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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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39. Hypervariable DNA is DNA showing a great deal of interindividual variation. A RFLP (restriction fragment length polymorphism) is a polymorphism (variation) that shows up after Southern blotting and probing of restriction digests. A VNTR (variable number of tandem repeats) locus is one that is hypervariable due to unequal crossing over among the tandem repeats. VNTR loci are a hypervariable subset of RFLPs. Sequence-tagged sites are unique sites in the genome that can be ampli ed with polymerase chain reaction. Microsatellite DNA, repeats of very short segments such as CA, forms VNTR loci that are usually examined by polymerase chain reaction (PCR) if primer sequences are known. Critical Thinking Question:
If this arrangement is correct, digestion with BglII, EcoRI, and PstI should yield 0.3 kb (BglII EcoRI), 0.7 kb (EcoRI PstI) 2.0 kb (Pst I BglII), and this is, in fact, observed. 29. The second possibility predicts that a 1.5-kb fragment should be seen in the double digestion. This fragment would result from digestion of the 3.5-kb EcoRI fragment with BglII. Since we don t see a 1.5-kb fragment in the double digest, the second possibility does not agree with the observed results. 31. Two of the three number 21 chromosomes present in the child came from the father, not the mother. Since the probe produces different bands in the mother and the father, all of these bands must also be present in the child. The intensity of a band is proportional to the amount of DNA present. The bands that are of paternal origin are more intense than the maternal bands. The father contributed two number 21 chromosomes. 33. For the steps in the dideoxy sequencing method, see gures 13.32 and 13.33. Use of uorescent dyes has allowed for the automation of the process and the elimination of radioactive tags. 35. The DNA can be inserted into the M13 general sequencing vector. 37. What will appear in the gel are fragments of the newly synthesized strand. Since DNA synthesis proceeds 5 3 , the 5 base will be T in the new strand. Proceed up the gel by indicating the base complementary to the sequence given.
1. Sticky ends would exist if the plasmid DNA had a 3 overhang of one nucleotide residue, while the foreign DNA had a 3 overhang of its complement. The exact number of bases would not need to be the same because repair enzymes plus ligase could close the gap. For example, the plasmid could have a 3 tail of thymines added, whereas the foreign DNA could have a 3 tail of adenines added. This method (see gure next page) was called the polydA/poly-dT technique. 14 Gene Expression: Control in Prokaryotes and Phages 1. a. inducible (wild-type); repressed; f. inducible b-d. constitutive; e. neither, super-
3. One mutant could fail to bind to operator DNA but could still bind the inducer (op , in ). This mutant would have constitutive transcription of the operon. The reverse situation could also be true; the repressor could bind the operator but not the inducer (op , in ). This mutant would be off all the time. A third mutant could fail to bind both (op , in ), being constitutive. 5. a. Operator and repressor. b. Make a partial diploid with wild-type; the operator mutant will make -galactosidase constitutively, and the repressor mutant will make it only in the presence of lactose. Four mutations are possible in the lac operon: mutations in the z gene or the promoter never make the enzyme. Mutations in the repressor always make the enzyme because the repressor cannot bind DNA. In operator mutations, a good repressor can never bind DNA. In a partial diploid, i o /i o , the wild-type repressor is trans acting and can bind to both operators, creating an inducible situation. In i o /i o , repressor cannot bind to o , and this DNA is always on, even though the wild-type DNA is off in the absence of lactose. 7. Cyclic AMP, combined with CAP protein, attaches to CAP sites enhancing transcription of nonglucose, sugar-metabolizing operons in E. coli. Glucose inhibits its formation by inhibiting adenylcyclase. 9. We must think about how these operons are controlled. Not only do they need inducer, but they also require the catabolite repressionactivation system.These mutants could be unable to make cAMP because the adenylcylase gene is defective. Alternatively, they could be making a defective catabolite activating protein (CAP). 11. b tryptophan synthetase gene, a operator, c repressor. Look rst for the single mutation that never gives enzyme; this genotype will give the letter of the structural gene.The genotype a b c ts this requirement, so b is the structural gene, and a and c represent control regions. Genotypes 4 and 5 tell us nothing. Look at genotype 6. The right DNA will never make the enzyme. If c is the operator, the left DNA should always make the enzyme, and this is not seen. Therefore, a must be the operator. Check these assignments with genotype 7. The right DNA will never make the enzyme. If a is the operator, the left DNA is always on. If a is the repressor, the right DNA makes a good repressor that will bind to the top DNA and regulate it.
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