Class Notes (806,741)
Canada (492,424)
Biology (6,676)
Biology 1002B (1,340)
Tom Haffie (863)

bio notes.docx

12 Pages
Unlock Document

Western University
Biology 1002B
Tom Haffie

Lecture 18: DNA Replication components necessary for DNA synthesis: substrates, template, primer and enzymes. Substrates: A,T, G, C base pairs Template: the original strand of DNA Primers: RNA primers base paired with the template to allow DNA polymerase to work Enzymes: all enzymes and proteins required to make the replisome. (to synthesize DNA) structure of DNA with respect to polarity and complementarity: DNA has a sugar- phosphate backbone. It has 4 nitrogenous bases: adenine, guanine, thymine and cytosine. Adenine and guanine are purines (double-ring structures). Thymine and cytosine are pyrimidines (single-ring structures). A base pairs with T, C base pairs with G. DNA is polar because it has a 5’ end and a 3’ end. 5’ and 3’ end of a DNA strand: The 5’ end of DNA has a phosphate group. The 3’ end has a hydroxyl group. DNA synthesis occurs in the 5’ to 3’ direction. relationship between DNA synthesis and chromosome duplication during S phase: DNA is synthesized during S phase, and therefore all the chromosomes (bundles of DNA) get duplicated. structure of replication fork and replication bubble: problem that arises during replication of the ends of linear chromosomes: DNA polymerase can only extend DNA from a free 3’ OH group. Because the lagging strand ends on a 3’ end, there is no way to finish the DNA even if a primer were there, as we would not be able to replace the primer because there would be no free OH group. This means that the chromosome would get progressively shorter and shorter if there were no mechanism to fix this. how telomerase addresses the above problem: the ends of chromosomes contain a G rich series of repeats, called a telomere. Telomerase recognizes the tip of an existing repeat series. Using an RNA template, telomerase elongates the parental strand in the 5’ to 3’ direction and adds additional repeats as it moves down the parental strand. The lagging strand is then completed by DNA polymerase alpha, which carries a DNA primase as one of its subunits. In this way the original information at the ends of linear chromosomes is completely copied into the new DNA, and the only thing lost is a bit of the telomere. Lecture 19: Gene Mutation characteristics of the Ames Test as a screen for chemical mutagens: the bacterium Salmonella typhimurium is used because there is a mutant that is unable to synthesize the amino acid histidine. Therefore when this mutant is grown on minimal medium, it will not grow. The Ames test takes possible chemical mutagens and puts them on minimal medium with the mutant bacteria. If the bacteria start to grow, it is assumed that the chemical caused a back mutation, allowing the bacteria to start to synthesize histidine. This means that the chemical causes mutations in bacteria and possibly other organisms and is a positive Ames Test. characteristics of mutation vs. DNA damage: a mutation is a heritable change in double-stranded DNA sequence such as base substitutions, insertions/deletions/reorganizations. DNA damage is a physical abnormality in the DNA. Damage can be recognized by enzymes and be correctly repaired. If it cannot be repaired then transcription, translation and replication will be blocked and/or the cell with go through apoptosis. mechanisms of spontaneous substitution mutation arising from errors in replication: Sometimes a base pair can undergo a tautomeric shift, which means that the electrons within the base pair shift and then it pairs with a different base pair. An example is thymine, which normally has 2 hydrogen bonds with adenine. Sometimes it undergoes a tautomeric shift and will base pair with guanine and have 3 hydrogen bonds. mechanisms of spontaneous indel mutation arising from errors in replication: Because of the sequence (same base pairs all in a row), one or the other of the strands can loop out. Therefore when the DNA is replicated, one of the sides is too short. mechanism by which base analogues, such as 5-bromouracil, increase mutation frequency: 5-bromouracil is indistinguishable from thymine and will be put in DNA instead of thymine. The bromine atom leads to a molecule that is tautomerically unstable, which is bad as it will always change which base it wants to pair with. likely effects of silent, missense and nonsense mutations on protein function: (all are caused by a change of a single base pair) Silent: the changed codon will specify for the same amino acid, due to the degeneracy of the genetic code. Missence: the changed codon will specify for a different amino acid. This may or may not alter the protein function in a big or small way. Nonsense: the changed codon will specify for a stop. This causes a premature termination of the protein during translation. This will almost certainly affect the protein in a big way, it will likely be partially functional at best. mechanism of frameshift mutation and likely effect on protein function: A frameshift is