BSC 114 BSC 114
Popular in Principles Of Biology I
verified elite notetaker
Popular in Biological Sciences
verified elite notetaker
This 19 page Bundle was uploaded by Ashley Bartolomeo on Sunday March 27, 2016. The Bundle belongs to BSC 114 at University of Alabama - Tuscaloosa taught by Edwin Stephenson in Winter 2016. Since its upload, it has received 32 views. For similar materials see Principles Of Biology I in Biological Sciences at University of Alabama - Tuscaloosa.
Reviews for BSC 114
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 03/27/16
Bio Test Number 3 Chapter 14 a. The Basics of Mendelian Genetics Genetic Nomenclature Monohybrid Crosses Dihybrid Crosses What Are Genes And What Do They Do? Gregor Mendel o “Father of Genetics” o Austrain Monk o ~1860: breeding experiments in monastery garden Genetic Nomenclature Mendel started with strains of peas with variable characters o Character: feature that differs between individuals. Example: flower color o Trait: differences in a character, eg., “purple flowers” and “white flowers” Starting strains were obtained from seed suppliers and were truebreeding o Truebreeding: all individuals are identical, and selffertilization gives rise to offspring like parents, generation after generation Peas will selffertilize, but can be made to hybridize* o Stamens (male organs) removed to prevent selffertilization o Pollen (male gametes) transferred from white to purple * Hybridize: produce offspring between genetically different strains. Hybrid: offspring of genetically different parents Monohybrid Crosses Monohybrid cross: a mating in which one character is followed Mating: P: purple x white F1: 100% purple flowers Nomenclature: o P (parental) generation o F1 (first filial): offspring of the P mating o F2: offspring of an F1 mating o “x” indicates a mating, or a cross P: purple x white F1: 100% purple F1 plants allowed to selffertilize F1: purple x purple F2: 75% purple, 25% white 3:1 purple : white ratio Six other characters gave same result: o One trait “disappears” in the F1, but reappears in the F2 o 3 : 1 ratio in F2 Mendel’s conclusion: phenomena are universal Mendel’s Explanation Every individual has two copies of each gene There are different variants (alleles) of the gene o Eg., a “purple” allele and a “white” allele In an individual with two different alleles, only one contributes to the character o The expressed trait is dominant o The nonexpressed trait is recessive Each gamete has only one of the two alleles The Genome, Genes and Alleles. An Analogy The genome contains the instructions (genetic code) to make proteins (and other molecules) A cookbook contains the instructions (recipes) to make delicious foods There are two genomes in every cell (except the X and Y chromosome genes in males) The human genome contains 20,000 – 25,000 genes, and two copies of each*. There may be variation between genes (each variant = allele), most of its harmless and part of normal genetic variation Some alleles are harmful because they result in nonfunctional proteins * Except the X and Y chromosome genes in males, and mitochondrial genes (both sexes) Phenotype: physical attributes; eg., purple and white flowers Genotypes: genetic makeup, eg., which alleles are present Let “P” represent the purple flower color allele Let “p” represent the white flower color allele o Since each individual has two alleles, there are three possible genotypes: PP (purple phenotype) (homozygous) Pp (purple phenotype) (heterozygous) Pp (white phenotype) (homozygous) o Homozygous: both alleles are the same o Heterozygous: two alleles are different Monohybrid Crosses Punnett Squares A Punnett square diagrams the outcomes of fertilization o Vertical: the 2 egg genotypes, P and p o Horizontal: the 2 sperm genotypes, P and p o 4 boxes represent the 4 possible fertilizations Probability Probability: branch of mathematics that deals with the likelihood of some event Probability that a particular result occurs: # of outcomes with a particular result / total # of possible outcomes Law of Segregation Mendel’s Law of Segregation o There are different variants (alleles) of some heritable factor* that causes a trait o In an individual with two different alleles, only one contributes to the characters (dominance / recessiveness) o During gamete formation, alleles segregate from each other, so that each gamete has one allele o Each individual receives 1 allele from each parent o *Gene: a heritable factor that causes a trait Testing The Law of Segregation if Mendel’s theory