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AGRY 320 Exam 1 Study Guide

by: Sean Anderson

AGRY 320 Exam 1 Study Guide AGRY 320

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These notes cover all the material we have covered thus far and includes detailed diagrams, images, and charts aiding in the understanding of the material.
Mohsen Mohammadi
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This 30 page Study Guide was uploaded by Sean Anderson on Sunday October 2, 2016. The Study Guide belongs to AGRY 320 at Purdue University taught by Mohsen Mohammadi in Fall 2016. Since its upload, it has received 2 views.


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Date Created: 10/02/16
Sean Anderson Professor Mohammadi AGRY 320 10/05/16 AGRY 320 Exam 1 Study Guide Transmission of Genes, Gene Actions, Mendel’s Laws  Learning objectives: o Learning how genes transmit from one generation to another o Learning mode of action (gene action) o Learning how genetics involves probability and how we draw the space of outcomes in genetic crosses o Learning first and second laws of Mendel  Terms we will cover: o Gene: a unit of heredity that is transferred from a parent to offspring and is held to determine some characteristic of the offspring. Gene is technically defined as a distinct sequence of nucleotides forming part of a chromosome, the order of which determines the order of monomers in a polypeptide or nucleic acid molecule which a cell (or virus) may synthesize o Genotype: the genetic constitution of an individual organism. o Locus: a particular position, point, or place o Allele: One of two or more alternative forms of a gene that arise by mutation and are found at the same place on a chromosome o Phenotype: the set of observable characteristics of an individual resulting from the interaction of its genotype with the environment o Dominance: the phenomenon whereby, in an individual containing two allelic forms of a gene, one is expressed to the exclusion of the other o Diploid: containing two complete sets of chromosomes, one from each parent o Haploid: having a single set of unpaired chromosomes o Homozygous: If you're homozygous, you've got a pair of matching alleles, which are the two genes that control a particular trait. If both your alleles that determine blood type are O, you're homozygous — and you've got type O blood o Homozygote: an individual having two identical alleles of a particular gene or genes and so breeding true for the corresponding characteristic o Heterozygous: If you're heterozygous for a gene of interest, you've got different alleles for that gene o Heterozygote: An individual having two different alleles of a particular gene or genes, and so giving rise to varying offspring o Codominance: A condition in which both alleles of a gene pair in a heterozygote are fully expressed, with neither one being dominant or recessive to the other  Gene action in inheritance o Dominant—A or B o Recessive—a or b o Co-dominant—AB o Additive—Aa or Bb  Transmission of genetic factors from parents to offspring 1. A1 x A1  A1A1 2. A2 x A2  A2A2 3. A1 x A2  A1A2 4. A1A2 x A1A  A1A1 or A1A2 5. A1A2 x A1A2  A1A1 or A2A2 or A1A2  We show the game of probability using Punnett square o Three genotypic outcomes  AA or Aa or aa o If dominance is the gene action, then two phenotypic outcomes  A or a  First Law of Mendel o The principle of segregation (First Law): The two alleles separate from each other in the formation of gametes. Half the gametes carry one allele, and the other half carry the other allele o A1A2 x A1A1  A1A1 or A1A2  TT x Tt  TT, TT, Tt, tt  Second Law of Mendel o Mendel’s second law is also known as the law of independent assortment o The law of independent assortment states that alleles of one gene sort into gametes independently of the alleles of another gene  4 possible outcomes for genotypic classes Monohybrid, dihybrid, and trihybrid   Learning objective: o Learning basics of monohybrid, dihybrid, & trihybrid o Understanding the independence of genes in inheritance o Learning Punnett square to derive genotypic and phenotypic ratios  Terms we will cover: o Monohybrid o Dihybrid o Trihybrid o Allele o Gametic space o Punnett square o Genotypic o Phenotypic ratios o Segregating locus o Testcross o The genetics of blood type  Monohybrid o Homozygous x homozygous  Aa x aa  Aa  Progeny: 100% Aa o Homozygous x heterozygous  Aa x aa  Aa and aa  Progeny: 50% Aa 50% aa o Heterozygous x heterozygous  Aa x Aa  AA, aa, Aa  Progeny: 25% AA, 50% Aa, 25% aa  Dihybrid o Homozygous x homozygous  AABB x aabb  AB x ab  AaBb  Progeny: 100% AaBb o Hetero/Homo x homozygous  AaBB x aabb  AB, aB x ab  AaBb or aaBb  Progeny: 50% AaBb, 50% aaBb o Heterozygous x heterozygous  AaBb x AaBb  AB, ab, Ab, aB x AB, ab, Ab, aB  A_B_, A_bb, aaB_, aabb  Progeny: 9:3:3:1  Trihybrid o Heterozygous x heterozygous  AaBbDd x AaBbDd  ABD, ABd, AbD, Abd, aBD, aBd, abD, abd x ABD, ABd, AbD, Abd, aBD, aBd, abD, abd   How may gametes can an individual produce? 