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3/12/04 Biol 207 Workshop 8 Answer Key 1. A true-breeding red snapdragon was crossed to a true-breeding white snapdragon. The F1 progeny were red. When F1 x F1 crosses were made, the following F2 progeny were observed: 1850 red a) - Give the F1 and F2 genotypes 670 pink b) - What phenotypic ratio would be expected from a testcross of an F1 831 white snapdragon? a) P: RR; WW and rr; ww F1: Rr; Ww F2: R-; W-, and rr; W- and -/-; ww b) 1red: 1 pink: 2white 2 . In sweet peas the allele C is needed for color expression (cc results in no color: white). The flower colour that is expressed is determined by the alleles R (red) and r (blue). A cross between certain red and blue plants resulted in progeny as follows: 3/8 red a) What were the genotypes of the plants crossed? 3/8 blue Rr;Cc x rr;Cc 1/4 white 3. In tetraploid wheat, a cross between true-breeding red-kernel and true-breeding white-kernel strains yielded F1 offspring with red kernels. When the F1 were intercrossed, the F2 plants had a ratio of 15 red-kernel : 1 white-kernel. A testcross of the red-kernel plants yielded 3 red-kernel : 1 white-kernel. a. What are the parental genotypes? b. What are the F2 genotypes and phenotypes? c. What conclusions can be made about the allelic and gene interactions? a. Their F1 parents are R/R; C/C (red) x r/r;c/c (white). b. R/_:C_ (red), r/r;C/_ (red), R_; c/c (red), r/r; c/c (white) c. The relationship between alleles of R and C could be either dominant suppression or, more likely, a duplication (e.g. tetraploidy). 4. In a hypothetical species of rat, C (colored) vs. c (albino) and B (black) vs. b (brown) are two autosomal pairs of alleles and show dominance as indicated (capitalized alleles are dominant). Rats from the true-breeding brown strain are crossed with albinos of genotype ccBB; when the F1’s were crossed with albinos of genotype ccbb, three phenotypes were produced: black 102; brown 198; albino 300 a. Explain why one can conclude that the two genes are linked. b. Calculate the percentage recombination between the two genes. c. If each of the 102 black offspring is used as a parent in a testcross, what phenotypes would you expect to appear in the progeny? Explain your answer. d. In what proportion would you expect the different phenotypes? a. Based on information in the question, there is a 2:1:1 albino: brown: black ratio expected from the testcross. The results does not match the expected ratios, therefore the alleles of B and C are not segregating independently and are linked. b. The recombinant genotypes in the progeny are CB/cb (black) and cb/cb (albino). The albino phenotype is produced by the parental combination cB/cb. Usually, the Rf is determined by dividing the total recombinant phenotypes by the total progeny. In this case, parent and recombinant genotypes are giving the same phenotype. Since the number of black progeny should equal the number of recombinant cb, then doubling the number of black progeny should equal the total recombinant progeny expected. Therefore the answer is 2x102/600x100=34 m.u.. c+d. The black (CB/cb) parents are heterozygous because they are crossed to a tester. The expected ratio is is 33% black, 50% white, and 17% brown. 5. While scoring some D melanogaster and you discover a fly with both female and male sexual characteristics. Cells from this fly are cultured and arrested during metaphase. Individual cells had either 1, 2, or 3 X chromosomes. a. Describe the event that could lead to this somatic aneuploidy? b. What is the relationship between sex chosen and X chromosome number? Biol 207 Workshop#8 Winter 2004 Page 1 3/12/04 a. Nondisjunction between sister chromatids during mitosis will lead to one daughter inheriting 3 X chromosomes and the other daughter inheriting 1 X chromosome. Since there are cells with 2 X chromosomes, this mitotic error must have occurred after other mitotic rounds of replication. The X/X cells are wild type cells that did not originate from the mitotic error. Since there is no Y chromosome, the fly must have inherited two X chromosomes and would have been female. b. In Drosophila, it is the ratio between X chromosome and autosomes that determines sex. If the ratio is 1:2 (X:A) then a male develops. In the above female the cells with a single X chromosome are developing as males. When the ratio is 2:2 (X:A) then a female develops. 5. While working in a lab, you are asked to do a restriction digest of a (circular) plasmid and determine where BamH I, EcoR I, and Hind III cut in relation to each other. In the following gel, there are single, double, and triple digests using these restriction enzymes. The lane labeled ladder has linear DNA fragments of a pre-determined size. a) Draw a restriction map indicating distance and order of the Hind III, BamH I, and EcoR I restriction sites. b) The uncut circular plasmid is 15 kb, but it runs a different distance than the 15 kb (linear) fragment in the ladder. In addition, sometimes there are three bands seen in the undigested sample. Explain these observations. c) Describe two qualities that a molecular biologist would desire in a plasmid to make a useful tool for cloning a segment of DNA. Well Distance in kb + a) d) Circular DNA and linear DNA run at different rates in an agarose gel. After purification, the plasmid can have three different conformations: supercoiled, relaxed circular, and linear. During Biol 207 Workshop#8 Winter 2004 Page 2 3/12/04 plasmid purification endonucleases sometimes cut the supercoiled circular plasmid producing the relaxed circular and linear forms of the DNA molecule. e) 1. A molecular biologist needs to be able to select for transformed bacteria. A plasmid with a selectable marker gene such as one that makes the bacteria resistant to an antibiotic is usually used. 2. Unique restriction sites into which DNA fragments can be cloned are desired. There are special fragments of DNA that have multiple sites in which many restriction enzymes cut called multiple cloning site (MCS). These especially construct fragments are cloned into plasmids used to clone DNA. 3. The plasmid must have an origin of replication (ori) or it will not replicate in the bacteria. Biol 207 Workshop#8 Winter 2004 Page 3