is correct, the F2 purple class should consist of: o 1/3 genotype PP o 2/3 genotype Pp What will the result be if each crossed to a white flowered plant (genotype pp)? Predict the outcomes of the following crosses: o 1. PP x pp 100% Pp (purple) o 2. Pp x pp 50% Pp (purple) 50% pp (white) Mendel found that 1/3 of the F2 purples gave the #1 result and 2/3 gave the #2 result o The theory has passed a test Laws and Theories A scientific theory is an explanation of a phenomenon that cannot be observed directly, accounting for the “inner workings” of the system Mendel’s Law of Segregation is actually a theory How do you prove that a theory is true? o You can’t. It is impossible to prove that something is true o You test predictions of the theory (eg., purple F2s should be 2/3 Pp and 1/3 PP). if the predictions are confirmed the theory is increasingly likely to be true o You show that all other “rival” theories are not true. Eventually one is left with only one surviving theory, which is widely accepted as correct. These are sometimes renamed as “laws” Dihybrid Crosses Previous crosses were monohybrid crosses: 1 character only Dihybrid cross: 2 characters P: yellow, round seeds x green, wrinkled seeds F1: 100% yellow, round phenotype (YyRr genotype) (yellow dominant to green; round dominant to wrinkled) F1s selffertilize F2: 9/16 yellow, round; 3/16 yellow, wrinkled; 3/16 green, round; 1/16 green, wrinkled 9:3:3:1 ratio Law of Independent Assortment Independent Assortment: segregation of alleles for one gene does not affect segregation of alleles of a second gene Ratios are consistent with complete independence: o ¾ of the F2 are yellow o ¾ of the yellow F2s are round o ¾ x ¾ = 9/16 of the F2 are both yellow and round Alleles for two independent genes segregate independently during gamete formation o Example. YyRr parent: gamete with Y can have R or r, with equal probability Probability, Genotypes and Phenotypes For independent genes, probability of any dihybrid phenotype/ genotype can be calculated from each monohybrid probability o Yy x Yy. What fraction of progeny have dominant phenotype? ¾ o Rr x Rr. Same question. ¾ o YyRr x YyRr. What fraction of progeny have both dominant phenotypes? ¾ x ¾ = 9/16 Same Principles Apple to Trihybrid and More Complex Cases Yellow is dominant to green, round is dominant to wrinkles, purple is dominant to white In a selfcross of a yellow, round heterozygote, what fraction of the progeny will have the phenotype: o Green, round ¼ green x ¾ round = 3/16 green, round In a selfcross of a yellow, round, purple heterozygote, what fraction will have the phenotype: o Yellow, round, purple ¾ yellow x ¾ round x ¾ purple = 27/64 o Green, round, white ¼ green x ¾ round x ¼ white = 3/64 B. Complex Phenotypes and Human Genetics Major Topics Complex phenotypes Human genetics Complex Phenotypes Some phenotypes do not conform to simple dominance / recessiveness o Incomplete dominance o Codominance o Multiple alleles o Epistasis o Polygenic inheritance Transmission of genes follows Mendelian rules; genotype: phenotype relationships are different Incomplete Dominance Incomplete dominance: heterozygote has an intermediate phenotype o Example: Snapdragon Segregation of alleles follows Mendelian rules White spotting in cats o SS cats are mostly white o Ss cats have white on tail, legs, chest, ears o ss cats have no white at all Codominance Codominance: the heterozygote has properties of both homozygotes o Example: human MN blood groups o Gene encodes glycoproteins expressed on surface of red blood cells. Two alleles M and N, encode two protein variants o MM individuals have only M protein; MN individuals have only N proteins; MN heterozygotes have both Multiple Alleles Multiple alleles = genes have more than 2 alleles o Example: three alleles of the I gene, Ia, Ib and i. i is recessive to the other two o IaIa and Iai have type A blood o IbIb and Ibi have type B blood o Ia and Ib are dominant to i o ii have type O blood o IaIb have type AB blood, that is Ia and Ib are codominant ABO Blood Types The I gene encodes and enzyme that synthesizes an extracllular carbohydrate o Ia enzyme produces “A” type carbohydrate o Ib enzyme produces “B” type carbohydrate o i enzyme produces no carbohydrate ABO Blood Types and Transfusion ABO blood group is important in transfusion. Immune system produces antibodies against foreign molecules, but not against self o A phenotype individuals have antiB antibodies o B phenotypes individuals have antiA antibodies o O phenotype individuals have both antiA and antiB antibodies o AB phenotype individuals have antibodies to neither A or B Epistasis Epistasis: genotype at one gene affects phenotypic expression of a second gene Shows up as a deviation from 9:3:3:1 ratio in dihybrid crosses Example: B and C genes affect mouse hair color. 