Chapter 19 Solutions Manual

*look at #38, 39, 40 and 41 in solutions manual*

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*look at #30 in solutions manual*

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10. Briefly explain how the polymerase chain reaction is used to amplify a specific DNA sequence.

First, the double-stranded template DNA is denatured by high temperature. Then, primers corresponding to the ends of the DNA sequence to be amplified are annealed to the single-stranded DNA template strands. These primers are extended by a thermostable DNA polymerase so that the target DNA sequence is duplicated. These steps are repeated 30 times or more. Each cycle of denaturation, primer annealing, and extension results in doubling the number of copies of the target sequence between the primers.

*35. A molecular biologist wants to isolate a gene from a scorpion that encodes the deadly toxin found in its stinger, with the ultimate purpose of transferring this gene to bacteria and producing the toxin for use as a commercial pesticide. Isolating the gene requires a DNA library. Should the genetic engineer create a genomic library or a cDNA library? Explain your reasoning.

Solution: A cDNA library, created from mRNA isolated from the venom gland. Bacteria cannot splice introns. If the engineer wants to express the toxin in bacteria, then he needs a cDNA sequence that has been reverse transcribed from mRNA, and therefore has no intron sequences. The venom gland must be the source of the mRNA for cDNA synthesis, so that the cDNA library will be enriched for toxin cDNAs.

12. How does a genomic library differ from a cDNA library?

Solution: A genomic library is generated by cloning fragments of chromosomal DNA into a cloning vector. Chromosomal DNA is randomly fragmented by shearing or by partial digestion with a restriction enzyme. A cDNA library is made from mRNA sequences. Cellular mRNAs are isolated and then reverse transcriptase is used to copy the mRNA sequences to cDNA, which are cloned into plasmid or phage vectors. Therefore a cDNA library only represents the genes expressed in the tissues of origin for the RNA.

37. In Figure 19.22 Bob and Joe are each homozygous for different restriction fragment patterns. How many bands would you expect to see on the gel if a person was heterozygous for the A and B patterns? Explain your reasoning.

Solution: A person heterozygous for the A and B patterns would exhibit four bands: two bands associated with pattern B, and the two additional bands represented by pattern A. This is because a heterozygous person would have alleles for both the A and B patterns. The DNA for the A pattern would be cut into three fragments. The DNA for the B pattern would be cut into two fragments, one of which was the same as that for the A pattern and one which was different. Thus, the heterozygote would have DNA cut into four fragments and would exhibit four different bands.

26. How often, on average, would you expect a type II restriction endonuclease to cut a DNA molecule if the recognition sequence for the enzyme had 5 bp? (Assume that the four types of bases are equally likely to be found in the DNA and that the bases in a recognition sequence are independent.) How often would the endonuclease cut the DNA if the recognition sequence had 8 bp?

Solution: Because DNA has four different bases, the frequency of any sequence of n bases is equal to 1/(4n). A five-bp recognition sequence will occur with a frequency of 1/(45), or once every 1024 bp. An 8-bp recognition sequence will occur with a frequency of 1 per 48, or 65,536 bp.

7. Give three important characteristics of cloning vectors.

Solution: Cloning vectors should have: (1) An origin of DNA replication so they can be maintained in a cell (2) A gene, such as antibiotic resistance, to select for cells that carry the vector (3) A unique restriction site or series of sites to where a foreign DNA molecule may be inserted

18. What is DNA fingerprinting? What types of sequences are examined in DNA fingerprinting?

Solution: DNA fingerprinting is the typing of an individual for genetic markers at highly variable loci. This is useful for forensic investigations, to determine whether the suspect could have contributed to the evidentiary DNA obtained from blood or other bodily fluids found at the scene of a crime. Other applications include paternity testing and the identification of bodily remains. The first loci used for DNA fingerprinting were variable number of tandem repeat (VNTR) loci; these consist of short tandem repeat sequences located in introns or spacer regions between genes. The number of repeat sequences at the locus does not affect the phenotype of the individual in any discernible way, so the numbers of repeats at each of these loci are highly variable in the population. More recently, loci with smaller repeat sequences of just a few nucleotides, called short tandem repeats (STRs), have been adopted because they can be amplified by PCR. The variable number of repeats creates PCR fragments of different sizes. Genotyping at 13 to 15 of these unlinked STR loci can identify one individual among trillions of potential genotypes.

