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61 Biol DNA & DNA Replication (Chapter 16)
2/23/09 & 2/25/09
An organism has a phenotype, which is the physical appearance/physical characteristics of that organism. The phenotype of an organism is determined by its genotype, the alleles of the various genes that an organism carries on its chromosome or chromosomes. We now know that genes are the instructions for making the various polypeptides (and functional RNAs, which well talk about for the next chapter) that are used in each cell of an organism. Now, of course, we know that each individual gene is comprised of a specific sequence of deoxyribonucleotides found along a portion of the DNA polymer that makes up a chromosome. But what evidence led us down the path to understanding what the hereditary material was in each cell and where it was located? You can read the first part of the chapter on your own to look at the experimental evidence for DNA as the genetic material. I will not be lecturing on these experiments, but they are described in my notes - read them over and understand them (you will see a couple of questions on them in the practice exams for Exam 2). In the 1930s Joaquim Hammerling discovered that the hereditary material resided in the nucleus of a eukaryotic cell. The nucleus contains chromosomes, but they are made up of both nucleic acids and proteins, so what was the important part? Many people predicted that proteins were carrying the genetic information because they seemed so much more chemically complex than DNA. In the 1940s Oswald Avery, Colin MacLeod and Maclyn McCarty showed that what was important was the DNA. They started with work carried out originally by Frederick Griffith. Griffith was conducting experiments with bacteria that can cause pneumonia Streptococcus pneumoniae. He had found that if you inject live S. pneumoniae bacteria from the smooth strain (or S strain) into a mouse, the mouse would die. When he injected the mice with a mutant rough strain of live S. pneumoniae bacteria (called the R strain), the mice survived. (The R strain made rough looking, irregular colonies when grown on media in a petri dish, while the S strain made nice smooth round colonies. It turns out the R strain lacked the polysaccharide coat that made the S strain smooth and dangerous as a pathogen something that causes a disease). If Griffith injected mice with killed (boiled!) S strain, they would live. However, when he mixed dead S bacteria with live R bacteria, he now found that the mice died. Griffith then found live S strain bacteria in the blood of the mice. Somehow, the dead S strain had transformed the live R strain into the dangerous, virulent S form. When boiled S strain extract was added to live R strain bacteria, the S strain extract was capable of producing a heritable change in the live R strain it had transformed the phenotype and genotype of the R strain into an S strain. In 1944, Avery and pals wanted to understand what this transforming principle was. The boiled S strain extract contained proteins, DNA, RNA, sugars, and lipid. For other reasons, no one thought the transforming principle was sugar or lipid, so they ignored those components. In their experiment they took dead S strain bacterial extract and treated it with different enzymes that could chew up particular compounds in the extract, and then they added this modified extract back to live R strain bacteria to see if the R strain could be transformed into the virulent S strain. 1. They treated dead S strain extract with proteases, which are enzymes that chew up proteins. They mixed this with R strain bacteria and found that they could still transform the R strain into S strain bacteria. 2. They treated dead S strain extract with an RNAse, which chews up RNA. They mixed this with R strain bacteria and found that they could still transform the R strain into S strain bacteria. 3. They treated dead S strain extract with DNAse, an enzyme that chews up DNA. They mixed this with R strain bacteria and found that they could NOT transform the R strain into S strain bacteria - only R cells were present after treatment. This showed that DNA was the transforming principle! Well it was pretty good evidence, but a lot of people didnt want to believe them. DNA is structurally boring its only made up of 4 different deoxyribonucleotides as opposed to the 20 amino acids that make up structurally complex proteins. In addition, proteins were already known to have many different functions, so many scientists thought that the hereditary material had to be protein. To tackle this question again, Alfred Hershey & Martha Chase (in 1952) used bacteriophage to again show that DNA was the hereditary material. A bacteriophage is a virus that attacks bacteria (phage means eat and so they eat bacteria). They infect the bacteria, reproduce, and then pop the bacteria (called lysis) to release the progeny phage. All this takes about 30 or so minutes. When a virus infects a cell it injects its genetic material into the bacteria. The result is that the phage has taken over the bacterial cell to make more viruses, and this infection can be viewed as a genetic change in the bacteria brought about by the activity of the phage genetic material. To find out whether this genetic material was DNA or protein, Hershey and Chase grew up bacteriophage in growth media