9:4:3 ratio. The white genotype cc prevents expression of the black/ brown phenotype Polygenic Inheritance Polygenic inheritance: multiple genes affect the same character and have additive effects on phenotype o Examples: Skin color Height Human Genetics Pedigree Analysis Traits are followed through families using pedigrees Key: o Circles = females, squares = males o Affected = filled, unaffected = unfilled Distinguishing dominant and recessive traits: o Dominant traits are always expressed in one (or more) of the parents o Recessive traits are often not expressed by either parent Genetic Diseases Most human genetic diseases are recessive, caused by genes that produce defective proteins: o Cystic fibrosis: chloride channel protein o TaySachs disease: enzyme that degrades lipids o Sickle cell disease: hemoglobin protein Some conditions and diseases are dominant: o Achondroplasia (dwarfism) o Huntington’s disease (nervous system deterioration) Genetic Counseling A couple’s first child has cystic fibrosis, although neither of the parents do. What is the probability that the couple’s second child will have cystic fibrosis? To solve the problem: o Determine the parents’ genotypes Each must be Cc, since their first child was cc, but neither parent has the disease. Both parents are carries o What is the probability that the two heterozygotes will produce a child with genotype cc? ¼ Prenatal diagnosis o Sample of fetal tissue is obtained: amniotic fluid (contains fetal cells), or from chorionic villus (fetal portion of placenta) o Karyotype analysis (detects wrong # or other chromosomal defects o DNA analysis to detect specific mutations Learning Goals Genetic nomenclature Dominance and recessiveness Monohybrid crosses and the Law of Segregation o Predict genotypes and phenotypes for monohybrid crosses o Use of Punnett squares Dihybrid crosses and the Law of Independent Assortment o Predict genotypes and phenotypes for dihybrid crosses Genotype: phenotype relationships for o Incomplete dominance o Codominance o Multiple alleles o Epistasis o Polygenic inheritance Human genetics o Symbols and practice of pedigree analysis o Detecting dominant and recessive traits by pedigree analysis o Predictions in genetic counseling Chapter 15 Chromosomes a. Sexlinkage Major Topics Mendel and chromosomes Sexlinked inheritance Mendel and Chromosomes Genes ala Mendel and chromosomes during meiosis: o Segregation Organisms have two copies of each gene, but gametes contain only one Somatic cells contain two copies of each chromosome, but gametes only one o Assortment Genes segregate independently of each other Nonhomologous chromosomes segregate independently o Chromosome Theory of Inheritance (early 20 century): Genes reside on chromosomes Alleles on chromosomes segregate according to Mendel Sexlinked Inheritance Drosophila as a genetic model Morgan chose Drosophila to examine genetic mechanisms: o Short generation time (~10 days) o Hundreds of offspring from one pair o Survives well in the lab First genetic variant: white eyes (vs. normal red eyes) Drosophila genetic nomenclature: o White eye allele w (recessive) o Wildtype allele w+ (dominant; red eyes) P: redeyed female x whiteeyed male F1: all red (conclusion?) o Red is dominant to white F2: (F1 brother x F1 sister) o Females: 100% red o Males: 50% red, 50% white Explanation: females and males are genetically different, XX and XY respectively. The Y chromosome is not a functional homolog of the X* o Males are haploid for genes on the X X and Y = sex chromosomes Other chromosomes = autosomes *Although the X and Y are not homologs (do not contain the same genes) they segregate during meiosis. Sperm cell has an X or a Y, never both Conclusion from previous results: the white eye gene is on the X chromosome o Provided the first conclusive evidence in support of the chromosomal theory of inheritance, that genes reside on chromosomes A. XY: (humans, fruit flies) B. XO: females have two X; males have one X (no Y chromosome) C. ZW: females are WZ, males are ZZ (birds, butterflies) D. Haplodiploidy: females are diploid, males are haploid (bees, wasps, ants) Humans Key feature of sexlinked traits: fathers transmit X chromosome to daughters, but never to sons o A. An effected