Bio 351 Final Exam Review
Bio 351 Final Exam Review BIO 351
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This 5 page Study Guide was uploaded by Anna Kretschmar on Thursday March 10, 2016. The Study Guide belongs to BIO 351 at California Polytechnic State University San Luis Obispo taught by Edward Himelblau in Winter 2016. Since its upload, it has received 108 views. For similar materials see Principles of Genetics in Biology at California Polytechnic State University San Luis Obispo.
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Date Created: 03/10/16
Bio 351 Final Exam Review 1. What is the advantage of performing a testcross in a genetic experiment? 2. How did Mendel’s experimental approach make it possible for him to elucidate the laws of inheritance? 3. Explain the relationship between the following terms: gene, allele, mutation, genotype, phenotype 4. Explain the formation of a “replication bubble” during DNA replication. 5. What is “exonuclease activity” and why is it important that polymerases have this activity? Distinguish between 5’to 3’ exonuclease activity and ‘3 to 5’ exonuclease activity. 6. DNA sequencing involves DNA polymerase, a primer, and nucleotides. What’s special about the nucleotides and how does this allow the DNA sequence to be determined? 7. How are RNA primers removed and the DNA double strand completed during DNA replication? 8. Distinguish between homologous chromosomes, identical chromosomes, sister chromatids. 9. Explain the relationship between histones and nucleosomes. 10. Be familiar with the events that occur during the major phases of mitosis and meiosis: prophase, prometaphase, metaphase, anaphase, telophase. 11. Compare metaphase of mitosis and meiosis. Distinguish between mitosis and meiosis based on chromosome alignment and pairing. 12. Outline the differences between female and male meiosis in humans. Focus on when meiosis begins and ends in each. 13. Approximately how far back (generations) must we go to reach the common pool of ancestors for all humans alive today? Explain. 14. Explain what we learn about human sex determination from observing people with Turner’s syndrome and Klienfelter’s syndrome. 15. Explain sex determination in Honeybee. 16. Diagram the role of SRY, SOX9, and MIH in human sex determination. Consider how gain and loss of function of these genes could influence sex determination. 17. Distinguish between incomplete dominance and codomiance. Solve genetics problems involving alleles that are incompletely or codominant. 18. How are penetrance and expressivity similar? How are they distinct? 19. Define epistasis and be prepared to explain how the following modified 9:3:3:1 ratios come about: 9:4:3, 12:3:1. 20. How do linked genes differ from genes that assort independently? Distinguish between complete linkage and incomplete linkage. 21. Explain why there is a maximum distance between linked genes that can be detected by genetic experiments (i.e. crosses (not DNA sequencing))? 22. Diagram the steps of recombination diagramed in class (breakage, invasion, migration, resolution). At which point is a Holliday Structure visible? 23. Distinguish between the results of horizontal and vertical resolution of the Holliday Structure. Which results in recombination between homologous chromatids? 24. Distinguish between a mismatch, an incorporated mismatch, and a mutation. 25. How are mismatches repaired? How are incorporated mismatches repaired in prokaryoptes? 26. What are the four ways that chromosomes can change? (duplication, deletion, inversion, translocation). For each, suggest how each type of change can lead to altered dosage (this is easy in the case of insertion and deletion, but what about inversion and translocation?...consider events during meiosis.) 27. Distinguish between Autopolyploids and allopolyploids. ANSWERS ON NEXT PAGE 1. A test cross is when something with an unknown genotype is crossed with a recessive expressing organism. This allows one to determine if the unknown dominant is homozygous or heterozygous for its dominant trait. 2. Because Mendel was able to manipulate the fertilization he performed test crosses and reciprocal crosses. He crossed male purple with female white and male white with female purple to show that parent sex did not play a role in the outcome out the F1 generation. Both crosses still had all purple F1. 3. A gene is the genetic information to make a particular protein. Alleles are the different versions of that gene. Two alleles per gene can determine the genotype of the gene. This is then expressed as a phenotype which is the physical trait associated with a certain genotype for a certain gene. Mutations are any changes in the DNA sequence of a gene, thus changing the info that encodes a protein. This can cause new alleles to arise which than can alter the genotype, and encodes for a different protein thus changing the phenotype. 4. The replication bubble begins at the origin of replication. This process begins when proteins recognize the start signal on a strand. Helicase then binds to the DNA to begin pulling the strands apart by breaking hydrogen bonds between base pairs. As this is occurring, gyrase/topoisomerase attaches further down the strand to help relieve the tension of the coiled DNA. Next RNA primase binds to the strand to provide an RNA primer. DNA polymerase III attaches and begins to add base pair nucleotides to the 3’ side of the strand. The sliding clamp in front keeps the polymerase tight on the strand. On the leading strand this is a continuous process. On the lagging strand it is discontinuous. Polymerase is attached at multiple places thus replicating DNA and creating Okazaki fragments. SO then the primers need to be removed so the fragments can be attached. Polymerase III goes back to the front and polymerase I binds, it is a 5’ to 3’ exonuclease that removes nucleotides from the end of a strand, thus removing the primer. A nick, or open space is then left so DNA ligase comes in and creates a phosphodiester bond to complete the strand. 