Unformatted text preview: KEY TO BICD100 Midterm 2 - 2/14/05
1.) This pedigree shows the pattern of inheritance for a rare trait that is not fully penetrant. a. Which is a more likely mode of inheritance for this trait, autosomal dominant or X-linked dominant, and why? In your explanation, refer as needed to the numbered individuals shown in the pedigree (8 pts). 1 2 3 8 4 9 12 5 10 13 6 11 Autosomal dominant. #10 must have the trait in order to pass it on to his daughter (since the trait is rare, we assume that #13 did not inherit the trait from her 7 mother). Incomplete penetrance explains why #10 does not express the trait. However, an X-linked trait cannot be passed from a father (#5) to his son (#10 - we assume that #10 did not inherit the trait from his mother since the trait is rare). Therefore the trait cannot be X-linked dominant. b. Assuming the mode of inheritance you indicated in part a, what is the maximum percent penetrance for this trait based on the data in this pedigree? (8 pts) 6/7 (this would be the correct answer for either autosomal or sex linked dominant) 2.) As discussed in class, coat color in Labradors is determined by alleles of two genes, B and E. The recessive allele b confers brown fur, the recessive allele e confers golden fur, and e is epistatic to b. Dogs with at least one wild type allele for both genes are black. A dog breeder finds that a mating between a golden lab (“Goldie”) and a brown lab (“Hershey”) results in a litter in which half the pups are brown and the other half are black. A second mating between a different golden lab (“Wheatie”) and a different brown lab (“Cocoa”) results in a litter in which half the pups are black and the other half are golden. a. If Goldie is mated with Cocoa, what coat colors do you expect to find among the progeny and in what proportions (8 pts)? All genotypes a golden lab could have: BB ee, Bb ee, bb ee All genotypes a brown lab could have: bb EE, bb Ee If you work out what progeny would result from all possible matings between golden and brown genotypes above, you find that only Bb ee x bb EE can produce a litter with 1/2 brown and 1/2 black pups. So Goldie is Bb ee and Hershey is bb EE. You also find that only BB ee x bb Ee can produce a litter with 1/2 black and 1/2 golden pups. Therefore, Wheatie is BB ee and Cocoa is bb Ee. Goldie (Bb ee) x Cocoa (bb Ee) 1/4 Bb Ee (black) 1/4 bb Ee (brown), 1/4 Bb ee (golden), 1/4 bb ee (golden). Overall: 1/2 golden, 1/4 black and 1/4 brown. b. If Wheatie mates with Hershey, what coat colors do you expect to find among the progeny and in what proportions (8 pts)? Wheatie (BB ee) x Hershey (bb EE) all progeny Bb Ee, so all progeny are black. 1 3.) Since the discovery of Morgan's X-linked white eye color mutation in Drosophila that was discussed in class, many other Drosophila eye color mutations have been isolated. The table below summarizes results of crosses between flies from lines that are true-breeding for each of the following recessive eye color mutations: white (Morgan’s original X-linked mutation), buff, rosy, carnation, and brown. For each cross that was done, the eye colors of male and female progeny are given. A “-“ in the table means that this cross wasn’t done. female parent > male parent V buff white rosy carnation brown buff buff males and females white white males, buff females white males and females rosy wild-type males and females wild-type males and females rosy males and females carnation carnation males, wild type females carnation males, wild type females carnation males, wild type females carnation males and females brown wild-type males and females wild-type males and females wild-type males and females wild type males and females brown males and females a. Why do male and female progeny of all crosses with carnation females have different eye colors? (2 pts) The carnation mutation is also X-linked, but affects a different gene from white, buff, rosy and brown. So you see complementation in the female progeny, but not the male progeny. b. Organize these mutations into complementation groups (groups of mutations affecting the same gene), clearly indicating how many groups there are altogether (8 pts). 