Evan Zhao DIS 7 Wks

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University of Illinois, Urbana Champaign *

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Biology

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Apr 3, 2024

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Evan Zhao MCB 250 Spring 2024 Discussion 7 / Work Sheet 7 March 7-8 & 18 1. Homologous Recombination (Moved homologous recombination to Week 7’s discussion so it is more relative and students won’t be confused about whether it will be on exam 2 or not. A. How are the repair of double strand breaks (DSBs) and production of DSBs for sexual reproduction connected? Double strand break repair and production are connected because DSBs can be created for recombination. Strand invasion will create a Holiday junction that facilitates recombination and repair. B. How does the general model of strand invasion, branch migration, and resolution of the Holliday junction work? First, a single-stranded DNA molecule invades the joined template double-stranded DNA molecule, which results in a region of heteroduplex DNA (strand invasion). After strand invasion, the branch migration takes place where DNA strands are unwound and rewind to enable the exchange of genetic information. After strand invasion, the branch migrates in both directions to stabilize the Holliday junction, and the Holliday junction is resolved by cleavage of the junction into two separate molecules. C. How do RecA and RecBCD function to promote recombination in E. coli? RecA is protein that functions by binding to the single-stranded DNA regions that were exposed during DNA damage. It searches for homologous sequences in the intact sister chromatid or another DNA molecule. On the other hand, RecBCD has many functions in the DNA recombination and possesses both helicase and nuclease abilities. RecBCD unwinds the double-stranded DNA to create a single-stranded DNA region and degrades the DNA to produce 3' single-stranded tails. D. What is the connection between nonhomologous end joining repair and cancer? Other disorders? Nonhomologous end joining repair introduces errors in the affected sequence which leads to cancer and other disorders like SCID, radiosensitivity, and microcelphaly. 2. Mendelian genetics In tomato plants, the gene R has two alleles that control fruit color: the R allele is dominant and produces red tomatoes; the r allele is recessive, and when homozygous produces yellow tomatoes. Gene T controls plant height: the T allele is dominant and produces tall plants; the t allele is recessive and when homozygous produces dwarf plants. Genes R and T undergo independent assortment. A. Parental tomato plants of genotype RRTT and rrtt are crossed. What genotype and phenotype do you expect in the F 1 generation? RrTt

Evan Zhao B. The F 1 tomatoes are interbred. Draw a Punnett Square to predict the genotypes and phenotypes that will be seen in the F 2 generation. Why is this only a prediction? RT Rt rT rt RT RRTT RRTt RrRR RrTt Rt RRTt RRtt RrTt Rrtt rT RrTT RrTt rrTT rrTt rT RrTT RrTt rrTT rrTt C. What fraction of the F 2 generation is expected to be purebreeding? Fraction of the F2 generation = 1/8 D. Would you expect to see the same or different results if the parental tomato plants had genotypes RRtt and rrTT ? I would expect different results. 3. Genetic screens Geneticists use the F 3 screen (see below) to isolate and characterize new mutations in a wide variety of animal and plants. Use your knowledge of mutagenesis and Mendelian inheritance to answer the following questions. A. Why is the F 3 screen necessary, i.e. why is it that most mutants cannot be identified in the F 1 generation? F3 screen is homozygous, so it can demonstrate the mutations in its phenotype. B. Would it work to mate the F 1 fish carrying a mutation to one of its mutant siblings? Why or why not? No, the mutation would be heterozygous instead of homozygous, demonstrating the wildtype allele as the dominant trait in the phenotype.

Evan Zhao C. A single zebrafish mating produces around one hundred offspring. In practical terms, a geneticist who performs the F 3 screen will separate out each individual F 1 offspring, and carry its descendants through the F 2 and F 3 generations independently of the others. Why is this necessary? Homozygous mutants are necessary to demonstrate the mutation as a phenotype.

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Evan Zhao 4. PCR - Choose the one best answer for each question. 3.1 What is the polymerase chain reaction (PCR)? A. A method to propagate a gene in bacteria B. A method to determine the sequence of bases in a gene C. A method to join two fragments of DNA together D. A method to amplify a fragment of DNA 3.2 What is one limitation of the PCR method? A. It generates very little product. B. It only works for bacterial DNA. C. You must know enough about the DNA of interest to make primers. D. You must know the amino acid sequence of the DNA to be amplified. 3.3 The Taq enzyme is a type of DNA polymerase that is used in PCR because of its ability to withstand which step of the PCR cycle? A. Denaturation B. Annealing C. Extension D. It needs to withstand each of the steps listed 3.4 How many DNA molecules would there be after four rounds of PCR if the initial reaction mixture contained four molecules of DNA (in other words, two pieces of double stranded DNA)? A. 64 B. 8 C. 32 D. 16 3.5 For which step in the PCR cycle are dNTPs required? A. Annealing B. Extension C. Denaturation D. All of these steps 3.6 During which step in the PCR cycle do primers form bonds with a single- stranded template? A. Annealing B. Extension C. All of these steps D. Denaturation

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