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containing either 1) a radioactive isotope of sulfur (35S), which would be incorporated into viral proteins (since cysteine and methionine have sulfur groups in them) or 2) a radioactive isotope of phosphorus (32P) that would be incorporated into the DNA backbone. Here was the experiment: - infect non-radioactive (cold) bacteria with radioactive (hot) phage and wait 15 minutes for the phage to infect the bacteria. After a few minutes, the phage have transferred their genetic material. They then put this mixture into a blender to separate the phage ghosts (the protein coats that are left on the surface after infection) from the bacteria, and centrifuged the mixture in a test tube to pellet the bacteria at the bottom and separate them away from the phage ghosts that remain in the supernatant. They set up the following two experiments:
1. infect cold bacteria with hot phage containing 35S, incubate, pulse in a blender, separate the phage
ghosts from the bacteria in a test tube, and then see where the 35S was located. Was all the 35S associated with the proteins of the phage ghosts or had some been injected into the bacteria upon infection? The answer was that all the 35S was associated with the phage ghosts. infect cold bacteria with hot phage containing 32P, incubate, pulse in a blender, separate the phage ghosts from the bacteria in a test tube, and then see where the 32P was located. Was all the 32P associated with the phage ghosts or was some injected into the bacteria upon infection? The answer was that all the 32 P was associated with the bacterial cells. Viral DNA was the hereditary material injected into bacteria upon infection!
2.
Many scientists still didnt understand how DNA could be the genetic material since they knew that DNA was only made up of 4 different deoxyribonucleotides. Some scientists thought DNA was a long polymer of simple, repeating nucleotides: GATCGATCGATCGATC If this was true, how could this simple polymer contain the information that could make a bacterial species different from a protist, or a mouse different from a human? Erwin Chargaff measured the amount of each kind of nucleotide base present in the total DNA of various organisms, and he found what is referred to as Chargaffs rules: the proportion of T present in DNA is always equal to the amount of A present, and the amount of G present = the amount of C present. But, the amount of A+T does NOT have to equal the amount of G+C, and different organisms have different ratios of A/T to G/C (for example, one organism could contain 15% A, 15% T, 35% G and 35% C, but another organism might have 22% A, 22% T, 28% G and 28% C). This suggested that DNA was not just a simple polymer containing an equal amount of all 4 bases, with the same structure in every organism. James Watson and Francis Crick, using Chargaffs data as well as data generated by Rosalind Franklin (whose borrowed data showed that the shape of a DNA molecule was helical, with a width of about 2 nanometers), came up with the correct structure that explained how DNA was organized the double helix. Remember that DNA is a polymer of deoxyribonucleotides (remember, the actual monomer is really a deoxyribonucleotide triphosphate, abbreviated dNTP). Each individual dNTP has a 5carbon attached to 3 phosphate groups and a 3OH in the ribose sugar. A DNA polymer is made by forming an ester linkage (-O-) between a single phosphate group (the one directly attached to the 5 carbon) of a dNTP monomer and the 3 OH of the last nucleotide in a growing DNA polymer. We can only add to the polymer at the 3OH end! As the polymer is built up, a long backbone is made that contains alternating sugar and phosphate groups connected by phosphodiester linkages, called the sugar/phosphate backbone (remember for what you learned about DNA structure: a phosphodiester linkage gets its name because the linkage involves a phosphate group making two ester bonds, which means the phosphate is connect by a -O- linkage to the 3C of one monomer and is attached by a -O- linkage to the 5C of the other monomer). To attach one dNTP to another is a chemically favorable reaction and the G is negative, because we remove two negatively charged, high energy phosphate groups from the dNTP in order to add the monomer to the growing polymer (and this results in a negative G). No matter how long the DNA polymer gets, it will always have one end with a free phosphate group (5end) and one end (3end) with a free hydroxyl group. Watson and Crick proposed that a chromosome is composed of two polymers of DNA aligned in opposite directions (anti-parallel), with one strand facing in the 53 direction, and the other in the 3 5 direction. The two strands are held together by hydrogen bonds between the different bases attached to the ribose sugars (these interactions are called base pairs). A always base pairs with T, while C always base pairs with G (this matching of a
