1 Biology 321 Spring 2013 Assignment Set #8 9th edition users should consult the 10th edition for this assignment set Required Reading Assignments: Reread article assigned earlier in the quarter: GENOMES UNZIPPED: The complicated relationship between genotype & phenotype http://fire.biol.wwu.edu/trent/trent/genepheneunzipped.pdf Google.doc assignment: Why Skinny Moms Sometimes Produce Heavy Kids http://fire.biol.wwu.edu/trent/trent/skinnymoms.pdf Optional Reading Assignment: Trauma and Stress: from child to adult http://fire.biol.wwu.edu/trent/trent/traumastress.pdf Reading and Problem Assignments Text Linkage & Molecular Markers: Chapter 4 • pgs 121-132 (linkage basics up to but not including three-point testcross) • Figure 4-13 on pg 136 • Section 4.3 pgs 137-141 (Mapping with molecular markers; be sure to look carefully at Figures 4-15 & 4-16) • Section 4.7 pgs. 148-150 (Using recombination-based maps in conjunction with physical maps) • For your personal enrichment: Section 4.8 (The molecular mechanism of crossing-over) Work solved-problem 1 & problems 12, 13, 17, 18, 20, 27, 29, 32, 34 (review), 63 Positional cloning: Chapter 10 • Section 10.5 pg 364-367 Haplotypes & Molecular Markers: Chapter 18 • pg 637-643 Inheritance of Complex Traits: Chapter 19 • pg 683-685; pg 717-723 • think about problem 18 Epigenetics Chapter 12 • pgs 426 – 432 Section 12.3: Dynamic Chromatin • Optional: Section 12.6 (pgs 441-444) Gender-specific silencing……..2 Additional Linkage Problems  Problem -1 Draw a clearly labeled picture to illustrate the following genotype in a 2n=4 cell: A/A; BC/bc • The cell should be in metaphase I of meiosis and should not show any crossing over events. • Label a pair of sister chromatids; a pair of homologs. • Indicate ALL copies of ALL alleles -- do not use any shorthand notation  Problem 0 Genes A and B are linked. On the average, there is one cross-over between these genes (involving two non-sister homologous chromatids) for every 25 meiotic divisions. What is the map distance between these genes? NOTE: The answer is not 4 map units  Problem 1 An organism of genotype AaBbCc is test-crossed. The genotypes of the progeny were as follows: Genotype # of progeny AaBbCc 90 AaBbcc 10 aabbcc 90 aabbCc 10 a. If these three genes were all assorting independently of each other, how many genotypic classes would you expect in the progeny of the testcross? b. If these three genes were so tightly linked that crossing-over never occurred between them, how many genotypic classes would you expect in the progeny of the testcross? c. What can you conclude from this data?  Problem 2.1 The following diagram is a map of the X chromosome of Drosophila: vermilion eyessable-bodyminiature wingscut wingssinged bristleslozenge eyes20.027.733.036.143.044.021.0garnet eyes3 Values are map units measured from the gene closest to the end of the chromosome. A true-breeding wild-type female is crossed with a vermilion-eyed, sable-bodied male. The F1 females are wildtype. If the F1 males and F1 females are crossed to each other, what proportion of their male progeny will be: a. wild-type b. vermilion-eyed and sable-bodied c. vermilion-eyed d. sable-bodied  Problem 2.2 Examine the map of the Drosophila X chromosome Check your work carefully before selecting your answer. Partial credit possible. (i) A female of genotype m+ g / m g+ is crossed with a phenotypically wild-type male. All female progeny are wild-type. The percentage of male progeny that will have miniature wings and garnet eyes is: a. 4% b. 8 % c. 46% d. 92% e. 50% f. 100% g. not enough info to determine h. none of the above (ii) A female heterozygous for both the sable body and garnet eye genes is crossed with a phenotypically wild-type male. All female progeny are wild-type. The percentage of male progeny that will have both sable bodies and garnet eyes is: a. 0.5% b. 1 % c. 49.5% d. 99% e. 50% f. 100% g. not enough info to determine h. none of the above (iii) A female of genotype sc+ lf+/ sc lf is crossed with a phenotypically wild-type male. All female progeny are wild-type. The percentage of progeny males that will be wild type for both traits is: a. 0% b. 16 % c. 25% d. 34% e. 50% f. 68% g. not enough info to determine h. none of the above4  Problem 3 Your lab partner has a strain of Drosophila which has bent wings (w) , a recessive mutation on the X chromosome. He also has a different strain which has stubbly bristles (b), a recessive mutation also on X and about 1 map unit away from the bent wing gene. He does the following experiment to generate a doubly mutant male: Cross: bent wing X stubby bristle Cross: F1 X wild-type  He scores 100 male progeny for bent wings and stubby bristles and finds no doubly mutant flies. He gives up in disgust and goes to see The Fly, hoping for an inspiration. Your lab partner thinks that his failure to bind a bent-winged, stubbly-bristled male means that the doubly mutant phenotype is lethal. But you look at the outline of his experiment and suggest that he redo it with one modification. What do you suggest and why? Be explicit about what you think he needs to do and why.  Problem 4 Drosophila melanogaster has one pair of sex chromosomes (XX or XY) and three pairs of autosomes, referred to as chromosomes 2, 3, and 4. A male fly with short legs was discovered in a wild-type stock of flies by an observant genetics student. Using this male, the student was able to establish a pure breeding stock. Next the student constructed a homozygous strain carrying the short mutation (sh), the mutation black body (b gene located on chromosome 2) and the mutation cardinal eyes (cd gene located on chromosome 3). A female from the short, black, cardinal eye stock was mated to a wild-type male. All the progeny were wild type. The F1 females from this cross were then testcrossed. The F2 progeny were as follows: cardinal eye, wild-type cardinal eye short, black short, black females 63 58 57 62 males 59