5. Exonuclease activity is the ability to remove one nucleotide from the end of a strand. 5’to 3’ is used to remove primers and create the nick that ligase will fill in. 3’ to 5’ is used as proofreading system. If an error is found then polymerase III is called back to correct the error. 6. The nucleotides used for sequencing are actually dideoxyribonucleotides, which means both the 2’ and 3’ carbons lost their oxygen. Each base has a specific tag that has its own florescent color. After DNA strands are sequenced they are put on a cell. The shortest ones run off the end first. When a piece runs off it is shot with a laser, which causes the tag to be excited and show its color. You can then ID that specific base and then determine what its pair would be and find out what the complementary strand is. 7. DNA polymerase I binds to the 3’ end of the strand. This includes an exonuclease that removes nucleotides so it cuts off the primers from the end of the strand. Ligase then comes in to form a phosphodiester linkage that is missing in the nick to complete the strand. 8. Chromatin is condensed to create a chromosome which is then replicated to get sister chromatids. Sister chromatids, also called identical chromosomes, are identical and carry the same versions of all their genes because one was produced as an exact copy of the other. These are held together by centromeres. Homologous chromosomes are the two chromosomes that make up a homologous pair, with one being inherited from each parent so there can be different versions of the same gene. They are where crossing over occurs. Sister chromatids are for cell division/DNA replication/mitosis and homologous chromosomes are for reproductive division. 9. A nucleosome is made up of histones with the double wrapped DNA around it. Histones are just groups of proteins. Groups of 9 histones cluster together and then are wrapped with DNA and connected to other clusters with linker DNA. 10. prophase – chromosomes condense; each chromosome consists of a pair of identical sister chromatids joined at the centromere. metaphase – chromosomes line up at the middle of the cell, along the plane of cell division, pushed and pulled by microtubules of the spindle apparatus anaphase – sister chromatids separate and migrate towards opposite ends of the cell telophase – chromatids cluster at opposite ends of the cell and begin to decondense cytokinesis – the membrane pinches in to divide the two daughter cells Centromere – point where sister chromatids are connected Metaphase plate – where the sister chromatids or homologs line up during metaphase Tetrad – also called a bivalent, when two homologous chromosome pairs connect during prophase I Crossing over – only occurs in meiosis, occurs during prophase as tetrads form Chiasma – site of crossing over, helps with the cohesion between two sister chromatid pairs 11. In mitosis the homologous chromosomes align end to end while in meiosis they align side by side. In mitosis and meiosis II it is sister chromatids that align, compared to meiosis I when homologous chromosomes align. In mitosis this is so one sister chromatid of each gene ends up in the two daughter cells that result, to create identical diploid copies. In meiosis the same chromatids go together so that there is one copy in each of the four different haploid daughter cells. 12. Oocyte is the starting cell for females. Meiosis I takes 13-55 years which is from start of ovulation to menopause. One cell becomes the polar body. Meiosis II is then very quick once the egg has been fertilized. Then one becomes the mature egg and the other forms the second polar body. For males, the spermatocyte is the start. Meiosis I takes 16 days and meiosis II takes 16 days as well. Then it takes 16 days to become mature sperm. Females produce one egg with 2 polar bodies and males produce four sperm. 13. If we go back 30 generations then we would have 1 billion ancestors, however, the world population was only 300 million at the time. This is because there are a lot of people who contribute multiple times to our gene pool. This is like when cousins or long distance relatives marry, so their common ancestors has now contributed multiple times to the gene pool. 14. These syndromes result from aneuploidy, or having one too many or one too few chromosomes. Nondisjunction results in ova that carry either two X chromosomes or none. The former results in Klinefelter syndrome when fertilized by a Y-containing sperm; these are males but are usually sterile, some breast development. The latter results in Turner syndrome when fertilized by an X-containing sperm, which are then sterile females. 