1. When an individual is homozygous in a locus, that locus will not produce variation in gametes  For example, an AA individual will only produce 1 type of gamete. Likewise, for the case of two loci, aaBB genotype will only produce aB gamete 2. A locus can only have a role in producing variation when it is in heterozygous state  For instance, an individual with Aa genotype is able to produce two types of gametes at equal frequency 3. When an individual is heterozygous in a locus we say that individual “is segregating for that locus / gene” 4. Look at the genotype formula given for an individual. Count the number of segregating loci”. Two to the power this number is the total possible types of gametes that this individual can ever produce  Example:  Aa – 2 = 1 2  AaBb – 2 = 4 3  AaBbDd –2 = 8  Blood types: ABO and Rh o Blood types: A, B, O, AB Mom Dad Genotype Blood Type A A AA A B B BB B O O OO O A O AO A B O BO B A B AB AB o 4 different variations but only 3 alleles o A and B is dominant over O Hypothesis Testing in Science and Chi-square Test   Learning objectives: o Mechanism of inference in science o Using probability in establishing inference o Essentials of hypothesis testing o Chi-squared test to evaluate deviations from expectations  Terms we will cover: o Hypothesis o Common outcomes under a hypothesis o Rare outcomes under a hypothesis o P-value o Expectation of ratios o Deviation from expectations o Mechanism of a Chi-squared test  How a chi-square test works? o (O-E)/E  (O-E) /E(amount of vi= (O-E) /E + (O-E) /E + (O-E) /E +…. 2 2 o X = Σ (O-E) /E  O = frequencies observed  E = frequencies expected  Σ = the sum of o Example:  Aa is crossed with aa and 120 progeny is produced. According to the segregation law of Mendel, 50% of progeny should be Aa and another 50% should be aa. The observed progeny were not exactly at the expected ratio of 50% : 50% (which would make 60 Aa: 60 aa). Instead, a ratio of 80 Aa versus 40 aa was observed. Table below shows how to calculate the Chi-square statistics which is named X . Aa aa Observed numbers (O) 80 40 Expected numbers (E) 60 60 O-E 2 20 -20 (O-E) 400 400 2 (O2E) /2 6.67 6.67 X = 6.67 + 6.67 = 13.34 Degrees of freedom: number or variable -1 2-1 = 1  The way we judge o If 1:1 is a true hypothesis, the amount of deviation should be really small and therefore, negligible o If instead we obtain an X statistics that is large enough this is a good indication that the 1:1 ratio is statistically violated  Chi-square test example o A wheat genotype is resistant to stem rust disease (R). Another genotype is susceptible (S). We cross R with S by emasculating S and fertilizing egg produced by S with pollen from R. If the genotype of R plant is A1A1 and that of S plant is A2A2, then the progeny resulting from R x S cross are all A1A2 o A1A1 (R) x A2A2 (S)  A1A2  A1A2 x A1A2 o Progeny: 25% A1A2, 50% A1A2, 25% A2A2 Epistasis between Loci Interactions   Learning objectives: o The distinction between “gene action” and “gene interaction” o Deviations from simple Mendelian ratios o Different types of epistatic deviations and their interpretation  Terms we will cover: o Allelic interaction o Gene action o Gene interaction o More than one gene involved in determination of a trait o Deviation from Mendelian ratios o Complementary action (Duplicate recessive epistasis) o Duplicate dominant genes o Duplicate genes with cumulative effect o Dominant epistasis o Recessive epistasis  The distinction between “gene actions” and “gene interaction” o Gene action is the interactions between alleles of ONE gene which often referred to as:  Dominance  Recessive  Co-dominance  Partial dominance  Additive o Gene interaction refers to the interrelation between the effects of TWO OR MORE genes  One gene involved in the control of trait o Intermediate product  substrate  final product trait o Enzyme activity 2  No enzyme (aa)  Yes enzyme (AA & Aa)  Gg x Gg  GG, Gg, gg  Complementary action (duplicate recessive epistasis) (9:7) o Two genes one trait relationship: two enzymes is needed in the pathway o Substrate  Intermediate productsubstrate final product  trait  Enzyme activity 1 Enzyme activity 2  No enzyme (aa) No enzyme (aa)  Yes enzyme (AA & Aa) Yes enzyme (AA & Aa) o The flower color is encoded by two loci CCPP  Key: Color is only produced if both genes have at least one dominant allele. Therefore, 9 out of 16 types of genotypes