cause by an insertion or deletion of a base pair. This alters the whole reading frame from there on. The resulting polypeptide is usually non-functional because of the significantly altered amino acid sequence. likely effects of mutations outside of coding regions: Some of the non-coding regions are used for transcriptional, translational, post-transcriptional regulations. These mutations would then affect the regulations. It really depends on which part is being affected. Lecture 20: Chromosome Mutation mechanism of action of DNA damage by non-ionizing radiation (UV): pyrimidines, in particular thymines, that are side by side can absorb photons of light and this causes a reorganization of electrons which causes covalent bonds and creates a thymine dimer. This distorts the helix and blocks replication and transcription. mechanism of action of DNA repair enzymes: Proofreading: if a newly added nucleotide is mismatched, DNA polymerase will reverse and use a a built-in deoxyribonuclease to remove the incorrect nucleotide. The enzyme then resumes working forward. (RNA polymerases cannot proofread). Reverse transcriptases can’t really proofread either. Excision repair: the part where the bad base pair is is cut out and then it is filled in. Photoreactivation: In most organisms thymine dimers can be repaired by photoreactivation. Photoreactivation is a repair process in which photolyase enzymes directly reverse thymine dimers via photochemical reactions. Lesions on the DNA strand are recognized by these enzymes, followed by the absorption of light wavelengths (particularly blue light). This absorption enables the photochemical reactions to occur, which results in the elimination of the pyrimidine dimer, returning it to its original state. mechanism of action of DNA damage by ionizing mutagenesis: When ionizing radiation passes through your cells, it creates reactive oxygen. This then damages everything, from proteins to DNA. mechanism of production of aneuploid (abnormal number of chromosomes) gametes during meiosis: This happens by nondisjunction, which is the failure of homologous pairs to separate during the first meiotic division or of chromatids to separate during the second meiotic division. difficulties of studying effects of radiation exposure on large populations: its hard to isolate large populations, and there are infinite amounts of other factors affecting the population, because it is so large and therefore more diverse. Lecture 21: Immunogenetics role of phagocytosis in immune response: This is the non-specific part of the immune system. It goes after bacteria and other antigens and engulfs them and breaks them down. characteristics of innate vs. adaptive immune response: The innate (non-specific) immune response goes after any sort of antigen it sees as abnormal. The adaptive immune response involves inherited mechanisms that lead to the synthesis of molecules such as antibodies that target pathogens in a specific way. mechanism of generation of diverse antibody genes: The receptors and the antibodies the cell will eventually produce are tetramers. There are two heavy chains and two light chains that make up this protein. On the light chain gene, there are V segments and J segments. V means variable, J means joining. There are several, sometimes hundreds of V segments and 4-5 J segments. As the B-cell matures, it purposefully causes mutations in order to build an immunoglobulin gene. The segments are like lego inherited from parents to make your own genes. This is called somatic recombination. On the heavy chain gene, it is similar but they also have D segments. D stands for diversity. Ds are put with Js, the DJs with Vs. structure of antibodies and corresponding B cell receptor (BCR): Each antibody molecule consists of 4 polypeptide chains: two identical light chains and two identical heavy chains. These form a Y structure, called a B-cell receptor (BCR). The BCR is different in every type of antibody because of the reasons stated above. role of chromosomal rearrangements of antibody sequences in activation of oncogenes: the DNA in your lymphocytes gets rearranged and sometimes it can mess up and you can have antibody genes (promoters, etc) controlling oncogenes. mechanism giving rise to immunological memory: some of the activated lymphocytes differentiate into memory cells that circulate in the body, ready to initiate a rapid immune system response on subsequent exposure to the same antigen. plausible explanations for why different cell lines in chimeras do not mount immune responses against each other: the sheep can tolerate the goat because the immune system forms after the embryogenesis of the geep. Therefore the immune system treats both the sheep and the goat as "self". Lecture 22/23: DNA Technologies I and II (ELYSIA) Recall that the surprising ability of the “solar slug” Elysia to maintain functional chloroplasts stolen while feeding on Vaucheria algal cells (kleptoplasty) is an unsolved mystery in science. One possibility is that genes required for plastid maintenance have undergone evolutionary lateral gene transfer from the nucleus of the algae to the nucleus of the slug. These two lectures were designed to analyze a primary research article and showcase how the authors used selected DNA technologies to generate data in support of their hypothesis that a particular photosynthetic gene (psbO) from the nucleus of Vaucheria is also present in Elysia. Overall, you should be able to I) identify the various types of data shown by this paper; II) explain how these data were obtained; and III) conclude whether each type of data does, or does not, support the hypothesis. I) 1. Picture of the amplified portion of the cDNA of the psbO transcript on agarose gel in Vaucheria and Elysia 5 months after algal feeding. 