Bio 311 Genetics Exam 1 Study Guide
Bio 311 Genetics Exam 1 Study Guide Bio 311
CSU Monterey Bay
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This 14 page Study Guide was uploaded by kaitlinvallin on Wednesday October 5, 2016. The Study Guide belongs to Bio 311 at CSU Monterey Bay taught by Charmaine Robinson in Fall 2016. Since its upload, it has received 49 views. For similar materials see Genetics in Science at CSU Monterey Bay.
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Date Created: 10/05/16
BIO 311: Genetics EXAM 1 STUDY GUIDE Highlight=Important People Highlight=Important Concept Highlight=Key Term Chapters 1, 9, 11, 18 What is a Gene? A gene is the physical structure of DNA that acts as an instruction manual to make proteins. What is the structure of DNA, and how does it differ from RNA? DNA o made up of nucleotides, which are comprised of nitrogen in the form of nitrogenous bases. 1. Adenine (A) 2. Cytosine (C) 3. Guanine (G) 4. Thymine (T) Adenine & Guanine are Purine…. Cytosine & Thymine are Pyrimidine A&T two H bonds G&C – three H bonds (harder to break) o Double stranded and antiparallel (directionality matters) 3’ 5’ o 5 carbon ring deoxyribose o Phosphate backbone Gives DNA a negative charge Consists of 3 phosphate groups Exergonic & high in energy Contains various covalent bonds Requires high amounts of energy to break o Found only in the nucleus o Function is to store genetic information that controls cell function RNA o Made up of nucleotides, which are comprised of nitrogen in the form of nitrogenous bases. 1. Adenine (A) 2. Cytosine (C) 3. Guanine (G) 4. Uracil (U) o Single stranded o 5 carbon ring ribose o Found both in the nucleus as well as outside the nucleus o Function is to transfer genetic code out of the nucleus and translate it into proteins o There are three major types of RNA 1. tRNA transfer RNA 2. mRNA – messenger RNA 3. rRNA – ribosomal RNA What is a model organism? And why do we use more than one kind? 4 main model organisms: 1. Yeast 2. Bacteria 3. Fruit Fly 4. Mouse (or chicken) Model organisms are used due to these common characteristics: o Fast generation time o Fairly simple in terms of housing, care, food, cost, and genome o There’s sequential data, hopefully known genetic material to build from o Manipulatable o Produce multiple offspring, which could be beneficial in capturing probabilities What is the difference between genotype and phenotype? An organism’s genotype is the set of genes that it carries. A phenotype is all of the organism’s observed characteristics that are influenced by both the genotype and the environment. Structure of a Gene Prokaryotic DNA Replication Origin of Chromosome Replication (OriC) o Only one OriC o Origin can initiate multiple times at once, where another round of replication begins before the first finishes. Initiation o OriC is A=T rich. o There’s a series of DNA “A” boxes across the origin of chromosome replication. o These DNA “A” boxes respond to the “A” recognition sequence that DNA protein A provides. There are several DNA A proteins located around this OriC. o The DNA eventually wraps around these DNA A proteins (even before helicase activates them) and the mechanical pressure begins to pull apart hydrogen bonds, leaving just enough room for DNA B and DNA C to fit in. o DNA B protein is actually called Helicase. Helicase serves as an ATPase domain and is required for DNA A to become active. o DNA C protein is called a Helicase loader. This is a helper protein that carries Helicase and places it onto DNA A. *There are two replication forks on DNA therefore, it is important to remember that there ate two sets of proteins involved that work in opposite directions. Elongation o DNA Polymerase I. Removes RNA primers and fills in gaps with DNA. II. Repairs DNA damage. III. Main replicating enzyme that links the phosphate backbone by covalent bonds in a 3’ to 5’ direction. IV. Repair enzyme. V. Repair enzyme. o Ligaselinks okazaki fragments and makes the covalent bonds to link the phosphate backbone. o Topoisomerase (type II)reduces supercoiling by breaking the backbone and causing it to unravel. o Single Stranded Binding Protein (SSB)coats parts of the DNA that are single stranded to keep the hydrogen bonds from coming back together. It also serves as protection from enzymes that degrade single stranded DNA. o Primasemakes RNA primers (~10 base pairs). Leading strand 1 Lagging strand – several Termination o TER termination sequence o TUS protein termination sequence utilization substance Physically blocks the advancement of the replication fork. Eukaryotic DNA Replication OriC o Multiple origins of replication. Huge genome For speed and efficiency They only initiate once Initiation o The OriC has a high A=T content. o Considered the ARS (autonomously replicating sequence). o Process is similar to prokaryotic as far as proteins involved and the roles they play in the initiation process. Elongation o DNA Polymerase 1. Alpha αworks with a primase to make primers. 