Practice Problems 4 Answers1. (a) trisomic; (b) tetraploid; (c) monosomic2. (a) inversion (of FED); (b) transposition (of C); (c) deletion or deficiency (of C)3. Remember that the rule is that homologous portions of chromosomes synapsewith each other. Each diagram shows first, the relevant chromosomes, andbelow that, the appearance of the synapsed chromosomes.4. Each chromosome will have replicated but the sister chromatids will not besegregated into separate cells. The cell and its mitotic descendants will be anallotetraploid: A A B B A' A' B' B'.5. There would be a mixture of ordinary molecules that are double-strandedthroughout their entire length and molecules that are partly double-strandedwith a single-stranded loop projecting out. The following diagram illustrates theexperiment; actually, in the EM one would not see the separate strands.6. Inserted in a gene, the element would certainly cause major changes in theamino acid sequence of the protein, and very likely shorten it, as does aframeshift mutation. Inserted upstream from a gene, the element might modifythe sequences that control the action of the gene. In either case, it would looklike a mutation in the gene itself that rendered the gene nonfunctional.7. All the resulting organisms will be triploid.(a) Two sperm, each N, fertilize an N egg; if the three nuclei fuse, the resultingnucleus is 3N.(b) The chromosomes in the germ line cell have replicated but there is noanaphase so the cell is 4N. This cell undergoes mitosis to produce a clone of 4Ncells. If one of these undergoes meiosis, the resulting egg will be 2N, and afterfertilization the zygote will be 3N.(c) The sperm will be 2N (but it will have two identical sets of chromosomes).Fertilization results in a 3N zygote.(d) The egg nucleus and polar body nucleus are both N, and fuse to produce a2N egg, which after fertilization becomes a 3N zygote. Again, the two sets ofchromosomes contributed by the egg are identical.8. (a) A crossover between direct repeats gives a deletion. If the centromere liesbetween the repeats, unlike the case shown here, the result is a centric ringchromosome and an acentric linear chromosome. The acentric is lostimmediately; the acentric ring is lost due to failure to migrate to the polesomewhat more frequently than a normal chromosome, but may be retained fora long time.(b) Unequal sister-strand crossing-over results in expansion and contraction ofseries of tandem repeats:(c) Intrachromosomal crossing-over between inverted repeats produces an inversion:(d) Crossing-over between repeats on different chromosomes can give reciprocaltranslocations:o 1 2 3 o 1 2 3 o X --> o 1 2 3 1 2 39. (a) 5/6 Su : 1/6 su; because only 1/6 of the gametes of the trisomic will be su,while the remainder will carry at least one Su.(b) 1/2 Su : 1/2 su.10. Locus a lies within 1, 3, and 4, i.e. in the only region where they overlap.Locus b lies within the overlap of 1 and 5. Locus c is covered by 1 and 4 but not3. Hence the locations are:Δ1 Δ2 Δ3 Δ4 Δ5 MAP b c a 11. You can write the cross using the knob as a marker, just as you do any kindof genetic marker. Then this becomes a two-factor cross, beginning withPr Pr 9 9 X pr pr 9k 9k and the F1 is Pr 9/pr 9k. If the pr and knob were unlinked,the four genotypes of F1 gametes would be in equal numbers, but they are not:the parental types (Pr 9 and pr 9k) are much more common than therecombinants. Hence the knob and pr loci are linked, and therefore they must beon the same chromosome.12. Each entry in the table represents a diploid genotype, e.g. ade1 ade3. If ade1and ade3 are in different genes, they complement each other; if they are in thesame gene, they will not because the diploid will have two mutant copies of thatgene.(a) The complementation groups are 1 and 2, 3 and 5, and 4.(b) Since there are three complementation groups, there are three genes.(c) Yes; No. If two mutant genes complement each other, the mutations are indifferent genes and they are not
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