4 groups: buff/white, rosy, carnation, brown c. If F1 progeny of the cross between a white female and a buff male are crossed together (i.e. white F1 male x buff F1 female), what eye colors do you expect male and female progeny to have? (8 pts). Let Xw represent the X chromosome carrying the white mutation and Xb the chromosome carrying the buff mutation (the table above tells you that these two mutations affect the same gene). Original cross between white female and buff male: XwXw x XbY 1/2 XwXb (buff females) and 1/2 XwY (white males) Cross between F1 progeny of original cross: XwXb x XwY • • Females half XwXw (white) and half XbXw (buff – you know this because they have the same genotype as female progeny of original cross) Males half XwY (white) and half XbY (buff) 2 From the brown female X rosy male cross in the table, F1 progeny were crossed together to produce an F2 population consisting of 35 flies with wild type colored eyes, 10 flies with brown eyes, 13 flies with rosy eyes, and 4 flies with mauve eyes. d. Do these results suggest that brown and rosy mutations affect genes acting in the same pathway or different pathways? (2 pts) Different pathways. Mauve must be the double mutant because it’s present at a frequency of 1/16 and is a new phenotype not seen before, which is intermediate between brown and rosy. As discussed in class, a double mutant whose phenotype looks like a combination of the two single mutants suggests that the two genes act in separate pathways. If they act in the same pathway, you expect to observe epistasis, which is not seen here. e. If an F2 female with mauve eyes is crossed to a male from a true-breeding brown eye strain, what color eyes do you expect the progeny to have? (If more than one, state the proportions of each type – 8 pts). Mauve F2 female: brown/brown rosy/rosy x brown/brown Rosy/Rosy All progeny brown/brown Rosy/rosy brown eyes Note: on this question many students were thrown off because they made the assumption that brown and rosy mutations are X-linked. However, the data in the complementation table shows that this is not the case. 4.) A viticulturalist is studying genes controlling grape characteristics. She is interested in three traits of importance to winemaking: green is recessive to purple, round is recessive to elongated, and dry is recessive to sweet. A plant from a true-breeding green, round, sweet fruit line was crossed to one from a true-breeding purple, elongated, dry fruit line. The F1 progeny were testcrossed to a true-breeding green, round, dry fruit line to obtain the following 1000 F2 progeny fruits: 19 purple, elongated, sweet 304 purple, elongated, dry 6 purple, round, sweet 169 purple, round, dry 172 green, elongated, sweet 21 green, round, dry 8 green, elongated, dry 301 green, round, sweet Construct a map showing the order of these three genes and the distances between them in cM (20 pts). elongated/round--------------------35.5 cM-----------------------purple/green----5.4 cM----sweet/dry 3 5.) An outbreak of food poisoning has been tracked down to a pathogenic strain of E. coli. Unlike the strains of E.coli that normally reside in the human gut, the pathogenic strain produces a toxic protein that makes people sick. As an initial step toward the isolation of the toxin gene, a microbial geneticist wants to map this gene on the E. coli chromosome. F plasmids are introduced into the pathogenic strain via conjugation with F+ bacteria, and four different Hfr strains are isolated (Hfr1 – Hfr4). Each of the pathogenic Hfr strains, which are strepS and wild-type for each of the other genes shown on the map of the E. coli chromosome shown below, is mated with an F- strepR arg- pro- phe- cys- met- strain. Time of entry mapping experiments yield the following results when disrupted mating pairs are plated out on streptoymycin: Hfr1 time gene first appears arg+ 5 min. pro+ 10 min. phe+ 35 min. arg+ 33 min. Hfr2 time gene first appears met+ 10 min. cys+ 25 min. Hfr3 time gene first appears phe+ 3 min. pro+ 28 min. Hfr4 a. On the E.coli chromsome map to the right, indicate the location of each F insertion (label them 1 to 4), and use an arrow to indicate the direction of transfer initiated by each F insertion (10 pts). Hfr4 time gene first appears cys+ 7 min. met+ 22 min. Hfr3 48 phe 45 cys 58 65
25 toxin gene 80 met
Hfr2 90 ( ( Additional time-of-entry mapping studies are used to determine the location of the toxin gene. Each pathogenic Hfr strain is mated with a non-pathogenic, F- strepR strain. Mating pairs are disrupted at a series of timepoints and plated on strep; strepR colonies are tested for production of the toxin. The toxin gene is never transferred to F- cells mated with Hfr2 and Hfr4. Results for matings with Hfr1 and Hfr3 are shown here: 10 20 15 0 strR
Hfr1 pro arg b. Indicate on the map above (in a) the location of the toxin gene in minutes (5 pts). See above c. Why isn’t the toxin gene transferred to F- cells mated with Hfr2 and Hfr4? (5 pts) Because the F insertions in Hfr2 and Hfr4 are too far from the toxin gene (>50 minutes) 4 ...
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This note was uploaded on 01/20/2009 for the course BIBC BIBC 100 taught by Professor Buehler during the Spring '09 term at UCSD.
- Spring '09