are colored and the remaining 7 are white  The ratio will therefore be 9:7 (as opposed to 9:3:3:1)  Duplicate dominant genes (15:1) o Intermediate product  substratefinal producttrait o Enzyme activity 1  No enzyme (bb) or (aa)  Yes enzyme (BB & Bb) or (AA & Aa) o In this scenario ONE trait is being controlled with two independent gene. Dominant alleles of only ONE gene IS ENOUGH for green phenotype and OTHERWISE (no dominant allele present) the phenotype is brown o Duplicate dominant Genes 15:1. From two loci, at least one loci needs to have dominant allele to express the trait. When both loci have recessive alleles the alternative phenotype is expressed  Dominant epistasis (12:3:1) o Dominant epistasis—When dominant allele ‘A’ masks the expression of ‘B’. In this case, ‘A’ is epistatic gene and ‘B’ is hypostatic gene. B expresses only when (A is not present) ‘aa’ is present. Therefore, in 9: 3: 3: 1 ratio both 9 and 3 are expressing the ‘A’ gene the ratio is now 12:3:1 o Y x y  Y = Yellow (hypostatic gene) & y = 3:1  Recessive epistasis (9:3:4) o Recessive alleles at one locus (aa) mask the phenotypic expression of other gene locus (BB, Bb or bb)  In Mouse coat color. A codes for color (a) for albino. B for dominant agouti color (b) for black. Interaction: homozygous albino is epistatic to agouti and black. Agouti 9/16 Black 3/16 Albino 4/16 o Y x y  Y = 3:1 (epistatic gene) & y = Brown o Example:  In Labradors, coat color is controlled by two sets of alleles  Black is dominant to brown, but yellow is epistatic, when two yellow alleles are present (ee), the yellow coat color is expressed, independent of the brown/black alleles present  Black: B_E_  Brown: bbE_  Yellow: __ee Mitosis and Meiosis   Learning objectives: o Learning basic concept of the chromosomal theory of inheritance and basic concept of DNA replication o Learning fundamentals of mitosis o Learning fundamentals of meiosis  Terms we will cover: o Chromosome o DNA replication o Sister chromatid o Daughter chromatid o Cell division o Interphase o G1 phase o S phase o G2 phases o Mitotic phase o Prophase o Metaphase, o Anaphase o Telophase o Cytokinesis o Gametogenesis o Meiosis o Tetral o Chiasmata o Crossing over o Recombination  Cell division and Mitosis o Parent cell  2 daughter cells o Interphase:  G1 phase—Cells grow and CHO, lipids, and proteins are assembled for the cell’s own use  S phase—DNA duplicates to form sister chromatids that are joined at the centromere  G2 phase—Cell grow slightly, and the proteins which are needed for the completion of mitosis are synthesized o Prophase:  Chromosomes gradually condense  The microtubules are formed between the two centrosomes and migrate to opposite poles  The nuclear membrane breaks down  4 chromosomes consisting of two sister chromatids o Metaphase:  Nuclear membrane disappears and sister chromatids attack to the microtubules through the kinetochore.  Kinetochore is a protein that surrounds the centromere of sister chromatids  Sister chromatids move toward the midpoint of the microtubules and align along a place called the metaphase plate  Chromosomes are now clearly visible under a microscope o Anaphase:  Separation of sister chromatids from each other caused by the protein complex cohesion  Once separated from each other they are now considered chromosomes  Microtubules shorten and pull the chromosomes toward the pole to which they are attached o Telophase:  Chromosome sets arrive at their respective poles and decondensation starts  The nuclear membrane reforms and separates the chromosomes from the cytoplasm  Thus two daughter nuclei are formed o Cytokinesis:  Daughter nuclei are now packed into separate cells  Cytoplasmic division follows the nuclear division  Meiosis o “Make the genetic materials to half, then double” o 2n x 2n (meiosis I) 1 sperm x 1 egg (meiosis II)  2n fertilized egg (mitosis)  Cell growth and division  The essence of meiosis o The simplest case scenario: An organism receives only one chromosomes from each parent  Chiasmata & crossing over  First division of meiosis o Formation of a tetrad Parent 1 Parent 2 (Reduction and recombination) Parent 1 recombinant (cell 1) Parent 2 recombinant (cell 2)  Second division of meiosis o Simple division of cell genetic material to half Cell 1 Cell 2 Cell 3 Cell 4  Mitosis vs Meiosis Genetics of Sex Determination   Learning objectives: o Learning basic concept of autosomal versus sex chromosomes o Learning basics of sex determination o Learning basics of inheritance of sex-related traits o Examples: Hemophilia and color deficiency  Terms we will cover: o Autosomal versus sex chromosomes o Heterogametic male o Heterogametic female o Consequence of Y chromosome o Genotypic