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big purine [A or G] with a smaller pyrimidine [T or C] not only explains Chargaffs rules but also why the helix has the same uniform diameter along its length). There are 2 hydrogen bonds that form between A and T, and 3 hydrogen bonds that form between C and G. It takes about 10 base pairs to make up a complete turn of the DNA helix. DNA also has alternating grooves of different widths on its outer surface: the major groove and the minor groove. These become important later when we look at various proteins that bind to DNA, since the major groove is a main contact point for many of these proteins. This structure readily lends itself to a model for how DNA can be copied during cell growth and division. The sequence of nucleotides found in one polymer of the DNA duplex complements the sequence found in the other polymer. If you know the sequence of one DNA strand, you can easily predict the DNA sequence of the second strand. Its easy to picture how one double-stranded helix of DNA could be copied into 2 new strands: unpeel the two polymers, and then use the single strands to copy off of to make new strands of DNA, reforming 2 doublestranded helices. However, there were 3 hypotheses for how DNA might replicate: 1) conservative replication, in which the parental DNA strands are copied into 2 new daughter strands, but the parental strands stay together; 2) semi-conservative replication, in which each parent strand is copied and stays associated with its new daughter strand; and 3) dispersive replication, in which the DNA was just all jumbled up together after it was copied. Matthew Meselson and Frank Stahl showed that DNA replicated in a semi-conservative manner that is, each newly copied DNA duplex contains one complete strand from its parent and one brand new strand. M & S grew E. coli bacteria in a heavy isotope of nitrogen (15N) for several hours so all the DNA was filled with this form of nitrogen (the DNA bases contain lots of nitrogen). They collected DNA from some of these cells and determined its density. After growing in media with just 15N available, all the DNA was heavy and formed one band of heavy double-stranded DNA when purified out of bacteria and centrifuged in a test tube. They then shifted the rest of the heavy 15N-containing cells to a new nitrogen source that only has the common, light form of nitrogen (14N) and let the cells grow for 1 round of DNA replication (about 20 min. for E. coli). When they measured the DNA density following this step, they again found only one band present, but this time it was intermediate in density between DNA containing only 15N and DNA containing only 14N, inconsistent with the model of conservative replication but consistent with the model of semi-conservative replication (since each DNA molecule after replication would contain one heavy strand and one light strand, and so all the DNA molecules would be the same density). However, this result was also consistent with the model of dispersive replication. If M & S let the DNA replicate one more time in 14N-containing media, they saw two bands present, one that contained DNA with two light strands, and one band that contained DNA with one light and one heavy strand. This result was consistent only with the model of semi-conservative replication of DNA. How is DNA actually replicated? Well start by describing how DNA is replicated in bacteria. Bacteria have a single, circular chromosome. The key enzyme in DNA replication is DNA polymerase. A DNA polymerase enzyme can add a deoxyribonucleotide (dNTP) to the 3 end of an existing polymer if it has 3 things: 1. a template strand of DNA to copy (and this needs to be single stranded) 2. dNTPs (without monomers, you cant build a polymer) 3. an existing 3OH to add nucleotides to (what we call a primer) Bacteria contain several different kinds of DNA polymerases, but only two will be important for replication: DNA polymerase I this enzyme is involved in lagging strand synthesis. DNA polymerase III this enzyme is used in both leading and lagging strand DNA synthesis. It is a very large protein with multiple different polypeptide subunits. DNA pol III has two very important activities: 1) it can act as a 53 polymerase, adding dNTPs to the free 3OH group of the last deoxyribonucleotide of an existing polymer (so the polymer grows by adding to the 3 end). A dNTP is NEVER added to the 5end of a DNA polymer. 2) DNA pol III also has a 35 exonuclease activity. enzyme The can check the sequence that it has just made and can backspace in the 35 direction to remove any deoxyribonucleotides that it added incorrectly.
