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| 1. Those two traits would not have assorted independently. Dihybrid crosses would not have produced 9:3:3:1 ratios. 2. Because he would not have been able to distinguish maternal and paternal members of a pair of homologous chromosomes to be able to follow how they assorted. 3. X and Z are at opposite ends with Y between them approximately 2/3 of the way toward Z. 4. Both of them would be linked to one or more other genes that resided somewhere in the middle of the chromosome. If cDNAs could be made for both genes and mixed together, sites at both ends of one chromosome should be labeled. 5. The DNA that codes for rRNA would reanneal in the second part of the curve with the moderately repeated fraction. The pure ribosomal DNA would reanneal rapidly, in the region where the satellite DNA fraction of the total DNA normally reanneals. 6. By measuring the number of nucleotide changes present between the two portions, particularly those that occur in a noncoding region whose sequence is not subject to natural selection. Nucleotide sequences diverge at a relatively constant rate and thus can be used as a molecular clock to determine times in the past that particular genomic changes occurred 7. Because sequences change over time, particularly those with a noncoding function. Eventually, older sequences become altered to the point that their origin is no longer recognizable. In the case of transposable elements, those that are derived from sequences that entered the genome of an ancestor more than about 200 million years ago no longer bear sufficient resemblance to the ancestral sequence to be identified as a retrotransposon. 8. The solution of double-stranded DNA would be considerably more concentrated (e.g., gram/ml) than that of single-stranded DNA because single-stranded DNA has a higher absorbance per unit weight than does double-stranded DNA as evidenced by the hyperchromic shift in Figure 10.16. 9. Labeled myoglobin mRNA would produce four spots, one on each half of each replicated chromosome of a homologous pair. The pattern with labeled histone mRNA would depend on whether the genes encoding histones were clustered or not. The number of sites would reflect the distribution among the cell's chromosomes. If they were all clustered, hybridization would produce four sites that were much more heavily labeled than those seen with myoglobin mRNA. 10. 2, 3, 4 are correct. 35 percent. 11. False. Any percentage from 0 to 70 is possible. However, 30 percent of the bases on the complementary strand will consist of Ts. 12. No. The T m is the point at which 50 percent of the base pairs have separated. This could occur in a population of DNA molecules where all of the DNAs remained in a duplex form being held together by the 50 percent of the bases that were still bonded to one another. 13. No. The genes from rodents and primates both have two introns, which indicates it was present in the gene prior to the evolution of primates. The second intron must have been lost during the evolution of the primate species being studied. 14. The DNA of individuals presumably does not change with age so it would appear, if these findings are correct, that persons with a greater number of trinucleotide repeats do not live as long as those with fewer repeats. In other words, larger repeat numbers may confer susceptibility to common diseases. |