notation of sex-related genes and traits o Hemophilia o Color deficiency  Autosome vs sex chromosomes o Autosomes: Pair of homologous chromosomes  One from mom, one from dad  Have the same genes arranged in the same order  Slightly different DNAs sequence o Sex chromosomes:  X –significantly larger than Y  More DNA (gene responsible for traits specific to females)  Y –significantly smaller than X  Y absence = female  Y presence = male o srY gene—testis determining factor o Less DNA (few genes responsible for male characteristics)  Notations o Somatic cells  Male: 44 + XY  Female: 44+XX  Heterogametic male & female system o Male—heterogametic  44+XY (Parents diploid) 22+X & 22+Y (Gametes haploid) o Female—homogametic  44+XX (Parents diploid)  22+X (Gametes haploid)  Determination of sex in human baby o 22+x or 22+Y (sperm) x 22+X (egg)  44+XX (female) or 44+XY (male)  Consequences of sex chromosomes in human genetics o For the case of autosomal chromosomes, the presence of an illness- causing allele (and usually recessive) on the chromosome coming from parent 1, may be compensated by a normal the allele (which is usually dominant) from parent 2 o If on the presence of the X-recessive chromosome, the female offspring will avoid the illness-causing allele due to having 2 X chromosomes, but the male offspring would pick up that trait since it only has 1 X chromosome and the Y chromosome is not dominant o Example:  In population or family  Women could be: o 44+X healtXhealthy o 44+X healtXill  Men could be: o 44+X healtYnull o 44+X Yilnull  If “healthy” is dominant over “ill”…  No women are ill  50% of men are ill  Hemophilia A o Hemophilia A, also called factor VIII (FVIII) deficiency or classic hemophilia, is a genetic disorder caused by missing or defective factor VIII, a clotting protein o H represents a healthy normal dominant allele o h represents the illness-causing recessive allele  Women:  44+X X Healthy H h  44+X X Healthy but carries the trait  44+X X Ill (extremely rare)— homozygous recessive  Male: H null  44+X Y Healthy  44+X Yh nulIll—hemizygous recessive  Color deficiency o Color blindness, also known as color vision deficiency, is the decreased ability to see color or differences in color o The most common cause of color blindness is due to a fault in the development of one or more of the three sets of color sensing cones in the eye o Males are more likely to be color blind than females as the genes responsible for the most common forms of color blindness are on the X chromosome o C represents a healthy normal dominant allele o c represents the color deficiency recessive allele  Women: C C  44+X X Normal vision  44+X X Normal vision but carries the trait c c  44+X X Color blind (extremely rare)—homozygous recessive  Male:  44+X YC nuNormal vision  44+X Yc nuColor blind –hemizygous recessive Pedigree Analysis   Learning objective: o Learning standard symbols in pedigree analysis o Learning how to ask and answer fundamental pedigree questions and deriving mode of inheritance from pedigree  Terms we will cover: o Standard symbols o Autosomal dominant o Autosomal recessive o X-linked dominant o X-linked recessive o Y-linked o Mitochondrial  Pedigree standard symbols  Autosomal vs X-linked o The first step in solving a pedigree is to determine whether the trait is autosomal or X-linked  When most of the males in the pedigree are affected, then the trait is X-linked  When a 50/50 ratio between men and women the trait is autosomal o The second step in solving a pedigree is to determine whether the trait is dominant or recessive  When the trait is dominant, one of the parents must have the trait  When the trait is recessive, the parents do not need to necessarily show the trait (they could be heterozygous and carrier)  Y-linked vs mitochondrial inheritance o All male offspring from parents are affected in Y-linked inheritance o Gene is passed on only by affected females o Mitochondrial inhertiance DNA structure  Learning objectives: o Learning DNA components o Learning DNA structure  Terms we will cover: o The ladder o Side-rail (phosphate-sugar) o Foot-steps (nitrogenous bases: adenine, thymine, guanine, and cytosine) o Polymer o Monomer o Sugar is pentose (5 carbons) o Chargaff's rules o Double helix  The ladder o Side rail: Sugar phosphate backbone o Steps: Nitrogenous bases  Nucleotide(top)/nucleoside(bottom)  Nucleotide o Monomers that make up DNA and RNA o Pentose (5’carbon) sugar  DNA—deoxyribose  2’ carbon –H  RNA—ribose  2’ carbon –OH o Nitrogenous base  Purines—2 rings, 3 H-bonds  Adenine  Guanine  Pyrimidines—1 ring, 2 H-bonds  Thymine (DNA)  Uracil (RNA)  Cytosine o Phosphate group attached to 5’ carbon  The connection of the “side rail” o 