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DNA replication begins at a site on the chromosome called the origin of replication. An origin of replication is a region of DNA that contains a particular sequence of nucleotides that attracts a complex of proteins that initiate DNA replication. The DNA is made accessible to DNA pol III (the two DNA strands of the helix are pulled apart and made single-stranded) by the actions of proteins at the replication origin. DNA is unwound by a protein called a helicase. This kind of protein can unwind the DNA helix so the bases are accessible on each individual single strand. The open single strands are then bound by single strand binding protein (SSB), which protects the DNA and keeps it from re-annealing, or coming back together. Now the bases in the DNA are accessible and can be used to copy a new strand (in other words, our template DNA is single-stranded). But how can DNA polymerase start working without an existing 3OH end to add nucleotides to? The answer is that DNA polymerase can add dNTPs to either an existing piece of DNA or RNA. At the site of the origin of replication, an enzyme called primase makes a short sequence of ~ 10 ribonucleotides of RNA called an RNA primer on both strands. Now DNA pol III can add deoxyribonucleotides to the available 3OH on the RNA primers to begin DNA synthesis (the RNA strand primes, or prepares, the DNA pol for action). As DNA is replicated, the replication bubble increases in size as more of the double stranded DNA is unwound by helicases present on both sides of the bubble. DNA pol III adds dNTPs onto the two RNA primers and can extend the DNA polymer from the origin to the site where DNA is being unwound by helicase, and the bubble grows larger as helicase unwinds more DNA at each end. We are now faced with three problems: 1) The DNA helix outside of the bubble gets tangled up as the bubble DNA gets unwound, and if it gets too tightly coiled replication will stop. 2) DNA is being made in a 53 direction from the RNA primers at the origin. This results in the synthesis of the leading strand only the two RNA primers placed at the origin by RNA primase can only initiate the polymerization of DNA in one direction (53), and as the replication bubble widens, how do we replicate the DNA behind each of these RNA primers? 3) There is a piece of RNA primer incorporated at the 5end of the newly synthesized DNA. This cant stay there, so how do we get rid of it? Starting with Problem #1 first: The action of the helicase as it unwinds the DNA generates knots (called supercoils) in the DNA sequence ahead of each end of the replication bubble (the DNA is more tightly wound up in front of the ends of the replication bubble). This winding pressure is relieved by the action of DNA gyrase, which cuts both strands of the DNA duplex (the double-stranded DNA) in front of helicase, lets the DNA spin around in order to relax the tension, and then gyrase hooks the DNA back together again. As long as helicase works, gyrase works to relieve the tension. Problem # 2. (For visualization, use the pictures of your replication bubble from lecture; or use the picture of a replication forks from the text but notice that a fork is only one half of the entire bubble, and you better understand the bubble!). Since DNA pol III can only add a nucleotide to an available 3 OH group and NOWHERE ELSE, this means that one strand on each side of the bubble (what we call the leading strand) will have no trouble being replicated since these new strands starts at the 3end of the RNA primer and can just keep growing as the replication bubble is opened. However, if we look at the region of DNA behind the 5 nucleotide of each RNA primer, we see that the DNA has to be made in the direction facing away from the ends of the replication bubble (the point where the DNA is unwound) in order to be made in the 53 direction. In fact, the DNA polymer being made on this lagging strand is made in short, discontinuous segments of about 1000-2000 nucleotides long that are called Okazaki fragments. How do these lagging strands get started? RNA primase not only puts an RNA primer at the site of the origin of replication, but as new sequence is unwound in the bubble, RNA primase can set a primer down at the open ends of each bubble. DNA pol III can then add dNTPs to the 3OH end of this newly made RNA primer and will add DNA until it reaches the 5 end of the first RNA primer at the origin. At this point, DNA pol III cant attach the DNA it has just made to the RNA of the primer, and so it falls off to be used again to make a new Okazaki fragment. This process keeps repeating itself as the replication bubble grows. Once an RNA primer from an older Okazaki fragment is reached, DNA pol III falls off and is now available to start a new Okazaki fragment.