father produces heterozygous daughters o B. A heterozygous mother produces 50% affected sons, but 100% normal daughters (although ½ of daughters will be carriers) Sexlinked traits are more often observed in males (because have only 1 X chromosome) o Duchenne muscular dystrophy o Hemophilia o Redgreen color blindness Mammals X Inactivation Female mammals inactivate one X, so that each cell has only one active X Inactive X = barr body Inactivation occurs early in development, is random, and pattern of inactivation is inherited thru subsequent mitoses Tortoiseshell and calico cat: alleles for black and orange fun on X. Each patch of black fur represents a “clone” of cells in which the orange allele X has been inactivated o White fur is due to a separate gene (epistasis) B. Linkage and mapping. Aneuploidy Major Topics Linkage, crossing over and gene mapping Aneuploidy and alterations in chromosome structure Mendel and Chromosomes Independent Assortment Mendel’s Law of Independent Assortment o P: YYRR x yyrr F1: YyRr 100% o F1 progeny are self crossed o F2: 9:3:3:1 phenotypic ratio What is the outcome if the F1 heterozygote is testcrossed? o Test cross: cross to an individual homozygous for recessive allele(s) A test cross: o F1 YyRr x yyr (yyrr = the “tester strain”) Much simpler since the tester strain produces only 1 gamete type Dependent Assortment What would happen if allele pairs C and D and c and d behaved as inseparable units (“dependent” assortment) P: CCDD x ccdd F1: CcDd 100% Cross F1 CcDd x ccdd F2: CcDd 50%; ccdd 50%; Ccdd 0%; ccDd 0% (eg., 1:1:0:0 ratio) Partially Dependent Assortment In practice, some genes behave as if intermediate, neither independent nor completely dependent P: EEFF x eeff F1: EeFf 100% Cross F1 EeFf x eeff F2: EeFf 40%; eeff 40%; Eeff 10%; eeFf 10% This was the result in early genetic experiments, but the exact numbers if the F2 progeny vary depending on the gene pair Linkage P: gray body, normal wings x black body, vestigial wings F1: all gray body, normal wings x black body, vestigial wings F2: 42% gray body, normal wings; 41% black body, vestigial wings; 9%gray body, vestigial wings; 8% black body, normal wings Explanation: the black and vestigial genes are on the same chromosomes Genes on the same chromosome do not sort independently but tend to be inherited together These genes are linked, specifically the b+ and vg+ alleles and the b and vg alleles are linked Linkage: genes are on the same chromosome Linkage, crossing over and gene mapping Mendel’s Law of Independent Assortment is true is genes are on different chromosomes For genes on the same chromosome: o Usually deviate from independence (no classes are 25%) due to linkage o Usually deviate from “dependence” (no classes are 50% or 0%) due to crossing over *Linkage: genes are on the same chromosome Crossing over: breakage and rejoining between nonsister homologous chromatids Occurs in prophase I F2 progeny classes are neither 25% (expected of Independent Assortment) nor 50% / 0% (expected of complete dependent assortment) The two most frequent F2 classes represent the phenotypes of the P generation, and are called the parental classes The two least frequent F2 classes are called the recombinant classes Morgan: the extent of crossing over is proportional to distance along the chromosome between the genes Can be used to map gene positions The recombinant (nonparental) classes are 391/2300 = 17% of the total 1% recombination = 1 map unit = 1 centimorgan Distance: black and vesitgal genes are 17 map units = 17 centimorgans apart Map distances are additive An example: genes A and B are 9 cM apart, and B and C are 3 cM apart. How far apart are A and C? It depends on the relative positions of the genes o 12 cM if the gene order is ABC o 6 cM if the gene order is ACB Experiments like the previous are used to construct a linkage map Genes that are very far apart on the same chromosome behave as if they are on different chromosome (eg., obey the Law of Independent Assortment) Example: o AaBb and aabb (parental classes): 55% of progeny o Aabb and aaBb (recombinant classes): 45% of progeny o Genes are 45 cM apart But suppose the genes were slightly farther apart: o AaBb and aabb (parental classes): 50% of progeny o Aabb and aaBb (recombinant classes): 50% of progeny o Recombinant classes never exceed 50% o Genes are neither >50 cM apart, or on different chromosomes Mendel and Chromosomes Why didn’t Mendel discover linkage? o Studied 7 traits, which are all on separate chromosomes, except two traits