HW 2 – BME 50A – Winter 2016 Question 1 Consider the following diagram of a bacterial genome (only a portion is shown). Note that there is a single origin of replication (Ori) and that directly to the right of Ori, there is a promoter driving the expression of a protein-coding gene (GENE1), whose start codon, ATG, is indicated. a) In the region where GENE1 is, which direction does the replication fork move during replication (left to right, or right to left)? b) In the region where GENE1 is, which strand acts as the template for leading strand DNA synthesis (A or B)? c) When GENE1 is transcribed into RNA from the promoter, which strand, A or B, acts as the template for transcription (A or B)? What direction does the RNA polymerase move during transcription (left to right, or right to left)? d) Highly expressed genes near GENE1 are usually oriented in the same direction as GENE1 (i.e., the promoter points in the same direction as that for GENE1). Why?Question 2 Let’s say I have two coding sequences, Seq1 and Seq2, below, that differ in one position. (The position where they differ is in bold.) Seq1: …AAA GAT TAC TTT GGC CGT CGT AGC ACG ATA ATG… Seq2: …AAA GAT TAC TTT GGC CGT CGG AGC ACG ATA ATG… Here is the genetic code for you to use. a) Please translate both sequences to amino acid sequences. What is the amino acid sequence of Seq1 and Seq2? b) Which of the two sequences, Seq1 or Seq2, do you think is more evolvable, defined as the likelihood that a mutation will result in a change in function? Why?Question 3 Recall the steps of homologous recombination: Biochemists have reconstituted this process in a test tube, which allows us to figure out the necessary and sufficient set of components for the reaction, since we control what’s in the test tube. Please predict if you will get to the final product (non-sister chromatids with crossover) under the following scenarios: a) You don’t add dNTPs. Final crossover product expected? Why or why not? b) You have the full set of components that make this reaction work and in addition, you add a nuclease that digests 3’ ends. Final crossover product expected? Why or why not? c) You have the full set of components that make this reaction work, but instead you replace the nuclease that digests 5’ ends with a nuclease that digests 3’ ends. Final crossover product expected? Why or why not? 652 CHAPTER 19 Sexual Reproduction and the Power of GeneticsCrossing-Over Occurs Between the Duplicated Maternal and Paternal Chromosomes in Each BivalentThe picture of meiotic division I we have just painted is greatly simplified, in that it leaves out a crucial feature. In sexually reproducing organisms, the pairing of the maternal and paternal chromosomes is accompanied by homologous recombination, a process in which two identical or very similar nucleotide sequences exchange genetic information. In Chapter 6, we discussed how homologous recombination is used to mend damaged chromosomes from which genetic information has been lost. This type of repair, uses information from an intact DNA double helix to restore the correct nucleotide sequence to a damaged, newly duplicated homolog (see Figure 6–30). A similar process takes place when homologous chro-mosomes pair during the long prophase of the first meiotic division. In meiosis, however, the recombination occurs between the non-sister chromatids in each bivalent, rather than between the identical sister chro-matids within each duplicated chromosome. As a result, the maternal and paternal homologs end up physically swapping homologous chro-mosomal segments in a complex, multistep process called crossing-over (Figure 19–10).Crossing-over is facilitated by the formation of a synaptonemal complex. As the duplicated homologs pair, this elaborate protein complex helps to hold the bivalent together and align the homologs so that strand exchange can readily occur between the non-sister chromatids. Each of the chromatids in a duplicated homolog (that is, each of these very long DNA double helices) can form a crossover with either (or both) of the chromatids from the other chromosome in the bivalent. The synapton-emal complex also helps space out the crossover events that take place along each chromosome.By the time prophase I ends, the synaptonemal complex has disassem-bled, allowing the homologs to separate along most of their length. But each bivalent remains held together by at least one chiasma (plural chiasmata), a structure named after the Greek letter chi, , which is shaped like a cross. Each chiasma corresponds to a crossover between two non-sister chromatids (Figure 19–11A). Most bivalents contain more than one chiasma, indicating that multiple crossovers occur between homologous chromosomes (Figure 19–11B and C). In human oocytes—the cells that give rise to the egg—an average of two to three crossover events occur within each bivalent (Figure 19–12). Crossovers that occur during meiosis are a major source of genetic variation in sexually reproducing species. By scrambling the genetic con-stitution of each of the chromosomes in the gamete, crossing-over helps to produce individuals with novel assortments of alleles. Crossing-over also has a second important role in meiosis. By holding homologous Figure 19–10 During meiosis I, non-sister chromatids in each bivalent swap segments of DNA. Here, only two of the four sister chromatids in the bivalent are shown, each drawn as a DNA double helix. During meiosis, the protein complexes that carry out this homologous recombination (not shown) first produce a double-strand break in the DNA of one of the chromatids (either the maternal or paternal chromatid) and then promote the formation of a cross-strand exchange with the other chromatid. When this exchange is resolved, each chromatid will contain a segment of DNA from the other. Many of the steps that produce chromosome crossovers during meiosis resemble those that guide the repair of DNA double-strand breaks in somatic cells (see Figure 6–30).5′TWO NON-SISTER CHROMATIDS IN A BIVALENTone of the maternal chromatidsone of the paternal chromatids3′5′3′5′3′5′3′5′3′5′3′5′3′5′3′5′3′5′5′5′3′3′3′5′3′5′3′5′3′5′3′STRAND EXCHANGEDNA SYNTHESISCAPTURE OFSECOND