphate residues so that DNA poly can fill the gap with nucleotides complementary to the other strand. DNA ligase then seals the new nucleotide into the backbone Nucleotide excision repair (NER) o DNA poly can’t continue to synthesize past a lesion: blocks replication. NER relieves replication and transcription blocks o Diseases caused by defects in NER Xeroderma pigmentosum (XP) Cockayne syndrome o Complex process requiring dozens of proteins 4 general phases of NER recognition of damaged bases assembly of multiprotein complex at the site cutting the damaged stran several nucleotides upstream and downstream of the damage site and removal of the nucleotides (30) between the cuts use of the undamaged strand as a template for DNA poly followed by strand ligation o two pathways for NER global genomic nucleotide excision repair (GC-NER) corrects lesions anywhere in genome and is activated by stalled replication forks Transcription-coupled nucleotide excision repair (TC- NER) Repairs transcribed regions of DNA Postreplication repair: mismatch repair o Reduced error rate to less than 10^-9 by recognizing and repairing mismatched bases and small loops caused by insertion and deletion of nucleotides in the course of replication o Mismatch repair systems have to do at least three things: Recognize mismatched base pairs Determine which base in the mismatch is the incorrect one Newly synthesized strand is unmethylated Excise incorrect base and carry of repair synthesis Error prone repair: translesion DNA synthesis o Some repair systems are a significant source of mutation o Appear to have evolved to prevent occurrence of portentially more serious outcomes such as cell death or cancer o Stalled replication fork can initiate cell death. Replication blocks can be bypassed by the insertion of nonspecific bases. o Process requires activation of SOS system (E. coli). Prevents cell death in the presence of significant DNA damage. o Translesion DNA synthesis: damage tolerance mechanism Eukaryotic error-prone repair system similar to SOS system. These bypass replication stalls Repair of double strand breaks o A double strand break mutation is a mutation in which both strands are damaged (both strands break at sites close together) Cause chromosomal aberrations resulting in cell death or a precancerous state if not repaired. o Two distinct mechanisms to repair double strand breaks Nonhomologous end joining (NHEJ): recognizes damage, prevents further damage to the ends, and recruits other proteins that trim strand ends to generate ends that are required for ligation. DNA ligase IV joins the two ends. Homologous recombination: occurs if strand breaks after replication of a chromosomal region in a dividing cell. Uses sister chromatids avaible in mitosis as templates to ensure correct repair. Lecture 10: Genetic Analysis and Complementation Tests Beadle and Tatum o Every gene encodes for one enzyme o Hypothesis has been refined o Experiment: Genetic and biochemical analysis of Neurospora crassa (red bread mold) Identified mutations and showed how each corresponded to a defective enzyme required in the pathway Confirmed that each mutation that generated nutrient requirement was inherited as a single gene mutation One gene one polypeptide Mutant genes and defective proteins o Minimal media v complete media N. crassa needs only inorganic salts, sugar, and a vitamin (biotin) to grow (minimal media) Complete media contains all nutrients o Induced mutations o Mutant screen o Test for required nutrients Concluded how many genes/steps/order of pathway required gene mutations complementation test for mutations in the same gene o complementation test identifies mutations that have a defect in the same gene by bringing them together in same cell o heterokaryon contains mutant form of both genes o complementation indicates mutations are in different genes o noncomplementation indicates mutations are in the same gene plus sign means mutants are complementary minus means mutans on same gene and fail to produce the same product Sample problems for complementation tests Lecture 11: Transmission Genetics: The Principle of Segregation Vocab: o Wildtype: nonmutant allele o Allele: alternative version of a gene o Homozygous: 2 copies of the same allele for a gene o Heterozygous: 2 different alleles for a gene o Dominant allele: expressed no matter what the second allele is o Recessive allele: must have 2 copies of the allele to be expressed o True-breeding: individuals that produce only progeny like themselves when allowed to self pollinate (homozygous) o Genotype: genetic makeup o Phenotype: physical or biochemical expression of a gene o Hybrid: “heterozygotes” produced from outcrossing plants that differ in one or more traits o P generation: parental generation o F 1eneration: 1 generation o F generation: 2 ndgeneration 2 Transmission genetics o Study of inheritance of traits Mendel’s Experiments o P (parental generation) True breeding (both parents homozygous) Looked at flower color (purple X white) o F1 generation (first filial) offspring purple Allowed to self fertilize (self cross) Purple X purple F1 hybrid express only dominant trait F2 (second filial) offspring purple and white 3 purple: 1 white showed white trait wasn’t lost, something masked it concluded dominant vs. recessive o homozygous vs. heterozygous o recessive trait reappears in F2 generation Reciprocal crosses o Used this to determine if sex influences gene expression for each of his characters. This type of cross switches the male and female phenotype in parental generation. o Female white male purple had 1:3 ratio o