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Problem #3: Leading strand synthesis starts at the first RNA primers at the origin and can extend rapidly in one long, unbroken polymer until the end of the chromosome is reached. However, each Okazaki fragment that is made is attached to a piece of RNA at its 5 end. In addition, when the lagging strand DNA reaches the RNA primer of an older Okazaki fragment, DNA pol III falls of and there is now a gap left between the newly synthesized polymer of DNA and the RNA primer that started that older Okazaki fragment. How can we fix this? Now the enzyme DNA polymerase I makes an appearance. DNA polymerase I has two of the same activities as DNA pol III: 1) it can act as a 53 polymerase, adding dNTPs to the free 3OH group of a deoxyribonucleotide of an existing polymer. A dNTP is NEVER added to the 5end of a DNA polymer; and 2) it also has a 35 exonuclease activity which means it can backspace in the 35 direction to remove any deoxyribonucleotides that it added incorrectly. But, DNA polymerase I has an activity that DNA pol III does not have! 3) DNA pol I has a 5 3 exonuclease activity, which means it can chew off nucleic acids from the 5 end of an existing strand. This means it can chew up the RNA primers present in the DNA of Okazaki fragments and replace them with DNA. After DNA pol III falls off the lagging strand, DNA pol I binds to the free 3OH of the last DNA nucleotide just before the RNA primer. DNA pol I then removes the RNA primer on the fragment in front (cutting ribonucleotides from the 5 end of that fragment) and fills the gap that it has made by adding new DNA monomers onto the 3OH of the DNA strand until it replaces the last RNA monomer with DNA. Restated, DNA polymerase I removes the RNA primers and extends the DNA of the Okazaki fragments into the region previously occupied by the RNA primer. At this point DNA pol I falls off, leaving a gap between the new fragment of DNA (the DNA that replaced the RNA primer) and the older piece of DNA from the Okazaki fragment in front of it. This gap is fixed by a protein called DNA ligase, which comes in and joins the two DNA polymers of the Okazaki fragments together (ligase being derived from the Latin ligare the action of binding or tying). Once the leading strands and the lagging strands have copied all the DNA and been properly ligated together, we have replicated our bacterial chromosome and now the bacteria is ready to divide into 2 daughter cells! DNA Replication creates mistakes DNA polymerases are pretty good at copying their templates, but they do make mistakes. In addition, various forms of radiation and chemicals can affect the DNA, leading to mutations. So cells have evolved a variety of enzymes that can be used to find mistakes (mismatch repair proteins) or damage to the DNA (such as nucleotide excision repair proteins) and fix it. Eukaryotic DNA Replication We have been talking about DNA replication in prokaryotes, so how is replication in eukaryotes different? 1. Eukaryotes have bigger genomes than prokaryotes, and they would take too long to replicate if they just used one origin. So eukaryotes have many origins of replication located along their chromosomes (humans have 3000 or so, and even then it still can take hours to replicate all the DNA in a cell). 2. The Okazaki fragments are smaller (100-200 bp) than those made in prokaryotes. 3. Eukaryotic chromosomes are linear, so they have ends to them. This last difference poses a problem for eukaryotes. If we consider two DNA strands being replicated, we can see that the leading strand will just continue along in the 53 direction until DNA pol III reaches the end of the chromosome, at which point it will just fall off and the leading strand synthesis is complete. However, the lagging strand is composed of a series of Okazaki fragments, each of which needs its own RNA primer to begin synthesis. At the two ends of the chromosomes being synthesized by lagging strand synthesis (the 5 end of the newly-made DNA strand), there is not enough room to put in an RNA primer. This means that at both newly synthesized 5 ends of the chromosome, a little bit of sequence information will be lost in the next round of division because it cant be copied. When the cell divides again, this same problem will result in the loss of an additional amount of DNA sequence from the ends of the chromosome. So each round of replication would remove a little more sequence information from the chromosome.
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In the example shown at the bottom of pg. 6, the double arrow shows you how much sequence information was lost in this exaggerated example as the chromosome labeled #1 in the figure is first replicated, and then as one of the daughter copies of this chromosome, labeled #3 in the figure, is replicated. This may not seem like a big deal, except that if an important gene is next to the ends of a chromosome and it gets eliminated, there could be a problem for the survival of the cell. Also, you can imagine that if this happened indefinitely for any eukaryote, ranging from little unicellular species of protist that reproduced asexually by division to sexually-reproducing species of mammals like ourselves, pretty soon all its chromosomes would get eaten away from the ends (well, only until an important gene or genes were eliminated, and then the cell or organism would just die). How could we maintain the genetic integrity of a species over millions of years if their chromosomes kept being chopped up? Eukaryotes deal with this problem by having telomeres at the ends of their chromosomes. Telomeres are short repeated sequences (in humans it is TTAGGG) that are added on to the 3 ends of each chromosome by an enzyme called telomerase. Telomerase will add dNTPs directly to the 3 ends of the chromosomes (using a template that it carries around with it), making sure there is plenty of room to build up Okazaki fragments at the ends of the chromosomes. It will add a TTAGGG, and then it will add another TTAGGG, and so on until there are 500-5000 copies of this little sequence added to the ends. The activity of telomerase is highest in the developing embryo and also high in the cells of the germline (sperm and eggs). The idea is that telomerase activity is very high in your germ cells so that the fertilized egg already gets to start off with chromosomes that have nice long telomeres. In addition, the cells of the developing embryo and young organism will have high telomerase activity because they are undergoing rapid division to generate all the tissues of the growing organism. Most cells of the adult have low to no telomerase activity, since few cell types still divide when you are an adult (and levels of activity of this protein may play a role in telling cells how old they are). A rare disease called progeria exists, in which children undergo premature aging (and early death), and in some forms of the disease children are born without telomerase activity (which may explain the effects of the disease). Cancer is just uncontrolled cell division, and it has been found again and again that various tumor cells have high levels of telomerase activity (unlike normal adult tissue cells), and this may be related to how or why tumor cells can divide uncontrollably compared to normal cells. It has been proposed that the reason telomerase activity decreases as we age is to help decrease the chance that cells will run amok and form cancers.
