hw4key - Genetics Homework 4 due in lecture Tuesday, Sept....

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Unformatted text preview: Genetics Homework 4 due in lecture Tuesday, Sept. 2 1. Six evolutionarily related but separate species each arose through a single inversion in a chromosome of its predecessor. The progenitor species had the chromosomal composition of ABCDEFGH. The following represent the chromosomes from the related species (not in order of how they arose): (1) AEBGHDFC (2) AEDHGFBC (3) AEBGFDHC (4) AEDCBFGH (5) AEDHGBFC Show how the 5 species arose from the original progenitor by indicating where the inversions occurred. ABCDEFGH AEDCBFGH AEDHGFBC AEDHGBFC AEBGHDFC AEBGFDHC 2. Wildtype phage T4 grows on both E. coli B and E. coli K12(), producing turbid plaques. The rII mutants of T4 grow on E. coli B, producing clear plaques, but do not grow on E. coli K12(). This host range property permits the detection of a very low number of r+ phages among a large number of rII phages. With such a sensitive system, it is possible to determine the genetic distance between two mutations within the same gene in this case, the rII locus. Suppose E. coli B is mixedly infected with rIIx and rIIy, two separate mutations in the rII locus. Suitable dilutions of progeny phages are plated on E. coli B and E. coli K12(). A 0.1ml sample of a thousandfold dilution plated on E. coli B produces 672 plaques. A 0.2ml sample of undiluted phage plated on E. coli K12() produces 470 turbid plaques. What is the genetic distance between the two rII mutants? The genetic distance is 0.07 mu. The plaques produced on E. coli K12() are r+, while those on E. coli B may be either r+ or r. Thus, the total number of progeny can be inferred from the number of plaques formed on B: Total number of progeny in 1 ml = (dilution factor) x (# progeny phage/ml) = 1000 x (672/0.1) = 6.72 * 106/ml. Since E. coli B is coinfected with rIIx and rIIy, the only way to obtain an r+ progeny phage is to have a crossover within the rII locus. The progeny resulting from a crossover would be r+ and rIIxy recombinants. The number of recombinant phage is twice the number of r+ phage, which can be assayed for by growth on E. coli K12(): Number of recombinant progeny in 1 ml = 2 x (number of r+ phage/ml) = 2 x (470/0.2) = 4700/ml. The map distance between rIIx and rIIy is [4700/(6.72 * 106)] x 100% = 0.07 mu. 3. Exploring Genomics, chapter 8, page 221222. Exercise I, part 4 (a) Familial melanoma is an inherited form of skin cancer of melaninproducing cells called melanocytes. It is often characterized by irregularly shaped moles. (b) Although there are different forms of melanoma, about 10% of cases show familial patterns of heredity. Linkage studies indicate that mutation of the CDKN2A gene on chromosome 9, p21 is involved in familial melanoma, suggesting that this gene is a melanoma susceptibility gene. CDKN2A (CDK stands for cyclindependent kinase) encodes a protein involved in controlling the cell cycle at the G1/S and G2/M checkpoints. Exercise II, part 1 (a) Pentasomy 21. The patient karyotype shows 5 copies of chromosome 21. (b) Gbanding karyotyping (Gbanding is described in Chapter 12) and fluorescence in situ hybridization (FISH; described in Chapter 10); bone marrow and blood (c) CML (chronic myelogenous leukemia) is a form of leukemia involving cells in the bone marrow that produce white blood cells. CML is one of the most common forms of leukemia in adults. (d) This case is considered atypical because CML is typically caused by a translocation between the long arm of chromosomes 9 and 22. This translocation fuses a gene (CABL) from chromosome 9 to a gene (BCR) on chromosome 22. The fused gene produces a fused protein called BCR/ABL that results in uncontrolled division of white blood cells, producing leukemia. In situ hybridization studies showed that the patient in this case study does not have the BCR/ABL fused gene. (e) Down syndrome (trisomy 21) ...
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This note was uploaded on 09/12/2008 for the course BICD 100 taught by Professor Nehring during the Summer '08 term at UCSD.

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