Then switched the sexes but always saw the same thing so sex didn’t influence Morphological and molecular phenotypes o Round v wrinkled seeds Wrinkled seed caused by transposable element inserted in gene Shape determine by starch branching enzyme I (SBEI) gene Gene can be interrupted by transposable element Insertion of transposable element into the gene prevents the round phenotype from being produced 2 copies of wildtype: smooth 2 copies of mutant: wrinkled when you have one of both, seeds are a mix of round and wrinkled. Punnett squares requirements o Define alleles o What is the cross (parental genotypes) o List gametes o Punnett square o Answer question Key features of single gene inheritance o Genes come in pairs o For each pair of genes, alleles may be homozygous or heterozygous o Reproductive cells contain only a single allele for each gene o Each gamete is equally as likely to include either allele o Alleles are reunited at random during fertilization Mendel’s first law: the principle of segregation o In formation of gametes, the paired hereditary determinants separate (segregate) in such a way that each gamete is equally as likely to contain either member pair o Round allele is actually incompletely dominant over wrinkled allele at microscopic level. Plant needs two functional alleles to produce a completely round seed Alleles of two genes segregate independently o During meiosis, single allele is segregated into each gamete. Since gametes are haploid, they contain a single allele of each gene Dihybrid cross o Parental generation is homozygous for both traits Mendel’s second law: the principle of independent assortment o Segregation of the members of any pair of alleles is independent of the segregation of other pairs in the formation of reproductive cells o Only applies to genes on nonhomologous chromosomes or far apart on the same chromosome. Genes on the same chromosome close together can’t independently assort Punnett square practice problems Lecture 12: Transmission Genetics: Sex-Linked Inheritance and Pedigrees Allele designations o Multiple alleles Name of gene is capital letter, name of allele is superscript (ex. Blood type) o Sex-linked X^R (dominant), X^r (recessive), Y (male sex chromosome) o Drosophilia Recessive phenotype is lower case; wildtype is lower case with a superscript + + eb (recessive mutant – ebony body color); eb (wildtype body color) x (wildtype eye color – sex-linked); x (white eyed mutant – sex-linked) sex-chromosome inheritance o males are more likely to express sex-linked genes than females. Because they only have one X chromosome o Males are hemizygous for a gene on the x chromosome. Females can be homozygous or heterozygous X-linked inheritance o Thomas hunt morgan first to describe x-linked inheritance o Studied white eye mutation in drosophilia o Reciprocal crosses produced different proportions Wildtype female X white eye male White eye female X wildtype male Characteristics of X-linked inheritance o Reciprocal crosses result in different phenotypic ratios in the sexes o Heterozygous females transmit each x-linked allele to about half of her daughters and half of her sons o Males that inherit a recessive X linked allele will exhibit the recessive trait Punnet squares with x-linked genes o You must indicate that the gene is on the x-chromosome o You must differentiate between male and female o Your ratios must indicate differences between males and females in the offspring Heterogametic females o Some species have heterogametic females and homogametic males (W, Z system) Nondisjunction o Failure of chromosomes to separate and move to opposite poles of the division spindle (3 chromosomes) Patau syndrome (trisomy 13) Edward’s syndrome (trisomy 18) Klinefelter syndrome (XXY males) XYY males XXX females Turner’s syndrome (XO) Only viable monosomy in humans Testcross o Experimental cross between an individual whose genotype is known (homozygous recessive) and an individual whose genotype is not known) Predict the expected phenotype and genotype ratios if the unknown parent is homozygous dominant Predict the expected ratios if unknown parent is heterozygous o Allows you to find out if something is homozygous or heterozygous Lecture 13: Transmission Genetics: Probability Probability o Allows you to determine probablility that a certain genotype/phenotype will result from a specific genetic cross Chance that the event will happen in future Random sampling error o Probability= # of times an event occurs/total # of events o Multiplicative rule If events A and B are independent, the probability they occur together (P(A and B)) is P(A) * P(B) o Additive rule If two events don’t overlap and want the probability of something happening or something else happening, P(A or B) = P(A) + P(B) o Binomial expansion probability Used to determine probability of different combinations of outcomes Boy or girl Diseased or healthy Mutant or wildtype Dominant or recessive o Represents all of the possibilities for a given set of unordered events o o J o P= probability that the unordered outcome will occur o n= total number of events o s= number of events in one category o t= number of events in the second category o p= individual probability of x o q= individual probability of the other category Used to answer the following types of problems. Key is that there are two alternatives and the order of events doesn’t matter Probability that x will exceed a particular value Probability that x will lie between two particular values Probability and binomial expansion practice problems
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