2. Epsilon ε leading strand synthesis. 3. Delta δ lagging strand synthesis. 4. Gamma γ mitochondrial DNA *There are at least 15 DNA polymerases BUT we only need to know these four for the exam. δ delta polymerase pushes primer up creating a yshape. Flap endonuclease cuts off primer flaps pushed by δ polymerase. Termination o End replication problem of linear chromosomes: there will be a single strand that will eventually be degraded by an enzyme and that genetic material will be lost. o Solution: Telomerase (both protein and RNA) adds nucleotides to the 3’ end. Protein telomerase reverse transcriptase “TERT” RNA telomerase RNA “TR” acts as a template sequence or repetitive sequence (no genes) Griffith’s Experiment1928 Griffith worked with mice and bacteria called Streptococcus pneumonia. + = S Strain mouse dead mouse “virulent” + = R Strain mouse live mouse “nonvirulent” + + = S Strain heat mouse live mouse + + = R Strain mouse dead mouse Heated S Strain o Virulentdeadly, fast, intense. o Viruses were heated only to a point where the cells broke down. o R strain and heated S strain were not deadly individually, but when combined this resulted in a dead mouse. Avery, MacLead, McCarty Experiment1944 These scientists took on the work previously done by Griffith and expanded upon it. They wondered which cellular component is capable of transforming the bacteria? CONTROL + + = R Strain mouse dead mouse Heated S Strain + lipase (destroys lipids) = dead mouse + + protease (destroys proteins) = dead mouse + RNAase (destroys RNA) = dead mouse R Strain + DNAase (destroys DNA) =live mouse Heated + analase (destroys carbohydrates) = dead mouse S Strain o When the DNAase was added, it destroyed the DNA, which actually destroyed the blueprint for the deadly component. Hershey, Chase Experiment – 1952 o Studied E. Coli and a phage virus. o The phage virus sits on the bacteria cell and injects its DNA inside. o Similar to an assembly line, more phages are made inside the bacteria cell until it bursts with more phages spreading everywhere. 35 o They grew 1 set of phages in S (found in proteindisulfide bridges), placed them on the E. Coli bacteria and allowed their DNA to be inserted. They blended to agitate and centrifuged. The liquid at the top contained phage capsids and the pellet at the bottom contained bacteria. o They grew a second set of phages in P (found in DNA phosphate backbone), placed them on the bacteria cell and allowed for their DNA to be inserted. They blended to agitate and centrifuged. The liquid at the top contained the ohage capsids as well while the pellet contained the bacteria o This proved that the DNA is where the blueprint is located since DNA is what’s inserted into the E. Coli and not a protein. Meselson, Stahl Experiment 1958 o The goal behind this experiment was to determine the mode of DNA replication between: conservative, semiconservative, or dispersive. Conservativesuggests that when an original strand of DNA is being replicated, two original strands up together and the two new strands end up together. Semiconservativesuggests that when an original strand of DNA is being replicated, the two resulting strands each consist of one old strand and one new. Dispersive suggests that when an original strand of DNA is being replicated, all strands have fragments from both the original and the new strands. o How did they do this? They14xposed E. Coli to two differ15t isotopes. One lighter in mass than the other. N“normal” or “light” and Nheavy. The template DNA was grown with E. Coli exposed to the heavy isotope. The heavy nitrogen was then removed. The following resulted from each of the trials: Conservative: For the first round of replication two strands were predicted to be present in the tube, one light and one heavy. For the second round of replication two strands were also predicted, but one was thicker than the other. SemiConservative: For the first round of