5’-Phosphate-sugar-3’-5’-phosphate-sugar-3’-5’-phosphate-sugar-3’ o Phosphodiester linkages holds the backbone together  The connection of the “footsteps” o Chargraff’s rules states that DNA from any cell of all organisms should have a 1:1 ratio (base pair Rule) of pyrimidine and purine bases (A = T and C = G) o Hydrogen bonds hold the bases together  Unit of measuring/counting DNA o Base pairs  C-G (1) & A-T (1) = 2 base pairs total  How much DNA do we have? o C-value is the amount (in pictograms) of DNA contained within a haploid (1n) genome in a eukaryotic organism o Different from organism to organism DNA Packing   Learning objectives: o Learning DNA and histones interaction o Learning nucleosomes and the nature of chromatin o Learning the basic structures of the 10 nm chromatin fiber and the 30 nm chromatin fiber o Learning that chromosomes are single pieces of DNA packaged into chromatin  Terms we will cover: o Double helix o Chromosome packing o Nucleosome o Histones o Core histones o Histone 1 o Beads-on-a-string o Linker DNA o Coiled nucleosome o Chromatin fiber o Dividing cell o Non-dividing cell o Looped fiber o Condensed loop o Mitotic chromosome  Double helix to chromosome: packing o 6x10 bp (diploid cells) in human genome 9 9 o 6x10 bp x 0.34 nm/bp = 2.04x10 nm = 2m/cell (length of DNA molecule)  Nucleosome o DNA double helix wraps around histone proteins o The DNA + histone complex is called “nucleosome” which is “beads-on-a-string” forms a chromatin, which is 10-11 nm thick o H1—locks histone complex o H2A, H2B, H3, and H4— core histones form an octamer o Each nucleosome bead is surrounded by 140 bp DNA and there are 60 bp in the linker region o Space between beads is about 14 nm  The packed (coiled) nucleosome o Chromatin forms a thicker coil called chromatin fiber and is 30 nm thick  In a non-dividing cell o In a non-dividing cell the nucleus is filled with a thread-like material known as "chromatin" o Chromatin is made up of DNA and proteins (mainly histones and some non-histone acidic proteins)  Higher levels of condensation o Level 1: DNA double helix (2nm) o Level 2: Chromatin packed with DNA and histone proteins (11nm) o Level 2: Chromatin fiber packed of nucleosomes (30nm) o Level 3: Chromatin fibers loops along protein scaffolds (300nm) o Level 4: Condensed section of chromosome (700nm) o Level 5: Metaphase chromosome (1400nm) DNA Replication  Learning objectives: o Learning proposed models of DNA replication o The proof that DNA replication follows semi-conservative model o The enzymes (workers) that do the job of replication o The building blocks of DNA synthesis o Errors and mismatch in DNA replication and repair mechanisms  Terms we will cover: o Models of DNA replication o Conservative o Semi-conservative o Dispersive o Helicase o Polymerases o Primase o Primer o Binding proteins o Meselson-Stahl experiment o Replication initiation sites called “The replication fork” o Unwinding the double-helix o Synthesizing two new strands o Enzymes and proteins o DNA polymerase III o Primase o Primer o Leading stand o Lagging strand o Okazaki fragments o Mismatch o Proofreading o 3’ end loss o Telomeric cap structure  Models of DNA replication o Meselson-Stahl experiment  DNA replication o Replication initiation sites o Unwinding the double-helix o Synthesizing two new strands o Enzymes and proteins involved o Initiation origins are where enzymes and proteins work and replicate DNA (can be many sites) o Helicase opens up the DNA strand, now called the replication fork o Single-stranded binding proteins bind to single stranded DNA to help stabilize the opened-up structure o Building blocks of replication are free single nucleotides o DNA polymerase III attaches the free nucleotides only to the 3’ end of the existing strand (cannot initiate a sequence of a polynucleotide, it can only add to the 3’ end, requires a primer to start) o An RNA primer that is synthesized by RNA primase on a single- stranded DNA to allow DNA polymerase III to initiate a polynucleotide sequence o Nucleotides are added by complementary base pairing with the template strand o DNA always reads from 5’ end to 3’ end for transcription replication o During replication, new nucleotides are added to the free 3’ hydroxyl on the growing strand o The nucleotides (deoxyribonucleoside triphosphates) are hydrolyzed as added, releasing energy for DNA synthesis. o On the strand that it is not possible to add nucleotides, the process of replication waits until there is sufficient replication is done at the first strand and then fill backward  This is why there is a lagging and leading strand o If you see progression of the fork on both directions, you will see that same strand