5. After DNA fragments have been separated by gel electrophoresis, how can they be visualized?

Solution: DNA molecules can be visualized by staining with a fluorescent dye, such as ethidium bromide, that intercalates between the stacked bases of the DNA double helix, and the dye-DNA complex fluoresces when irradiated with an ultraviolet light source. Alternatively, they can be visualized by attaching radioactive or chemical labels to the DNA before it is placed in the gel.

17. What is the purpose of the dideoxynucleoside triphosphate in the dideoxy sequencing reaction?

Solution: Dideoxynucleoside triphosphates (ddNTPs) act as a substrate for DNA polymerase but cause termination of DNA synthesis when they are incorporated, because they lack the 3′-OH for the addition of the next nucleotide. Mixed with regular dNTPs, fluorescently labeled ddNTPs generate a series of DNA fragments that have terminated at every nucleotide position along the template DNA molecule being sequenced. These fragments can be separated by gel electrophoresis. Because each of the four ddNTPs carries a different fluorescent label, a laser detector scanning near the end of the gel can distinguish which base terminates each fragment. Reading the fragments from shorter to longer, an automated DNA sequencer can determine the sequence of the template DNA molecule.

19. How does a reverse genetics approach differ from a forward genetics approach?

Solution: Forward genetics begins with mutant phenotype and proceeds toward cloning and characterization of the DNA encoding the gene. Reverse genetics begins with the DNA sequence of a gene, then generates specific mutations within that gene and requires the reinsertion of the modified gene back into cells or the whole organism to characterize the functions of the gene through new mutant phenotypes.

4. Explain how gel electrophoresis is used to separate DNA fragments of different lengths.

Solution: Gel electrophoresis uses an electric field to drive DNA molecules through a gel that acts as a molecular sieve. The gel is an aqueous matrix of agarose or polyacrylamide. DNA molecules are loaded into a slot or well at one end of the gel. When an electric field is applied, the negatively charged DNA molecules migrate toward the positive electrode. Shorter DNA molecules are less hindered by the agarose or polyacrylamide matrix and migrate faster than longer DNA molecules, which must wind their way around obstacles and through the pores in the gel matrix.

23. What is gene therapy?

Solution: Gene therapy is the correction of a defective gene by either gene replacement or the addition of a wild-type copy of the gene. For this to work, enough of the cells of the critically affected tissues or organs must be transformed with the functional copy of the gene to restore normal physiology.

20. Briefly explain how site-directed mutagenesis is carried out.

Solution: In oligonucleotide-directed mutagenesis, an oligonucleotide containing the desired mutation in the sequence is synthesized. This mutant oligonucleotide is annealed to denatured target DNA template and used to direct DNA synthesis. The result is a double-stranded DNA molecule with a mismatch at the site to be mutated. When transformed into bacterial cells, bacterial repair enzymes will convert the molecule to the mutant form about 50% of the time.

14. Briefly explain in situ hybridization, giving some applications of this technique.

Solution: In situ hybridization involves hybridization of radiolabeled or fluorescently labeled DNA or RNA probes to DNA or RNA molecules that are still in the cell. This technique can be used to visualize the expression of specific mRNAs in different cells and tissues and the location of genes on metaphase or polytene chromosomes.

21. What are knockout mice, how are they produced, and for what are they used?

Solution: Knockout mice have a target gene disrupted or deleted ("knocked out"). First, the target gene is cloned. The middle portion of the gene is replaced with a selectable marker, typically the neo gene that confers resistance to the drug G418. This construct is then introduced back into mouse embryonic stem cells and cells with G418 resistance are selected. The surviving cells are screened for cells where the chromosomal copy of the target gene has been replaced with the neo-containing construct by homologous recombination of the flanking sequences. These embryonic stem cells are then injected into mouse blastocyst-stage embryos and these chimeric embryos are transferred to the uterus of a pseudopregnant female mouse. The knockout cells will participate in the formation of many tissues in the mouse fetus, including germ-line cells. The chimeric offspring are interbred to produce offspring that are homozygous for the knockout allele. The phenotypes of the knockout mice provide information about the function of the gene.