that are very far apart on the same chromosome Aneuploidy and Alterations in Chromosome Structure Aneuploidy: too few or too many copies of a chromosome o Monosomy (1 chromosome) o Trisomy (2 chromosomes) Caused by error in chromosome distribution during meiosis (nondisjunction). Affected gamete will produce an aneuploidy zygote Most common autosome aneuploidy: Trisomy 21 = down syndrome Aneuploidies for other autosomes are very rare. Why? Aneuploidies for larger chromosomes are embryonic/ fetal lethal Sex chromosome aneuploidy. Tolerated better, because of inherent mechanisms to adjust gene dose o XXY Klinefelter Syndrome. Male sex organs, some female development. Testes undeveloped and sterile. Mental retardation common o XYY tall, but otherwise normal o XO Turner Syndrome (only viable monosomy in humans) Sterile females o XXX normal female Deletion: portion of chromosome is missing. Many/ most have dominant mutant effects o Cri du chat syndrome: missing end of chromosome 5. Individuals have strange catlike cries as infants. Mental retardation, facial abnormalities, etc. Translocation: ends of two chromosomes are exchanged. Break in chromosome may disrupt important gene o Philadelphia chromosomes causes a form of leukemia Learning Goals The relationship between chromosomes behavior and genes are predicted by Mendel The Genetics of sex determination and sexlinked traits Mechanism and an example of X inactivation Linkage o Exceptions to Independent Assortment and the evidence that genes reside on chromosomes o How linkage is detected in crosses o How quantitative analysis of linkage is used to construct genetic maps o Why distant genes on the same chromosome seem to be unlinked Human genetic traits caused by aneuploidy and alterations of chromosome structures Chapter 16 The Molecular Basis of Inheritance Major Topics Genes are made of DNA DNA structure DNA synthesis DNA and chromosomes Genes Are Made of DNA Genetic character: bacterial pathogenicity o 1. One strain, “S” of bacterium Streptococcus causes a lethal infection o 2. Another strain, R does not o 3. Heat killed S is not dangerous o 4. Exposure to extract from S (extract = nonliving “soup” of cell contents) transforms the harmless R strain into a pathogenic strain (transformation) A scientific theory to explain the previous results (theory = an explanation of underlying principles): o Strain S has the “pathogenicity” allele. Strain R has the “nonpathogenicity” allele (bacteria are haploid, so each has only 1 allele) o The pathogenicity allele in the S extract replaces the nonpathogenic allele in the living R cells, making them pathogenic By destroying various molecules in the extract and testing for pathogenicity, DNA was identified as the genetic material DNA Structure Components and Polynucleotides Each nucleotide contains 3 components: o Base (T, A, G or C) o Deoxyribose sugar o Phosphate Adjacent nucleotides are joined into polynucleotides through sugarphosphate bonds Each DNA strand has a polarity based on sugar orientation, 5’ to 3’ Double Strands Two polynucleotide strands pair side by side to make a doublestranded DNA molecule Base Pairing Parallel strands are held together by hydrogen bonds between complementary bases: o AT o GC Pairing is always 1 purine and 1 pyrimidine Antiparallel The two strands have opposite 5’ to 3’ polarity (are antiparallel) Helical Structure Two strands wind around each other in 3D helical structure, the double helix Watson and Crick’s structure was based on two types of data: o Chargoff’s Rules. Analytical chemistry shows that amount of A = T and amount of G = C o Xray diffraction experiments of Rosalind Franklin, showed helical structure, spacing of bases, and position of sugar and phosphates on outside Measurements DNA molecules are measured in base pairs One base pair = a pair of complementary nucleotides DNA molecules are huge: average human chromosome = 1 x 10^8 bp o Kilobase (kb) = 1000 bp o Megabase (Mb) = 10^6 bp DNA Synthesis Overview Simplified overview: 2 strands separate and each strand acts as the template for synthesis of the complementary strand o Old strand in dark blue. New strand in light blue o Result: 2 identical doublestranded molecules Biochemistry Nucleotide to be added is a triphosphate, eg., dTTP dTTP dTMP (added to DNA) + PPi Addition occurs only at the 3’ end of the growing strand Catalyzed by enzyme DNA polymerase (enzyme that adds nucleotides) Replication Fork The two “old” strands separate in local regions, producing a replication “bubble” Each bubble has