is leading on one direction and lagging on the other direction. o The process on the lagging strand is like making several lagging fragments called Okazaki fragments o DNA polymerase I (exonuclease) removes RNA primer and inserts the correct bases o Then ligase joins the Okazaki fragments and seals other gaps in the sugar-phosphate backbone  Proofreading and repair o Mismatch: DNA polymerase III attached the wrong nucleotide base potentially causing mutations at any one gene o DNA polymerase I and II proofread the strands and removed (extension) any incorrect base pairings and replace them with the correct base pairings  3’ end erosion o 3’ end loss due to loss of RNA primer (telomere shortening) o Telomerase anneals to 3’ overhang o Lengthening of the 3’ overhang  Elongation  Translocation  Elongation o Replication of complementary strand  RNA primer is synthesized (primase)  DNA polymerase III fills the gap  The RNA primer is removed and ligase seals the gap  Telomeres o Eukaryotic chromosomal DNA molecules have at their ends nucleotide sequences, called telomeres that postpone the erosion of genes near the ends of DNA molecules. If the chromosomes of germ cells became shorter in every cell cycle essential genes would eventually be missing from the gametes they produce. An enzyme called telomerase catalyzes the lengthening of telomeres in germ cells We Owe Our Traits to Proteins   Learning objectives: o Defining trait and phenotype o Explain what proteins do and how they do it o Basics of protein structure and function o Basics of genetic differences between individuals  Terms we will cover: o Trait o Phenotype o Protein o Amino acid structure o Amino acid function  Let’s define trait and phenotype o Trait is a feature of an organism, either inherited or determined environmentally, but typically occur as a combination of the two o Phenotype is the state of a trait  The trait of eye color has a phenotype of brown, blue, green...  Proteins are: o Structural o Storage o Enzymes o Transport o Defense o Signaling o Receptors o Regulatory  Proteins are shapers and workers offering a lot of roles o Each protein is defined as a sequence of amino acids o There are 20 amino acids that make up proteins, each with different side chains and the key to the shape is the forces that side chain of amino acids have on each other  These forces may change from one environment to another o These forces lead to a certain structure for a given AA sequence in a given environment o A protein’s structure determines its role and function o Individual organisms may be different in the sense that one individual may expresses the natural allele of one protein and another individual may expresses the mutated allele of the protein o Individual organisms may also be different in the sense that one individual may expresses the natural allele faster or in higher quantity than another individual and that this difference even in timing may result in a different phenotype Basics of Gene Expression and Genetic Code  Learning objectives: o Different cell types, different gene expression o Different times, different gene expression o Different environmental conditions, different gene expression o Temporal and Spatial patterns of gene expression o Key differences between replication and transcription o The language of genetic: the genetic code  Terms we will cover: o DNA o Messenger RNA o Gene expression patterns  The journey from genes to traits o Often called the Central Dogma in biology o DNA RNA Protein  Transcription o Transcription is a process where a message is copied from the gene on the DNA that give a message to the cell which protein should be synthesized  Differences in transcription and replication o Replication happens for all nucleotides on DNA whereas transcription only intends to make a copy of genes that are needed to be expressed o In replicating DNA four free nucleotide bases ACGT are used whereas in transcription four nucleotide bases ACGU are used o That means when DNA template is A, in replication the nucleotide T is used. In transcription however, the nucleotide U is used  The messenger RNA o Again, transcription is the step during which a message is copied from the gene on the DNA that describes to the cell which protein and how should be synthesized o Therefore, this type of RNA is called messenger RNA (mRNA)  Nucleotide combinations o How many combinations of 3 nucleotides can we make with 4 nucleotides?  How many options do we have (out of 4 ACGT)?  