2. Northern blot analysis of Vaucheria and Elysia psbO transcript (RNA) 3. RACE-amplified sequence of psbO gene in DNA of Vaucheria, egg Elysia, and adult Elysia. 4. Translation of the MSP amino acid sequences in Vaucheria and Elysia. They are identical between the two and have a tripartite targeting signal. II)1. Total RNA was isolated in Vaucheria and then was reverse transcribed into cDNA (complementary DNA, half RNA, half DNA). Degenerate primers were created for the psbO gene from other published sequences and the cDNA was amplified using PCR. The amplified cDNA was then electrophoresed. 2. a Northern blot analysis was done on the cloned psbO transcript. This is done by gel electrophoresis on the RNA and then detected with a hybridized probe. 3. Homologous primers were designed from the RACE-amplified sequence of the Vaucheria psbO cDNA. By PCR, these primers amplified the genomic DNA of Vaucheria, Elysia eggs and adult Elysia tissue as well as adult Elysia cDNA (created by RT-PCR). 4. The amino acid sequences were analyzed by phylogenetic methods, based on amino acid sequences of 25 mature proteins and carried out using maximum parsimony III) 1. Supports the hypothesis because we would not normally expect psbO transcription 5 months after feeding as the plastids cannot produce it themselves. 2. Somewhat supports the hypothesis as the probe reacted with both species but there was not picture of a negative control to really prove the hypothesis 3. Supports the hypothesis because it shows the gene in Elysia egg, proving Elysia doesn’t need to eat algae to get the gene. 4. Does not support the hypothesis because although the sequences are identical, the tripartite targeting sequence can only be used in chloroplasts with 4 membranes, like Vaucheria, and the chloroplasts Elysia retains only have 2 membranes (the chloroplast endoplasmic reticulum isn’t retained in the engulfed plastids). location of PsbO gene in photoautotrophic organisms: It is found in the nuclear genome. location and role of PsbO gene product in photosynthetic electron transport: PsbO gene product is found on the lumenal side of the thylakoid membrane. Found in complex 2 of the electron transport chain. purpose of molecular size markers in electrophoresis: A molecular weight size marker is used to identify the approximate size of a molecule run on a gel, using the principle that molecular weight is inversely proportional to migration rate through a gel matrix. Therefore when used in gel electrophoresis, markers effectively provide a logarithmic scale by which to estimate the size of the other fragments (providing the fragment sizes of the marker are known). These markers can be composed either of different proteins of known size, or nucleic acids of different sizes. Markers are loaded in lanes adjacent to samples lanes before the commencement of the run. interpretation of agarose gel data: Because a larger molecular weight means a slower migration rate through the gel, it is easy to compare many things this way. For example, seeing if two samples have the same gene (Vaucheria and Elysia) mechanism of polymerase chain reaction: PCR uses repeated cycles of heating and cooling to make many copies of a specific region of DNA. First the temperature is raised to near boiling (94℃), causing the double stranded DNA to separate, or denature into two single strands. Then the temperature is decreased (45-65℃), and short DNA sequences called primers bind/anneal to complementary matches on the target DNA sequence. The primers bracket the target sequence to be copied. At a slightly higher temperature (72℃), the enzyme Taq polymerase binds to the primed sequences and adds nucleotides to create the second strand on each individual strand, thus doubling the amount of DNA. This cycle is repeated many times to make multiple copies of the sequence. role of thermal cycling in polymerase chain reaction: The changes in temperature allows the DNA to separate, the primers to anneal and the Taq polymerase to add nucleotides. role of primers in polymerase chain reaction: The primers (one for the bottom and top strand) are about 16-25 base pairs in length. They anneal to certain parts on the DNA, thus specifying the region that will be multiplied. role of Taq polymerase in
More Less

Related notes for Biology 1002B

Log In


Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.