#1
5 3 5 3 3
3 5 3 3 5
DNA replication #2 #3 5 5
DNA replication of Chromosome #3 5 3 3
3
3 5
Chromosome Organization Bacterial chromosomes are not just naked pieces of double-stranded DNA, but are bound by a number of different kinds of proteins (which we will talk about in more detail later). The haploid human genome is 3 x 109 base pairs, almost 2000x larger than the bacterial genome (and many plant species have genomes that are much larger than this), and this much DNA has to be organized within the nucleus, a problem bacteria dont have. All this DNA also has to be able to efficiently separate during meiosis and mitosis, which is why eukaryotic chromosomes form an
6
extensive DNA-protein complex that we call chromatin. Bacteria dont have chromatin, but eukaryotes have highly organized DNA, and the way they organize their DNA can strongly influence gene expression. Some of the most important proteins that organize DNA into chromatin are the histones. The histones are a family of proteins (including H2A, H2B, H3 and H4) that associate into a complex (a little protein ball). This histone complex organizes DNA into nucleosomes. A nucleosome has a histone center with a piece of DNA wrapped around it ~ 2 times. Nucleosomes will be found stretched all along a chromosome: some DNA will wrap around the histones to form nucleosomes, and some is found histone-free between each nucleosome. DNA can also be wrapped up in even more tightly wound higher order complexes: 30 nm fibers organize the DNA in nucleosomes, 300 nm fibers are loops made up of the 30 nm fibers, and finally these loops can be organized into the 700nm structures that make up the condensed sister chromatids during M phase. In interphase cells, the DNA is wrapped into 30-nm fibers and then organized into radial loops (the 300 nm fibers). Eukaryotes can actually shut down transcription from an entire chromosome by winding it up this way even in interphase. A Barr body is an example of this: human females inactivate one of their X chromsomes in order to be equal with males, which only have 1 X to begin with. Highly wound up DNA is called heterochromatin, and it is not transcribeable. Less compacted DNA, the DNA available for transcription within radial loops, is called euchromatin. We will discuss chromatin further when we hit Chapter 18 and begin studying eukaryotic gene regulation.
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FUNGIReading: Chapter 31, pp. 636-641, 646-647 A. INTRODUCTION TO KINGDOM FUNGI 1. Phyla Chytridiomycota (not monophyletic; under revision) Zygomycota (not monophyletic; under revision) Glomeromycota Ascomycota BasidiomycotaFig. 31.11 What features of t
University of Victoria - BIO - General Bi
INTRODUCTION TO EVOLUTIONReading: Chapter 22 A. WHAT IS EVOLUTION? 1. The reality of evolutionary change is the central fact of biological science. "Nothing in biology makes sense except in the light of evolution." Theodosius Dobzhansky (1900 1975)2. Ev
University of Victoria - BIO - General Bi
EVOLUTION OF POPULATIONSReading: Chapter 23 A. POPULATIONS AND GENES 1. Population: members of the same species in the same location; interbreeding. It is populations that evolve, not individuals. Individuals develop and senesce.2. Gene pool: the collec
University of Victoria - BIO - General Bi
SPECIATIONReading: Chapter 24; Chapter 25, pp. 523-525 A. HOW ARE SPECIES DEFINED AND DISTINGUISHED? Wherever possible, biologists mark species boundaries by the biological species concept: "Species are groups of actually or potentially interbreeding nat
University of Victoria - BIOL - 190A