Place 1: 4x  Place 2: 4x  Place 3: 4x o 4x4x4=64 o Theoretically we can make 64 possible codons o With 64 possible codons and only 20 amino acids available, that means that we may have more than one codon per amino acid o Stop codons: UAA, UAG, and UGA—stop o Start codons: AUG –methionine Gene Model and Transcription  Learning objective: o A lot of non-coding DNA too little of coding DNA o Describing a gene model in Eukaryotic system. o The journey from pre-messenger RNA to processed RNA o Describing differences between replication and transcription o The concept of sense, antisense, and template o Steps of transcriptions  Terms we will cover: o Gene model o Promoter o Intron o Exon o Untranslated regions (UTR) o Messenger RNA o RNA processing o Sense o Antisense o Template o Cis-acting o Trans-acting o Initiation o Elongation o Termination  Eukaryotic gene model o Only 1.5% of the genome codes for protein-coding genes o The Promoter provides instructions as to when a gene must be turned on or off, more of a regulatory instruction  It is important because without it the process does not recognize where to start transcription of genes and making messenger RNA o Transcription starts at the promoter-exon border all the way through the end of the exons, including all the non-coding introns  The product is a long mRNA that contains genetic codes and a lot of non-coding DNA.  This long messenger RNA is called pre-messenger RNA o All of DNA will get transcribed but not all will get translated, an those that aren’t are referred to as untranslated regions (UTR)  5’ UTR regions  3” UTR regions o The long mRNA undergoes maturation through a process called RNA processing  The product is a smaller mRNA that does not contain all the non-coding regions (introns) and featured with additional components  Replication vs Transcription Replication Transcriptions Template Double strands Single strand Substrate dNTP NTP Primer Yes No Enzyme DNA polymerase RNA polymerase Product dsDNA ssRNA Base pair A-T, G-C A-U, T-A, G-C o Transcription is a process similar to replication that occurs in the nucleus o The process of transcription results in transcribing messenger RNA from a gene template o mRNA is synthesized from DNA using base pairing  Journey from genes to traits o The whole genome of DNA needs to be replicated, but only small portion of genome is transcribed o The synthesis of RNA molecules using DNA strands as the templates so that the genetic information can be transferred from DNA to RNA is called transcription o The process is directional from 5’ 3’  Sense-antisense-template o DNA is double-stranded. One strand has the copy of the gene (sense). The other strand is complementary (antisense) o In order to make mRNA that is able to code for the same gene, the antisense strand is used as template  Transcription process o Initiation  Transcription initiation needs promoter and upstream regulatory regions  The cis-acting elements are the specific sequences on the DNA template that regulate the transcription of one or more genes  The trans-acting factors are the proteins that recognize and bind directly or indirectly cis-acting elements and regulate its activity  The -35 region of TTGACA sequence is the recognition site and the binding site of RNA-polymerase  The -10 region of TATAAT (TATA box) is the region at which a stable complex of DNA and RNA-polymerase is formed trans-actingelement element Cis- Cis- Cis- acting acting acting element element CAAT box TATA box o Elongation  RNA nucleotides added in 5’3’ direction o Termination  Transcription is then proceed through elongation until reaching to the “termination sequence” which is “AATAAA followed by GT repeats”  The sequence of AAUAAA is a cleavage signal that will be cleaved by specific endonuclease RNA Processing  Learning objectives: o That pre-mRNA undergoes processing before it is ready for translation o That processing consists of processes including 5’ cap, 3’ polyadenylation, and splicing  Terms we will cover: o 5’ cap o 3’ polyadenylation o Poly-A tailing o Splicing o Methylated guanine triphosphate (GTP) o Poly A signal (AAUAAA) o Poly A polymerase  RNA processing o The long messenger RNA (that contains coding (exons) and non-coding regions(introns)) undergoes maturation through a process called “RNA processing” o The product is a smaller mRNA that does not contain all the non-coding regions and featured with additional components o Processing involves:  5’ cap  3’ polyadenylation  Splicing  Editing  The 5’ cap o 5’ capping occurs in the nucleus. The capping occurs prior to the splicing. Modification of the 5'-ends of eukaryotic mRNAs is called capping. The cap consists of a methylated guanine triphosphate (GTP) linked to the rest of the mRNA by a 5' to 5' triphosphate "bridge" o This guanosine is methylated on the 7 position directly after capping in vivo by a methyltransferase. It is referred to as a 7- methylguanylate