hw1key - Genetics Homework#1 Answer Key 1 Normally a...

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Unformatted text preview: Genetics Homework #1 Answer Key 1. Normally, a diploid meiocyte divides during meiosis I to yield two haploid secondary meiocytes. These undergo another division (meiosis II) to produce four haploid gametes. Imagine that you are studying the products of meiosis in a novel insect species. You see that instead of a haploid secondary meiocyte, these insects produce a diploid secondary meiocyte. The gametes are still haploid, however. Diagram the process of meiosis in this novel insect by drawing the primary meiocyte, the secondary meiocyte, and the gametes produced. Your insect has 2 pairs of chromosomes, one long and metacentric with the marker A on one copy, a on the other copy; the other chromosome pair is short and telocentric with the marker B on one copy, b on the other. Contrast this with a similar diagram of normal meiosis sideby side. Show only one possible combination for the A and B markers. Normal meiosis has two divisions, first one reductional, second one equational. In this insect, however, the two divisions are reversed. We start with a 2n cell containing 4 chromosomes, 8 DNA molecules, and it undergoes an equational division to yield 2 cells, both still 2n, each with 4 chromosomes and 4 DNA molecules. These then undergo the reductional division to yield 4 cells total, each n, with 2 chromosomes and 2 DNA molecules each. (10 points) 2. For the three experiments that helped to prove that DNA is the hereditary material (section 10.3), describe the experimental methods, and explain what controls were (or should have been) used. Don't give too many details; I want you to give the big picture, so no need to list what strains were used, for example. Just say if one strain caused death under normal circumstances and the other didn't. Griffith (1927) used different strains of strep (one virulent, normally caused death, the other avirulent, normally had no effect) to infect mice. Virulent strain alone caused death, avirulent strain alone had no effect, heatkilled virulent strain had no effect, but heatkilled virulent strain plus avirulent strain injected together caused death, and live virulent bacteria were found in the blood. This demonstrated that there was some cellular component present in the cells that was able to "transform" the avirulent cells into virulent ones. Avery, MacLeod, and McCarty (1944) demonstrated the nature of the transforming principle by heatkilling virulent cells, and then treating the extract with either protease (which destroys proteins), RNase (which destroys RNA), or DNase (which destroys DNA). When they tested the treated extracts for their abilities to transform avirulent cells into virulent ones, they saw that only the DNase treated extract lost its ability to transform. Therefore, DNA was the transforming principle. Hershey and Chase (1952) demonstrated that DNA was the molecule responsible for directing the formation of new virus particles by using radioactive labels to label either the DNA or the proteins of phage T2. When these phages infect bacteria, they saw that the labeled DNA ended up in new phages, but never labeled proteins. This demonstrated that it was the DNA that entered the bacterial cell and directed the production of new phage. (10 points) 3. Complete the "Exploring Genomics" exercise from the end of chapter 10 (p. 272273). Include the answer to each question. Exercise I, part 6 (a) The top three alignments were matches with variants of the human breast cancer gene 1 (BRCA1). (b) Each of the top three matches showed 100% identity to the query sequence, thus there were no gaps. (c) BRCA1 encodes a nuclear protein that suppresses tumor formation. Mutations in BRCA1 are responsible for a significant percentage of inherited breast cancers and ovarian cancers. Exercise I, part 7 This is the accession number for a BRCA1 variant. Exercise II, part 1 (a) A bull terrier (Canis familiaris) gene involved in polycystic kidney disease, an autosomal dominant condition causing formation of multiple, fluidfilled cysts in the kidney that can compromise kidney function and eventually lead to kidney failure. (b) Homo sapiens polycystic kidney disease 1 gene (PKD1). The identity is 83% with 3% gaps in sequence. (c) 83% identity with 2% gaps (d) human immunodeficiency virus 1 (HIV1) POL (polymerase) gene (e) mtDNA from the woolly mammoth (Mammuthus primigenius). What living animal has a similar sequence? This sequence is similar to a sequence from Elephas maximus (Asian elephant). (f) Human OB (or LEPTIN, LEP) gene. The top alignment shows a match with a genomic DNA sequence on chromosome 7, the locus for the human LEP gene. (g) human DMD (Duchenne's muscular dystrophy) gene Exercise II, part 3 (a) rdgB, retinal degeneration B gene (b) PITPNM2, phosphatidylinositol transfer protein, membraneassociated 2 (c) 69% identical (d) Retinal degeneration. In humans it is involved in a retinal degeneration disease condition called retinitis pigmentosa. Exercise II, part 4 This sequence aligns with the BRCA1 protein. (5 points) ...
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