replication, one strand was predicted to be present, at about the half way point with a mixture of colors. For the second round, it was predicted that two strands would be present, one consisting of the lighter nitrogen and the other at the half way point consisting of both isotopes. This turned out to be the correct model. Dispersive: For the first round of replication, on strand was predicted to be at the half way point just like in the semiconservative model. For the second round it was predicted that there would be one single strand somewhere between the half way point and the very top, implying that it would consist more of the lighter isotope. Chapter 2 Mendel’s Law of Independent Assortment Mendels law of independent assortment states that in a dihybrid cross, it is equally likely that the chromosomes will line up as either of the following probabilities. When would a test cross be performed? o A test cross could be performed in the event that there’s a fully recessive genotype and a heterozygote genotype. o Due to the fact that the recessive parent can only give one type of gamete, it allows the meiotic event in the dominant parent to be tracked. o The nomenclature for the test crosses leaves off the recessive alleles. Monohybrid vs. Dihybrid Crosses A monohybrid cross is the crossing of one gene. 3:1 Parent: PP (purple) x pp (white) F1: Pp (purple) F2: PP P ¾ purple p ¼ white Pp pp A dihybrid cross is the crossing of two genes or traits from parents. 9:3:3:1 Parent: PP(purple) RR (round) x pp (white) rr (wrinkled) F1: Pp (purple) Rr (round) F2: *Do not draw punnet square, just remember the ratio 9:3:3:1 9 purple, round P_R_ 3 purple, wrinkled P_rr 3 white, round ppR_ 1 white, wrinkled pprr Chapter 4 – Extensions to Mendel’s Inheritance These extensions still follow Mendel’s laws but they vary in ratios. Incomplete Dominance 1:2:1 Incomplete dominance involves one gene with three separate phenotypes. 4 O’Clock flowers Parent: r r (red) x rr (white) F1: r r (pink) F2: + + + ¼ red r r r r 2/4 pink r r rr ¼ white Expressivity Expressivity describes the degree to which every given allele is actually expressed at the phenotypic level (it’s intensity). Example 1: a beagles spot being either small or large Example 2: a chicken with polydactyl (varying number of digits) Penetrance Penetrance refers to the percent of individuals having the allele who actually exhibit the phenotype. Some of the carriers may not exhibit the phenotype due to The environment Influence of other genes Subtlety of phenotype Example: polydactyl Overdominance The example of sickle cell anemia in humans is used for this mendelian extension. Hemoglobin carries oxygen through the body by means of red blood cells (RBC). A normal RBC is crammed full of hemoglobin. A mutant RBC is easily ruptured and has a small life span. This leads to showing signs of anemia. A Hb wild type hemoglobin Hb mutant hemoglobin A A Hb Hb A Hb Hb S HbAHb Swild type A S S S Hb Hb heterozygous, less fit allele present Hb Hb Hb Hb Hb Hb diseased Even though Hb is less fit than the wild type allele, it is present in high numbers in humans throughout parts of the world where malaria is prevalent. Diseased individuals symptom intensity can vary, which makes this also a good example for expressivity. CoDominance The example used for codominance is blood type. Blood Types: A, B, AB, and O Genotypes “Blood Type” Phenotype I I , I i A B B B I I , I i B I I AB *codominance because both alleles determine phenotype ii O O is a universal donor “blank cell” An ABO blood type test was used before more sophisticated genotypic markers were used in paternity testing. It couldn’t point to a father but it could make exclusions. Autosomal Inheritance Male and female progeny show same proportions of trait 3:1 for both genders. SexLinked Inheritance (most of the time = xlinked) The example used to show sexlinked inheritance is the eye color of a fruit fly. W red (wild type) W – white (mutant) Parent: X X W+(red) x X y (white) F1: ½ hemizygous ½ heterozygous F2: ½ red ½ white A reciprocal cross would be when a male and a female trait are crossed. Lethality The example used to show lethality is the coat color of a mouse. Brown: wild type Yellow: mutant Yellow x brown 1:1 **just because it’s a mutant allele, doesn’t mean its recessive yellow yellow x yellow y y y y y 2/3 yellow A A A A A A dead yellow embryo A A AA A A – yellow 1/3 brown AAbrown So we know that if its either A A or AA, its an adult mouse. A