two replication forks where new nucleotides are added At the replication fork new nucleotides are added one at a time Each new strand is synthesized 5’ to 3’ At each replication “fork” one strand is synthesized toward the fork, and one away from the fork As synthesis continues, more “old” DNA becomes singlestranded, and the fork moves Because DNA is synthesized only 5’ to 3’, one strand is synthesized discontinuously, that is, in short fragments o Leading strand = continuous strand o Lagging strand = discontinuous synthesis. Short fragments are called Okazaki fragments Chromosome Level Replication forks move apart until the entire molecule is replicated o Circular chromosomes (bacteria): single origin of replication o Linear chromosomes (eukaryotes): many origins of replication Each chromatid contains 1 dsDNA molecule (1 old strand; 1 new strand) DNA replication occurs during S phase: 2 DNA molecules = 2 chromatids After mitosis, each chromosome = 1 chromatid = 1 DNA molecule DNA Repair Mispaired or damaged nucleotides result from: o Chemical or light damage, especially UV light o Mistakes in DNA synthesis Excision: a nuclease recognizes and removes the mispaired region Repair: DNA polymerase fills in the gap by the synthesis mechanism DNA and Chromosomes Each eukaryotic chromatid = 1 ds DNA molecule o Average chromosome contains a 4 cm long DNA molecule. All DNA per diploid G1 cell = 2 meters o DNA is highly compacted by coiling and packaging with proteins Chromatin: combination of DNA & proteins o Histones: small basic proteins (associate via opposite charge to acidic DNA molecule) o Nucleosome: complex of 8 histone proteins o DNA is wound around nucleosomes: “beads on a string” Nucleosome fibers are further wound into 10nm and 30nm fibers and arranged in loops in the chromosome Learning Goals The evidence that DNA is the genetic material The structure of nucleotides and polynucleotides Double stranded DNA structure o Complementary base pairing; polynucleotide polarity; antiparallel nature of the DNA double helix The mechanism of DNA replication o Why the complementary structure of DNA allows its replication o What happens at the replication fork o Replication at the level of chromosomes DNA repair Chromatin structure and packing into chromosomes Chapter 17 Gene Expression a. Transcription and RNA processing Major Topics Genes and proteins Transcription RNA processing Genes and Proteins The properties of a protein (shape, amino acid composition, folding, catalytic function, etc) depends on the sequence of amino acids that compose the protein. How is that determined? Protein function (shape, structure, enzyme specificity, etc) depends on… …The sequence of amino acids, ie., the primary structure of the protein (translation) The primary structure of the protein is determined by the sequence of a messenger RNA (transcription & RNA processing) The sequence of a messenger RNA is determined by the sequence of a gene “One Gene : One Protein” Experiment Beadle and Tatum Wildtype* fungus Neurospora grows on minimal medium (medium lacks arginine) Conclusion: cells can make their own essential compounds (including arginine) *Wildtype: the normal strain, ie., one that is found in nature Several mutant* strains are identified that cannot grow on minimal medium Conclusion: mutants are defective in making some essential compound *Mutation: defect in a gene. A mutant strain contains a mutation in a gene The 3 mutant strains survive on minimal medium supplemented with arginine Conclusion: the mutants are not able to synthesize arginine. The mutant defect is the inability to synthesize arginine Some of the mutant strains survive on media supplemented with ornithine and citrulline (molecules similar to arginine) Differences between the strains define a biochemical pathway Interpretation: arginine is synthesized from precursor molecules, as follows: Unnamed precursor ornithine citrulline arginine Each step () is catalyzed by an enzyme, which is represented by a class of mutants; each class is in a different gene Since the mutants differ in growth on ornithine and citrulline, it is possible to determine the order in which the genes act Therefore, each gene encodes one enzyme “One gene : one enzyme”* encode: contains the instructions for * Later expanded to “one gene : one protein” Transcription and Translation Flow of genetic information: DNA RNA protein o Transcription: synthesis of RNA using DNA template RNA processing: RNA molecules are modified o Translation: synthesis of protein using RNA