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23. An XO/XYY mosaic can occur by nondisjunction of the Y chromosome in a cell during early cleavage in an XY individual. An XX/XXY mosaic can come about if one of the cells during early cleavage in an XX zygote is fertilized again by a Y-bearing sperm. Trisomy 21 usually comes about from an egg with two copies of the chromosome; the egg had two copies because of meiotic nondisjunction. 25. The father. The allele for color blindness can only come from the mother. If meiosis in her is normal, an egg could get the X chromosome carrying the mutant allele. The daughter has only one X chromosome, so the sex chromosomes failed to separate in the man, and a sperm with neither X nor Y fertilized the egg. 27. The rst meiotic division in the father is normal, producing cells with either two X or two Y chromatids. During the second meiotic division in the cell with the two Y chromatids, both Y chromatids move to the same pole and end up in the same sperm cell.
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In fact, most single-stranded molecules have some regions that are complementary. 11. 19.1 billion base pairs. One base takes up 3.4 . Since 1 is 1/10,000,000 millimeter, there are 10,000,000/3.4 2.94 million bases per millimeter. Multiply by 1,000 to get to meters, and nally by 6.5 for 6.5 meters: 1.91 1010 bases.
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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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A-11
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Critical Thinking Question: 1. One way to study mutations that are generally lethal is by isolating temperature-sensitive mutations. These mutations involve amino acids that disrupt the functioning of the enzymes at some critical temperature but are phenotypically normal at other temperatures. Thus, the mutant organisms can be kept alive by growing them at one temperature (the permissive temperature) but their mutant effect can be studied at the temperature in which the protein function is disabled (the restrictive temperature). (For additional discussion of these mutations, see chapter 12.)
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15. A primosome is a helicase plus a primase; it opens the DNA and creates RNA primers on lagging strands and is part of the replisome. A replisome includes a primosome plus two copies of DNA polymerase III; it coordinates replication on both the leading and lagging strands at the Y-junction. 17. See gure 9.28. 19. At one time molecular swivels, presumably protein in nature, located periodically along the DNA, were suggested. 21. RNA RNA DNA
10
Gene Expression: Transcription
1. See gure 10.3. Complementarity is achieved between messenger RNA and ribosomal RNA and between messenger RNA and transfer RNA. 3. Transcription has higher error rates. Errors of DNA polymerase tend to become permanent, whereas errors of RNA polymerase do not. 5. A consensus sequence is made up of the nucleotides that appear in a signi cant proportion of cases when similar sequences are aligned. A conserved sequence consists of nucleotides found in all cases when similar sequences are aligned. For example, the Pribnow box ( g. 10.6) is the consensus sequence TATAAT. 7. See gure 10.8 for a promoter and gure 10.10 for a terminator. The transcript starting from the promoter would be 5 -CUUAUACGGU....The transcript from the terminator is shown in gure 10.10.
host Insert into host:
host
9. A stem-loop structure can form when a single strand of DNA or RNA has a double helical section (see g. 10.10). An inverted repeat is a sequence read outward on both strands of a double helix from a central point (see g. 10.10). A tandem repeat is a segment of nucleic acid repeated consecutively; that is, the same sequence repeats in the same direction on the same strand: 5 -TCCGGTCCGGTCCGG-3 3 -AGGCCAGGCCAGGCC-5
RNA This question really asks how double-stranded DNA can be formed from single-stranded RNA. First, we could synthesize a complementary DNA strand, then begin to make a second DNA strand complementary to the rst. While synthesizing the second DNA strand, we begin to degrade the RNA. The double-stranded DNA molecule now inserts into the host DNA. In order to get more viruses, a single-stranded RNA molecule must be made from the DNA. We would need, at a minimum, an enzyme to make a singlestranded DNA to form a hybrid with the RNA; an enzyme to degrade the RNA and to make the second, complementary DNA strand; an enzyme to cut the host DNA to allow the viral DNA to insert itself; an enzyme to ligate the two molecules; and an enzyme to make more viral RNA. The earlier functions are performed by one viral enzyme, reverse transcriptase. 23. Finding small pieces or fragments of DNA suggests the Okazaki pieces are only slowly, if at all, joined, a function of DNA ligase.The fact that not many long DNA molecules are seen also suggests that the DNA is being broken, implicating a nuclease as well. 25. It is unlikely that bases are added faster in developing embryos. So we must look for another mechanism. If there are more replicons, and hence more origins of DNA replication, each replicon will be shorter and be able to duplicate faster. Alternatively, and more likely, the process is regulated to slow down adult division.
A DNA sequence with a seven-base inverted repeat is 5 -ATTACCGCGGTAAT-3 3 -TAATGGCGCCATTA-5 11. Footprinting is a technique in which DNA in contact with a protein is exposed to nucleases; only DNA protected by the protein is undigested. Promoters could be isolated by protection with RNA polymerase in the absence of ribonucleotides the polymerase will not move and then sequenced. 13. The superscripts of the sigma factors refer to their molecular weights (e.g., 70 is 70,000 daltons). Different sigma factors usually recognize different prokaryotic promoters. 15. 3 - G G T A G T A C T G T C T G G G A A C G A T T G C G -5 5 - C C AT C AT G A C A G A C C CTT G CT A A C G C - 3 Begin by writing the strand that is complementary to the RNA.This will be the transcribed strand. Remember, U in RNA pairs with A in DNA. Since transcription proceeds 5 3 , the 5 end of the RNA is opposite the 3 end of the DNA. 17. The bottom strand is transcribed and the molecule is arranged as 5 _________________3 3 _________________ 5 Begin by writing the RNA that could be transcribed from each strand. Since the DNA represents the beginning portion of the gene, the RNA must have an AUG to start protein synthesis. Unfortunately,
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