15. Males come from unfertilized eggs that are haploid with one set of chromosomes. Females come from the fertilized, or diploid, eggs. However, some diploid males were found. It was discovered that those that were fertilized and homozygous at the sex determining locus were males and those that were heterozygous were females. This is when the complementary sex determiner gene, CSD, was found. It is necessary for the development of females. When it is inactive then a switch to male form is made. 16. Sox9 and MIH are present in males and females, however, are only active in males. Females have an RSPOI protein that inhibits the activation of SOX9. Because these are never expressed, then the mullerian develops into female gonads and it inhibits the growth of wolffian tissue which dies out. However, in males the SRY is present from the Y chromosomes. SRY then turns on SOX9 which then turns on MIH which then binds to the mullerian receptors to kill it off. This leaves the wolffian tissue to develop into male gonads. (Page 40). LOF of SRY or SOX9 or MIH could allow XY to develop into females. GOF of SOX9 or MIH could allow XX to develop into males. 17. Codominance is when the heterozygote has its own unique phenotype, such as white and black and then spotted. This creates a 2:1:1 ratio with the heterozygote as the 2. DNA gel would be normal heterozygote but show a different phenotype. Incomplete dominance is when the heterozygote is a mix/blend of the two homozygotes, such as a white and yellow creating a pale yellow. RNA gel depends on function of protein. 18. Penetrance is when the phenotype does reflect the genotype. Expressivity is the range of phenotypes that are expressed for a genotype, ranging from mild to severe. Two people could have a genotype for a disease that is completely penetrant. However, one could have very mild symptoms of the disease while the other has a severe case of it. Their expressivity would be higher. 19. Epistasis is when the alleles at one locus prevent you from seeing the phenotype of a different locus. When one is epistatic to another in a recessive way, that one will cover the other regardless if it is homozygous recessive, to get a 9:4:3 ratio. When one is epistatic to another in a dominant way, that one will cover the other regardless if it has a dominant allele, to get a 12:3:1 ratio. 20. Linked genes are found close together on the same chromosome and can be predicted to be found with the same version of the other gene as seen in the parent. Independently assorted genes are random found in the offspring. Complete linkage means there would only be parental offspring, while incomplete linkage would result in parental and recombinant offspring. 21. Once genes 50 or more map units apart then the offspring have the same percentages as independent assortment so it is hard to tell what type of genes were involved. Also, once the genes get that far apart crossing over usually happens multiple times so the offspring end of in the normal/parental state in the end. 22. Recombination starts with a single strand break. This allows for strand invasion where the genes cross over. RecA helps facilitate but helping the single strands find pairs. Then there is branching migration where the exchange region moves down the strand with the help of helicase. It is at this point that the holliday junction is visible. There is one of two resolutions which then give you your final strands. 23. Horizontal resolution results in just one section switching between the two strands to create a heteroduplex, but not recombinants. Vertical resolution results in recombinants because both ends of one set of strands are the same and continue in one direction while the other end continues with the other set, with some parts as a heteroduplex. 24. Mismatch is when there are non-complementary pairs. An incorporated mismatch is one when synthesis continues so the mismatch is then included in the strand to create a warped effect. A mutation is when the mismatch is not repaired and is replicated thus becoming a mutation that can be passed on. 25. Mismatches are repaired with a 3 to 5 exonuclease that proofreads. An incorporated mismatch is fixed with a certain system. An endonuclease removes the incorrect section from the strand to create a gap that is then filled with DNA polymerase and the nick is fixed with ligase. The endonuclease know which region to remove because DNA methylase is added to A nucleotides so after DNA replication the old strand has more methyl groups. The endonuclease then removes the region from the less methylated strand. 26. Duplication which is a form of insertion changes dosage by increasing it. Deletion changes dosage by decreasing it. Inversion just flips the order of genes so it has the same dosage, however dosage can differ in the offspring. Some strands could have multiple centromere while others don’t have any, which means the chromosomes won’t segregate normally in meiosis. Translocation doesn’t change the dosage but is when there is crossing over between nonhomologous chromosomes. Reciprocal is when the ends switch and robertsonian is when two combine and the tips are lost. This also only causes problems in the offspring. There can be some WT but there will also be ones with monosomy or trisomy which are typically lethal. 27. Autopolyploids make extra copies of each chromosome (not a mutation like aneuploidy) to change the ploidy from 2n to 3n to 4n and so on after a whole genome duplication with no mitosis/errors in mitosis. Allopolyploids end up with 2 whole genomes in one organism following whole genome duplication because of unreduced gametes/errors in meiosis. These create new species.
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