template Transcription and translation are similar in prokaryotes and eukaryotes except: o In eukaryotes, the original RNA copy is often modified before becoming a messenger RNA (RNA processing) o In eukaryotes the mRNA must exit the nucleus (thru nuclear pores) Transcription copies a limited region of one DNA strand o Gene = a segment of DNA that is transcribed o Each gene has a start signal and stop signal o Some DNA segments are not part of any gene and are not transcribed (“spacer DNA”) RNA Structure RNA is usually single stranded o Contains ribose instead of deoxyribose sugar o Contains uracil (U) base, instead of thymine (T) Transcription Transcription Units Chromosome = long piece of DNA, 10^6 – 10^9 base pairs in length Transcription units (= genes): shorter segments that are transcribed Spacer DNA is not transcribed RNA polymerase: the enzyme that carries out transcription Initiation Each transcription unit is defined by a start signal (promoter) and a stop signal (terminator) Each signal is a particular DNA sequence Other proteins are required for transcription initiation: bind to promoter and nearby sequences Very complex in eukaryotes: cluster of many accessory proteins forms at promoter Simpler in prokaryotes: one or a few proteins Elongation Short segments of DNA (1020 nts) become transiently single stranded and act as template for RNA synthesis RNA molecule is extended one nucleotide at a time, using nucleoside triphosphates, analogous to DNA synthesis Direction of RNA synthesis: always 5’ to 3’ Termination Transcription ends when (or just after) the RNA polymerase transcribes the terminator sequence RNA Processing In prokaryotes, the product of transcription is the mRNA, often used without further modification More complex in eukaryotes o Product of transcription is a premRNA or primary transcript undergoes RNA processing o mRNA is exported from nucleus (1) Capping and (2) Polyadenylation Capping: a modified nucleotide is added to the 5’ end of the RNA o Purpose: necessary for ribosome attachment. May also affect mRNA stability Polyadenylation: 50200 adenosine nucleotides are added to the 3’ end o Purpose: increases mRNA stability (3) Splicing RNA splicing: internal segments of RNA molecules are removed, and adjacent parts joined back together o Removed segment = intron o Retained segment = exon Splicing continued o Organelle where splicing takes place = spliceosome Purpose: introns encode meaningless sequences and must be removed before translation o But why do genes contain introns? Summing Up Only 1.5% of the human genome consists of proteincoding sequences The non proteincoding portion of the genome consists of: o ~26% intron sequences o ~72% spacer DNA and genes that encode RNA products B. Translation, the Genetic Code and Mutations Major Topics Translation Genetic Code Mutations Translation Translation: protein synthesis by ribosomes, based on sequence of bases in messenger RNA Participants: o Messenger RNA: produced by transcription (plus RNA processing in eukaryotes) o Ribosome: molecular “machine” that carries out translation o Transfer RNA: “Adapter” molecule in translation Which is Transcription and Which is Translation? Transcribe: same language converted from one medium to another. An interview is transcribed, it., converted from oral to written o Transcription: DNA to RNA (different forms of the same nucleic acid “language”) Translate: one language to another., eg., Spanish to English o Translation: nucleic acid “language” to protein “language” Translation Participants mRNA Messenger RNA: RNA that is “read” by ribosome to make a protein Each set of 3 nucleotides specifies 1 amino acid o Group of 3 nucleotides in mRNA = codon Genetic Code Codon: a 3 nucleotide sequence in an mRNA molecule o 61 codons each specify one of 20 amino acids o Methionine (Met) is usually the first amino acid; AUG is usually the start codon o 3 codons (UAA, UAG and UGA) have no amino acid; act as “stop” signals (stop codons) Ribosomes Ribosomes: molecular “machine” that carries out translation, ie., synthesizes proteins o Complex of several dozen proteins and 3 RNA molecules o 2 ribosomal subunits, large and small Transfer RNAs Transfer RNAs: act as “adapter” molecules during translation o RNA molecules of ~80 nucleotides o Folded due to intramolecular base pairing o Amino acid attached to 3’ end by enzyme aminoacyl tRNA synthase o One or more tRNAs for each maino acid o Anticodon: a 3 nucleotide region that base pairs with the codon on mRNA during translation Mechanism Initiation: