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Berkeley MCELLBI 140 - MCB 140 Second Practice Problem Set

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SECOND PRACTICE PROBLEM SET MCB140 SPR 07 page 1 A: Which approach to overcoming the problems of pleiotropy in mutant screens would be better able to recover mutations in genes that are NOT cell autonomous in their action: sensitization or mitotic recombination? Explain briefly. B: Shown below are eight different genotypes depicting a homologous pair of autosomes, with the following symbol conventions: FRT sites (#); centromeres (o); and a loss-of-function Minute mutant allele (M). Minutes are cell-autonomous haploinsufficient genes that are essential for cell survival and growth. Since cells that are heterozygous for Minute loss-of-function alleles (M/+) grow more slowly than their +/+ neighbors, mitotic recombination in a M/+ heterozygote can produce a M+/M+ clone that will have a growth advantage over its M/+ neighbors. M mutations are often included in clonal genetic screens, since if the newly induced recessive alleles to be screened are induced on the M+ chromosome, the M+/M+ clones induced will not only be homozygous for any new mutant alleles, they will also have a potential growth advantage over their M/+ neighbors. This growth advantage may make the clones larger and hence easier to score than would be the case in a background with no M mutation present. Moreover, the growth handicap of the cells outside the clone can reduce the competative pressures among cells and thereby allow homozygous mutant cells to survive that would otherwise be unable to compete with +/+ neighbors due to adverse effects of the new mutation on growth. Below if no M is shown, assume that the allele present is wildtype. xxx(#)xxx(M)xxxxx(#)xxx(o)xxxxxx A xxxxxxxxxxxxxxxxxxxxxxx(o)xxxxxx xxx(#)xxxxxxxxxxx(#)xxx(o)xxxxxx B xxxxxxxxx(M)xxxxxxxxxxx(o)xxxxxx xxx(M)xxxxxxxxxxx(#)xxx(o)xxxxxx C xxxxxxxxxxxxxxxxx(#)xxx(o)xxxxxx xxx(#)xxxxxxxxxxx(M)xxx(o)xxxxxx D xxx(#)xxxxxxxxxxxxxxxxx(o)xxxxxx xxxxxxxxxxxxxxxxx(M)xxx(o)xxxxxx E xxxxxxxxxxxxxxxxx(#)xxx(o)xxxxxx xxx(#)xxxx(M)xxxxxxxxxx(o)xxxxxx F xxxxxxxxxxxxxxxxx(#)xxx(o)xxxxxx xxxxxxxxx(M)xxxxx(#)xxx(o)xxxxxx G xxx(#)xxxxxxxxxxxxxxxxx(o)xxxxxx xxxxxxxxx(M)xxxxxxxxxxx(o)x(#)xx H xxxxxxxxxxxxxxxxxxxxxxx(o)x(#)xx B-A: Which, if any, of these arrangements of FRT sites and M alleles could be used for an efficient mitotic recombination screen to identify carriers ofSECOND PRACTICE PROBLEM SET MCB140 SPR 07 page 2 recessive mutations on the left arm of the chromosome shown that affect the orientation of cell divisions? (no explanation necessary) B-B: Can you deduce which parent was mutagenized to produce the genotype you chose in part A -- the one carrying the Minute mutation or the one that did not? Explain briefly. B-C: What additional genetic element or combination of elements is necessary to make the genotype you choose work for a mutant screen? C: Drosophila geneticists try to minimize the work needed to maintain the many lines of flies they investigate. They like to establish genetic stocks in which the flies in a culture bottle randomly mating with each other produce progeny that are just like their parents, generation after generation, with no further intervention by the geneticist. Because females homozygous for a strict maternal-effect lethal (mel) mutation are sterile (their offspring die during embryogenesis, regardless of their genotype with respect to mel), such mutations can be propagated only when females are heterozygous. A particular mel mutation on the second chromosome (an autosome) is kept as a heterozygote with a balancer chromosome carrying the Cy mutation, which is recessive (embryonic) lethal but also produces a dominant Curly-wing phenotype. C-A: When this balanced stock reaches equilibrium, do you expect there to be more CyO/mel males than mel/mel males? Explain briefly. C-B: When this balanced stock reaches equilibrium, do you expect there to be more mel/mel males than mel/mel females? Explain briefly C-C: What kinds of flies would you select from the culture bottle to study the phenotypic effect of the maternal-effect mutation in detail and how would you recognize them? D: The snarg is a hypothetical model diploid genetic organism that normally has seven legs in both sexes. Snargs have heteromorphic sex chromosomes, with the male being the homogametic sex. Male snargs homozygous for the amorphic (null) allele leggy12 (lgy12) have twelve legs, while males that are heterozygous (lgy12/+) have nine legs. Snarg females only carry a single copy of the lgy gene. Somewhat surprisingly, hemizygous lgy12 mutant female snargs have fifteen legs. The lgy12 mutant phenotypes are not affected by temperature. A dominant, partially temperature-conditional allele, lgyD, was discovered that, whether heterozygous, homozygous, or hemizygous, produces animals with three legs when those individuals are grown at 30oC, but five legs if they are grown instead at 20oC, regardless of sex. lgyD/lgy12 males have the same phenotype as lgyD/+ males. D-A: Does the snarg appear to have a ZZ/ZW or instead an XX/XY system of sex determination? Explain briefly. D-B: Is there any indication that Muller's ratchet has been operating in the snarg over evolutionary time? Explain briefly. D-C: The fact that lgy12/+ males are not wildtype shows that the lgy gene belongs to a rather exclusive (relatively few members) category of genes. What is this category? D-D: What, if any, evidence argues for or against snargs having a system of sex-chromosome dosage compensation? Explain, briefly.SECOND PRACTICE PROBLEM SET MCB140 SPR 07 page 3 D-E: What category of gain-of-function allele does lgyD appear to be? Explain briefly. D-F: Although lgyD does not show the kind of ideal temperature-sensitive behavior we might like to see, which growth temperature can we say is the more "restrictive" (or "non-permissive")? D-G: How many legs would a lgyD/+ male snarg have when grown at 20 C if this dominant allele were displaying the most clear-cut kind of temperature-sensitivity that we like to work with? D-H: If we did a series of single temperature shifts (that is, start growth at either the restrictive or permissive temperature, then shift to the opposite temperature once during the development of the snarg), what specific change in leg phenotype would signal the beginning of the temperature-sensitive period for lgyD/+ males? D-I: What phenotype would you expect for the


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Berkeley MCELLBI 140 - MCB 140 Second Practice Problem Set

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