second example would be the limbless chicken, which is a recessive developmental disease. A cross between two carriers (Ll x Ll) results in the normal 3:1 ratio except the recessive allele is lethal so ll does not survive. Temperature Sensitive Mutants Environment plays a large role for this type of organism. When in a lab, a “conditional lethal allele” can be developed and controlled by means of temperature. A growing organism @ Permissive temperature = wild type Restrictive temperature = mutant allele Pleiotrophy Pleiotrophy describes a single gene having multiple effects on phenotypes. o This depends on the type of cell that gene is expressed in. o Depends on the role of the genes product (kinase, transport protein, etc.) o Depends on the time during development that the gene is expressed. Examples: in mice A is yellow coat color for survival, in chickens the frizzle gene determines feather shape and metabolic differences. Epistasis Epistasis is the act of one gene overriding the expression for the product of another gene. It can be thought of as “sitting on top of” or “standing upon” another gene. Epistatic refers to the gene doing the masking, the one that is overriding the other gene. Hypostatic the overridden gene, the gene that is being covered up. ** Four forms of epistasis that we should know for the exam. 1. Recessive Epistasis 9:3:4 Example is in Labrador Retrievers coat color. eeyellow F1: BbEe x BbEe bchocolate brown B black F2: 9 B_E_ black 3 bbE_ brown ee no pigment is deposited 3 B_ee yellow E_ deposits pigment 1 bbee yellow ** even if the cells are making the pigment, it doesn’t actually make it into the fur due to the epistatic gene. 2. Dominant Epistasis 12:3:1 Example in squash color F1: WwGg x WwGg Wwhite wcolor of some type Gyellow F2: 9 W_G_ white ggreen 3 wwG_ yellow 3 W_gg white 1 wwgg green Dominance series: W>w>G>g **this series shows which allele is more dominant over another allele when in combination with eachother. 3. Recessive Suppression Epistasis 13:3 ** one gene cancels the expression of the “abnormal” gene Example in fruit fly eye color Pd_ red (wild type) F1: PdpdSusu x PdpdSusu pdpd – purple (mutation) susu – suppress F2: 9 Pd_Su_ red Su_ no suppression 3 Pd_susu red 3 pdpdSu_ purple 1 pdpdsusu red 4. Dominant Suppression Epistasis 15:1 Example in fly genes Pd_ red (wild type) F1: PdpdSusu x PdpdSusu pdpd – purple (mutation) susu – no suppression F2: 9 Pd_Su_ red 3 Pd_susu red 3 pdpdSu_ red 1 pdpdsusu purple Su_ suppression Complementation 9:7 Complementation refers to when 2 genes are involved in a chemical pathway. Functional products from both genes work to get end product. Precursor Intermediate Final Product enzyme enzyme Example in sweet pea flower color F1: CcPp x CcPp F2: 9 C_P_ colored 3 ccP_ white 3 C_pp white 1 ccpp white Four Types of Inheritance for Pedigree Analysis 1. Autosomal Dominant Should (not always) be equal #’s b/w male and female Affected individuals in every generation 2. Autosomal Recessive Should (not always) be equal #’s b/w male and female Trait skips generations 3. SexLinked Dominant Predominant in women When xlinked, more males are affected as they get their X from their mother 4. SexLinked Recessive Can see affected offspring from normal parents When xlinked, more males are affected as they get their X from their mother How to Attack Human Pedigree Problems o A circle usually refers to a female, and a rectangle to a male. o If a person is a “carrier” = heterozygous. o If a person is affected, their shape should be filled in. o Always draw out the pedigree and ensure that the multiple generations are spatially correct. o Assume a trait is rare unless stated otherwise. o Assume that if someone has married/mated in, they are not carriers. o When checking if a sexlinked trait its important to remember that males always get their X from their mother, so if mom is affected and son is not, it is not a sex linked trait. In this case it would be autosomal. Chi Square (O−E) 2 x =∑ E where O = observed values E = expected values Df= n1 **reject 0.05 & smaller Linkage Mapping We use genetic recombination to create genetic maps. The further apart the 2 genes are, the more likely it is for a crossover to happen between them. In order to determine the relative distance between sets of genes look at the frequency of recombination events. Higher % freguency = further apart Lower % frequency = closer together
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