Biomolecules - 1 -BIOMOLECULESCONCEPTS CHECK LIST INTRODUCTION There are four major classes of biomolecules: carbohydrates, lipids, proteins, and nucleic acids. Monomers can be polymerized into long polymers. Polymers can be classified as homopolymers
University of Victoria - BIOL - 190A
Cell Structure and Function -1-CELL STRUCTURE and FUNCTIONCONCEPTS CHECK LIST INTRODUCTION The cell theory states that the smallest unit of life is the cell, that every cell can be considered an elementary organism, and that cells can only be produced
University of Victoria - BIOL - 190A
Membranes and Membrane Transport -1-MEMBRANES and MEMBRANE TRANSPORTCONCEPTS CHECK LIST MEMBRANES The plasma membrane is a selectively permeable barrier that separates a cells cytoplasm from the external environment. The basic structure of a biological
University of Victoria - BIOL - 190A
Bioenergetics & Enzymes - 1 -BIOENERGETICS & ENZYMESCONCEPTS CHECK LIST INTRODUCTION An organisms metabolism refers to all the enzyme-catalyzed reactions that it performs. An organisms metabolic events can be considered as a variety of stepwise enzyme-
University of Victoria - BIOL - 190A
Aerobic Respiration - 1 -AEROBIC RESPIRATION CONCEPTS CHECK LIST[NOTE. This section of BIOL 190A is presented in classes in a manner such that the assigned text readings can be considered as reference material. you should consult your text as needed, bu
University of Victoria - BIOL - 190A
DNA & DNA Replication - 1 -DNA and DNA REPLICATIONCONCEPTS CHECK LIST MODELS OF DNA REPLICATION In cells, DNA is known to be replicated in a semiconservative manner. Older models of DNA replication include the conservative model and the dispersive mode
University of Victoria - BIOL - 190A
Gene Expression - 1 -GENE EXPRESSIONCONCEPTS CHECK LIST INTRODUCTION In a prokaryotic cell, transcription and translation take place in the same compartment and translation of an mRNA can begin even before transcription has finished making an mRNA. In
University of Victoria - MATH - 151
University of Victoria - MATH - 151
SCORE:/34NAME: _ STUDENT NUMBER: _Math 151 A03 & A04 Assignment #1 DUE: FRIDAY, FEBRUARY 5th, 2010 At the START of CLASS NO EXCEPTIONS!ANSWERS ALONE WILL EARN YOU NO CREDIT. ALL RELEVANT WORKINGS MUST BE SHOWN IN FULL. COMPLETE EACH QUESTION AS IF YOU
University of Victoria - MATH - 151
SCORE:/25NAME: _ STUDENT NUMBER: _Math 151 A03 & A04 Assignment #2 DUE: WEDNESDAY, FEBRUARY 24th, 2010 At the START of CLASS NO EXCEPTIONS! * UNSTAPLED ASSIGNMENTS WILL NOT BE ACCEPTED*ANSWERS ALONE WILL EARN YOU NO CREDIT. ALL RELEVANT WORKINGS MUST
University of Victoria - MATH - 151
University of Victoria - MATH - 151
University of Victoria - MATH - 151
SCORE:/25NAME: _ STUDENT NUMBER: _Math 151 A03 & A04 Assignment #3 DUE: WEDNESDAY, MARCH 10th, 2010 At the START of CLASS NO EXCEPTIONS! * UNSTAPLED ASSIGNMENTS WILL NOT BE ACCEPTED*ANSWERS ALONE WILL EARN YOU NO CREDIT. ALL RELEVANT WORKINGS MUST BE
University of Victoria - MATH - 151
University of Victoria - MATH - 151
SCORE:/25NAME: _ STUDENT NUMBER: _Math 151 A03 & A04 Assignment #4 DUE: WEDNESDAY, MARCH 17th, 2010 At the START of CLASS NO EXCEPTIONS! * UNSTAPLED ASSIGNMENTS WILL NOT BE ACCEPTED*ANSWERS ALONE WILL EARN YOU NO CREDIT. ALL RELEVANT WORKINGS MUST BE
University of Victoria - MATH - 151
University of Victoria - MATH - 151
SCORE:/20NAME: _ STUDENT NUMBER: _Math 151 A03 & A04 Assignment #5 DUE: WEDNESDAY, MARCH 24th, 2010 At the START of CLASS NO EXCEPTIONS! * UNSTAPLED ASSIGNMENTS WILL NOT BE ACCEPTED*ANSWERS ALONE WILL EARN YOU NO CREDIT. ALL RELEVANT WORKINGS MUST BE
