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Tamarin: Principles of Genetics, Seventh Edition
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Back Matter
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App. A: Brief Ans. to Selected Exercises, Problems, and Critical Thinking Ques.
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Appendix A Brief Answers to Selected Exercises, Problems, and Critical Thinking Questions
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Critical Thinking Question: 1. Adenovirus attacks normal cells by binding to p53; it also attacks cancerous cells that lack p53. If we remove the gene for the protein that binds p53, the E1B gene, then the modi ed adenovirus will not be able to attack normal cells but will be able to attack cancerous ones lacking p53. This modi cation of adenovirus as a potential tool in treating cancer was published in 1996. 17 Non-Mendelian Inheritance
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yields M1M1m2m2 or m1m1m2m2 in a 1:1 ratio). In other strains, one-fourth of the offspring will lose mu after autogamy, indicating that two unlinked loci were segregating (autogamy in M1m1M2m2 yields M1M1M2M2, M1M1m2m2, m1m1M2M2, or m1m1m2m2 in a 1:1:1:1 ratio). 19. Human mitochondrial DNA does not have introns. Finding an intron would suggest that the mitochondria had acquired a nuclear gene. 21. a. All type 1, 1.5 and 3.7 kilobases. b. All type 2, 2.5 and 6.0 kilobases. Recall that the chloroplast DNA from mt cells does not appear in progeny. 23. Conjugation produces exconjugants with the same genotype, but which haploid micronucleus survives in each cell is random.Therefore, one-fourth of the time, the genotypes of the exconjugants are expected to be KK, one-half of the time Kk, and one-fourth of the time kk. The sensitive exconjugant remains sensitive, regardless of which of the genotypes is present. Autogamy does not affect the genotype of homozygous exconjugants, but it does affect the heterozygote. Among the killer exconjugants we expect:
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1. Persistence of an environmentally induced trait into later generations is known as dauermodi cation and does not imply genetic control. After a suitable number of generations, the phenotype returns to normal, indicating an environmental rather than genetic response. 3. Maroon-like must affect the cytoplasm of the egg. The rst cross is unusual and alerts us to maternal inheritance. A true Mendelian factor should produce one-half maroon-like males and females.The genotypes of the F1 females are ma-1 /ma-1 and ma-1/ma-1; half of the females should be of each genotype. If the wild-type allele produces wild-type cytoplasm for the progeny, regardless of the genotype of the progeny, any female that is ma-1 /ma-1 will produce all wild-type progeny. 5. Although the genetic scheme predicts shell coiling perfectly, one could do experiments involving the injection of cytoplasm into eggs to test the viral hypothesis. 7. Female parent: Dd. Male parent: d . (A): dd. Since sel ng produces only sinistral snails, individual (A) must be homozygous for sinistral coiling, dd. The individual must get one d allele from each parent, so each parent must be at least heterozygous. Since (A) has dextral coiling, the mother must be heterozygous. The father could be either D/d or d/d. 9. By looking at different species, it is clear that few genes for oxidative phosphorylation are found in all mitochondrial genomes. 11. The rule of thumb is that suppressive petite mitochondria will dominate a cell, whereas neutral petite mitochondria will be lost in a competitive situation. Therefore segregational petite segregational petite segregational petite neutral petite neutral petite segregational petite segregational petites neutral petite segregational petites suppressive petite suppressive petites
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Killer Exconjugants 1/4 KK 1/2 Kk
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Autogamous Products all KK 1/2 KK 1/2 kk all kk
Phenotypic Ratio 1/4 killer 1/4 killer 1/4 sensitive (Kappa are lost) 1/4 sensitive (Kappa are lost)
1/4 kk
25. Use the striped plant as the egg parent, and get pollen from a plant with the following genotype: IjIj jj. If the striped plant is iojap (ijij JJ), all progeny will be heterozygous for both genes and will also contain iojap cytoplasm. The F1 plants will segregate green, striped, and white sections within the plant. If the original plant is IjIj jj (and thus japonica), all F1 plants will have striped leaves. 27. a. 2 normal:2 petite b. 0 normal:4 petite c. 4 normal: 0 petite. A nuclear gene should segregate 2:2 for each allele. Cytoplasmic factors will produce four spores with identical cytoplasm. Critical Thinking Question: 1. We believe there are two mechanisms to ensure the distribution of cellular organelles during cytokinesis: stochastic and ordered inheritance. Stochastic inheritance simply means that no real mechanism exists; rather, the cell depends on the large number of the organelles to ensure an even distribution during the dividing of the cell. Ordered inheritance requires the even distribution of organelles in small numbers. This can be accomplished by special structures that divide a large organelle (e.g., a single, large chloroplast), or by other mechanisms that insert part of the mitochondrial system into new buds in budding yeast. 18 Quantitative Inheritance
neutral petite neutral petites suppressive petite suppressive petites suppressive petite suppressive petites
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