ribosome begins translation at 3 nucleotide sequence AUG, which codes the amino acid nethionine Elongation: ribosome moves 3 nt, reading the next 3nt sequence and attaching the 2 amino acid to Met o Continues over and over until termination occurs Termination: occurs at a 3nt sequence that does not specify any amino acid Simplified Example mRNA: 5’ AACGUGCAUGCCUGAAUUCUGAGCGAU 3’ Initiation: ribosome / MettRNA finds an AUG codon. Two different ways o Prokaryotes: recognizes and binds to a sequence close to the AUG codon o Eukaryotes: binds to the 5’ end of mRNA and scans to find the first AUG codon Elongation: ribosome moves 3nt to the next codon CCU, ProtRNA recognizes CCU codon, and ribosome attaches Pro to Met Elongation: ribosome moves 3nt to the next codon GAA, GlutRNA recognizes GAA codon and ribosome attaches Glu to MetPro Elongation: ribosome moves 3nt, tRNA recognizes Phe codon and attaches Phe to Met ProGlu Termination: ribosome moves 3nt to UGA, terminates translation and releases protein MetProGlu_phe How Ribosomes Work 3 functional sites in the ribosome, E, P and A 1. P site: tRNA with growing polypeptide attached. tRNA with its amino acid enters the A site 2&3. peptide bond between the polypeptide and the amino acid tRNA in the A site. GTP hydrolysis provides energy. Polypeptide now attached to A site tRNA 4. Ribosome moves 3 nucleotides down the mRNA (GTP GDP). Polypeptide tRNA now in the P site. tRNA exits from the E site Odds and Ends Protein folding: protein folds into its 3D shape during synthesis, and afterward Polyribosomes: a single mRNA may be translated by more than one ribosome at the same time Intracellular targeting: proteins destined for rER, nucleus, mitochondria, chloroplast, etc. have targeting signals (ie., “address labels”) as part of the amino acid sequence Transcription/ Translation Summary Types of RNAs Messenger RNA: “template” for translation, nucleotide sequence specifies amino acid sequence Transfer RNA: accessory molecule in translation Several others not discussed: o Ribosomal RNA, small nuclear RNA, SRP RNA All RNAs are synthesized by transcription Initiation and Termination Initiation/ termination of transcription and translation are at different sequences (RNA processing omitted for simplicity) Mutations Mutation: abnormal genetic information Types of mutations o Large scale: aneuploidy, deletions and translocations o Small scale: point mutations (one or few nucleotides) Point Mutations Substitutions Substitution: one nucleotide substituted for another. Types of substitutions, based on effect on protein o 1. Missense mutation: change in 1 amino acid (example: sicklecell disease) More substitutions o 2. Nonsense mutation: codon changed to a stop codon (UAA, UGA or UAG), causing premature termination of the protein o 3. Silent mutation: nucleotide change has no effect on amino acid sequence Frameshift Mutations Frameshift mutation: 1 or 2 base insertions or deletions o Protein reading “frame” is based on translation start site Silly example: o The big red bug ate our mom Delete first b: o The igr edb uga teo urm om Insert new letter: o The big red bux gat eou rmo m Disruption in reading frame causes wrong amino acid at all points “downstream” (= frameshift) o Most sever if at 5’ (beginning) end of gene/ mRNA Causes Mutagen: a chemical or condition that produces mutations. Do so by breaking DNA, or damaging bases o Examples: chemicals in cigarette smoke, UV light, Xrays, etc. Multiple Definitions of “Gene” Mendel: a gene is a discrete heritable unit that produces a phenotype Beadle and Tatum: one gene contains the instructions to produce one enzyme (or protein) Modern view: a gene is the DNA sequence that encodes a protein or RNA product Learning Goals Experimental evidence in support of the one gene – one enzyme theory The mechanism of transcription o Promoters and terminators o RNA polymerase o The biochemistry of nucleotide addition RNA processing Capping – what it is and what it does Polyadenylation – what it is and what it does Splicing – what it is and what it does o Introns and exons o Organelles spliceosome The identification and basic structure of the components of translation: messenger RNA, ribosome, transfer RNA Use the codon table to determine the protein sequence of a sample mRNA o Special codons for initiation & termination Mechanisms of translation Mutation: terminology and consequences of mutations o Substitution: missense, nonsense, silent o Frameshift o Mutagens
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'