University of Victoria - MATH - 151
University of Victoria - MATH - 151
SCORE:/18NAME: _ STUDENT NUMBER: _Math 151 A03 & A04 Assignment #6 DUE: Wednesday, March 31st, 2010 At the START of CLASS NO EXCEPTIONS! * UNSTAPLED ASSIGNMENTS WILL NOT BE ACCEPTED*ANSWERS ALONE WILL EARN YOU NO CREDIT. ALL RELEVANT WORKINGS MUST BE
University of Victoria - MATH - 151
University of Victoria - MATH - 151
University of Victoria - MATH - 151
University of Victoria - BIOL - 190B
Introduction to animal diversityIntroduction to Animal Diversity1. What are the characteristics that define an animal? 1.1 Eukaryotes - Membrane bound nucleus & organelles1.2 Heteroptrophs - Obtain energy & carbon sources by ingesting other organisms1
University of Victoria - BIOL - 190B
Chordates part 1Chordates part 1Readings Chapter 34, pp 698 713 Self quiz questions 1, 2 & 3 1. Derived characteristics (Fig. 34.3) (a) Notochord(b) Dorsal hollow nerve cord (c) Post anal tail(d) Pharyngeal gill slits(e) Endostyle2. Lancelets (Clade
University of Victoria - BIOL - 190B
Chordates part 2Chordates part 2Readings: Chapter 34, pp 712- 720 Self quiz questions: None for these sectionsTetrapods started in previous notes 1. Evolution of limbsNew material 2. Amphibians (clade Amphibia) (a) There are three groups(b) The skin
University of Victoria - BIOL - 190B
Chordates part 3Summary of readings for Introduction to animal diversity, invertebrates & chordates Animal diversity Readings: Chapter 18: 370-371, Chapter 21: 445-447, & Chapter 32 Invertebrates Readings: Chapter 33 Self quiz questions 1 5 Chordates: Ch
University of Victoria - BIOL - 190B
Some comments about classifying organisms 1. Traditional taxonomy is based primarily on anatomical characteristics. - It gives no information about how the organisms are related. - There is no definition for the groupings (class, family etc.).2. Phylogen
University of Victoria - BIOL - 190B
Invertebrates -2: Platyhelminthes, Mollusca & AnnelidaFlatworms (Phylum Platyhelminthes)1. General characteristics (a) Dorso-ventrally flattened (b) Bilateral symmetry, triploblastic & acoelomate (c) Diverse habitats - Marine - Fresh water - Damp terres
University of Victoria - BIOL - 190B
Invertebrates 3 Nematodes, arthropods & echinodermsNematodes (Phylum Nematoda) 1. General characteristics (a) Called roundworms (b) Bilateral, non-segmented, pseudocoelomic ecdysozoans (c) Free living(d) Parasitic2. Locomotion (a) They have a tough cut
University of Victoria - BIOL - 190B
Invertebrates - Sponges & cnidariansInvertebratesThe phyla we will discuss are indicatedSponges (Phylum Calcarea & Phylum Silicea) 1. General characteristics2. Structure (a) No plane of symmetry(b) Adults are sessile & immobile(c) Pores (ostia)Page
University of Victoria - BIOL - 190B
PhotosynthesisPhotosynthesisReadings: Chapter 10, pp 185 -203 Self quiz questions: 1 7 Background 1. Plants are autotrophs (a) They produce organic molecules from CO2 & other inorganic molecules. (b) They use the energy from the sun.Location 1. It occu
University of Victoria - BIOL - 190B
Plant responses to internal & external signalsPlant responses to internal & external signalsReadings: Chapter 39 Self quiz: Questions 2, 4, 5, 6, 7, & 9 How do plants receive signals? 1. Signal-transduction pathway has 3 main parts (a) The signal (stimu
University of Victoria - BIOL - 190B
Plant structure, growth & developmentPlant structure, growth & development Readings: Chapter 35: pp 738-754 Self quiz 1-8 Plant structure 1. Plants have cells grouped into tissues and tissues grouped into organs (a) Tissues(b) Organ2. There are three p