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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? B. How does the general model of strand invasion, branch migration, and resolution of the Holliday junction work? C. How do RecA and RecBCD function to promote recombination in E. coli? The repair of double-strand breaks (DSBs) and the production of DSBs for recombination are connected through the process of homologous recombination. Homologous recombination uses a homologous DNA molecule as a template to repair DSBs, with strand invasion as a key step. It involves the invasion of single-stranded DNA, branch migration along the homologous DNA, formation of Holliday junctions, and their subsequent resolution to exchange genetic material. RecA and RecBCD proteins play important roles in promoting recombination. RecA facilitates strand invasion, while RecBCD processes DNA ends and aids RecA loading onto single-stranded DNA.

D. What is the connection between nonhomologous end joining repair and cancer? Other disorders? 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? 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? Nonhomologous end joining repair, an error-prone DNA repair mechanism, can introduce mutations, chromosomal rearrangements, and genomic instability, contributing to cancer development and therapy resistance. Homologous recombination is a precise DNA repair mechanism involving key steps like strand invasion and Holliday junction resolution, while NHEJ repair can be error-prone and is linked to cancer development. Y A X ? Genotype : RutE Phenotype : RT ROTEx RUTE RT RE UT Of J B/c it was a RT ART PRI ROT RUTE Rt R Rat RUS6 Rot mix amongst T RTT Rr56 rust or + themselves . * ↑ rustRret rust or bt

C. What fraction of the F 2 generation is expected to be purebreeding? D. Would you expect to see the same or different results if the parental tomato plants had genotypes RRtt and rrTT ? ° 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? B. Would it work to mate the F 1 fish carrying a mutation to one of its mutant siblings? Why or why not? In the F1 generation, most mutants cannot be readily identified because individuals are typically heterozygous for a mutation. The presence of one wild-type allele often masks the effects of the mutant allele, resulting in a normal phenotype In genetic studies, the F3 screen is a crucial tool used to identify and characterize mutations in organisms. This process involves multiple generations of breeding and careful observation Mandel's 1st Postulate RRTT and rott & ↑ Genotype of 51 = RuTE individual is sam I As geoffa os a e

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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? ° 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? Mating F1 individuals (heterozygotes) with their mutant siblings (also heterozygotes) is ineffective for identifying recessive mutations. This is because their offspring (F2 generation) will also be heterozygous and exhibit a normal phenotype. D

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 - · ↑ 24 = 16 O · G

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