cap, abbreviated m7G o Capping occurs very early during the synthesis of eukaryotic mRNAs, even before mRNA molecules are finished being made by RNA polymerase II. o Capped mRNAs are very efficiently translated by ribosomes to make proteins  The 3’ Polyadenylation (Poly A) o Modification of the 3'-ends of eukaryotic mRNAs is called polyadenylation (this poly A tailing process does not depend on the template) o The tailing process occurs prior to the splicing and in the nucleus o Polyadenylation is the addition of several hundred A nucleotides to the 3' ends of mRNAs o All eukaryotic mRNAs destined to get a poly A tail contain the sequence AAUAAA about 11-30 nucleotides upstream to where the tail is added  AAUAAA is recognized by an endonuclease that cuts the RNA, allowing the tail to be added by a specific enzyme, Poly A polymerase  mRNA splicing o Splicing: Eukaryotic genes are often interrupted by sequences that do not appear in the final RNA o Alternative splicing: cleaves out introns and some exons [E1][E2] [E3][E5] or [E3][E4][E5] o The intervening sequences that are removed are called introns o The process by which introns are removed is referred to as splicing o The sequences remaining after the splicing are called exons o Matured mRNAs are shorter than the DNA templates o Spliceosome is a large and complex molecular machine found primarily within the splicing speckles of the cell nucleus of eukaryotic cells  The spliceosome is assembled from snRNAs and protein complexes  The spliceosome removes introns from a transcribed pre- mRNA, a kind of primary transcript  RNA processing in one slide  Mature mRNA get out of nucleus Only mature mRNA is exported from nucleus Protein Synthesis  Learning objectives: o That mature mRNA will be used as information / intelligence / substrate to synthesize proteins o Processes involved in polypeptide synthesis o The mechanism of ribosomal machinery in protein synthesis  Terms we will cover: o Initiation o Elongation o Termination o Synthesis of a polypeptide o Information content of DNA o Genetic code o mRNA o tRNA o Ribosomal RNA o Small ribosomal subunits o Large ribosomal subunits  The central dogma of biology o The information content of DNA is in the form of specific sequences of nucleotides along the DNA strands o The DNA inherited by an organism leads to specific traits by dictating the synthesis of proteins o The process by which DNA directs protein synthesis, gene expression includes two stages, called transcription and translation  Translation is the actual synthesis of a polypeptide, which occurs under the direction of mRNA and occurs on ribosomes  Stages of translation o Initiation o Elongation o Termination  Overview of translation o Translation requires :  mRNA (which provides genetic code that defines the order of amino acids)  Ribosomes - Ribosomal RNA (which is a workhouse and assembly line that protein is manufactured)  Transfer RNA (is a type of RNA which does the manual job of carrying amino acids here and there and put them in the right place)  Genetic coding – codons (are the guidelines for order of amino acids) o Genetic information is encoded as a sequence of non-overlapping base triplets, or codons o The gene determines the sequence of bases along the length of an mRNA molecule  Ribosomes o Messenger RNA (mRNA) will bring the message of DNA code to a ribosome  A ribosome is made up of ribosomal RNA (rRNA) + proteins  rRNA is a single strand 100 to 3000 nucleotides long and globular in shape and made inside the nucleus  Its function is to associate with proteins to form ribosomes, a workplace for assembly of amino acids  Proteins are large and small subunits of ribosomal proteins  The genetic code—anticodon o A codon in messenger RNA is either translated into an amino acid or serves as a translational start/stop signal  This is the language of communication on the DNA and RNA side o The anticodon interprets and reads the codes on the amino acid side  The Codon chart The Anticodon chart  tRNA—transfer RNA o The translator o The interpreter o The carrier o Mutations in genes o Frameshift mutations—leads to allele differences  Insertion—inserting a single nucleotide or entire codon  Deletion—deleting a single nucleotide or entire codon o Missense (point) mutation—Swaps out one nucleotide for another changing the amino acid o Synonymous mutation—Swaps out one nucleotide for another not changing the amino acid (also known as wobble paring) o Duplication—a copy of the previous codon is replicated in front of it CCG CCG TAG ACT o Nonsense—Swaps out a nucleotide for another causing a stop codon to terminate the amino acid sequence prematurely


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