9. Briefly explain how an antibiotic-resistance gene and the lacZ gene can be used to determine which cells contain a particular plasmid.

Solution: Many plasmids designed as cloning vectors carry a gene for antibiotic resistance and the lacZ gene. The lacZ gene on the plasmid has been engineered to contain multiple unique restriction sites. Foreign DNAs are inserted into one of the unique restriction sites in the lacZ gene of plasmids and the plasmids are transformed into E. coli cells lacking a functional lacZ gene. Transformed cells are plated on a medium containing the appropriate antibiotic to select for cells that carry the plasmid. The medium also contains an inducer for the lac operon, so the cells express the lacZ gene, and X-gal, a substrate for beta-galactosidase that will turn blue when cleaved by β-galactosidase. The colonies that carry plasmid without foreign DNA inserts will have intact lacZ genes, make functional β-galactosidase, cleave X-gal, and turn blue. Colonies that carry plasmid with foreign DNA inserts will not make functional β-galactosidase because the lacZ gene is disrupted by the foreign DNA insert. They will remain white. Thus, cells carrying plasmids with inserts will form white colonies. This is known as blue/white selection.

22. How is RNA interference used in the analysis of gene function?

Solution: RNA interference is one potential reverse genetics approach to analyze gene function, by specifically repressing expression of that gene. Double-stranded RNA may be injected directly into a cell or organism or the cell or organism may be genetically modified to express a double-stranded RNA molecule corresponding to the target gene.

3. What normal role do restriction enzymes play in bacteria? How do bacteria protect their own DNA from the action of restriction enzymes?

Solution: Restriction enzymes cut foreign DNA, such as viral DNA, into fragments. Bacteria protect their own DNA by modifying bases, usually by methylation, at the recognition sites.

6. What is the purpose of Southern blotting? How is it carried out?

Solution: Southern blotting is used to detect and visualize specific DNA fragments that have a sequence complementary to a labeled DNA probe. DNA is first cleaved into fragments with restriction endonucleases. The fragments are separated by size via gel electrophoresis. These fragments are then denatured and transferred by blotting onto the surface of a membrane filter. The membrane filter now has single-stranded DNA fragments bound to its surface, separated by size as in the gel. The filter is then incubated with a solution containing a denatured, labeled probe DNA. The probe DNA hybridizes to its complementary DNA on the filter. After washing away excess unbound probes, the labeled probe hybridized to the DNA on the filter can be detected using the appropriate methods to visualize the label. For radioactively labeled probes, the bound probe is detected by exposure to X-ray film. Other probe labeling methods detect bound probe using enzymatic reactions that generate luminescence or color.

13. How are probes used to screen DNA libraries?

Solution: The DNA library must first be plated out, either as colonies for plasmid libraries or phage plaques on a bacterial lawn for phage libraries. The colonies or plaques are transferred to membrane filters. A nucleic acid probe can be used to identify colonies or phage plaques that contain identical or similar sequences by hybridization.

32. A geneticist uses a plasmid for cloning that has the lacZ gene and a gene that confers resistance to penicillin. The geneticist inserts a piece of foreign DNA into a restriction site that is located within the lacZ gene and uses the plasmid to transform bacteria. Explain how the geneticist can identify bacteria that contain a copy of a plasmid with the foreign DNA.

Solution: The geneticist should plate the bacteria on agar medium containing penicillin to select for cells that have taken up the plasmid. The medium should also have X-gal and an inducer of the lac operon, such as IPTG or even lactose. Cells that have taken up a plasmid without foreign DNA will have an intact lacZ gene, produce functional β- galactosidase, and cleave X-gal to make a blue dye. These colonies will turn blue. In contrast, cells that have taken up a plasmid containing a foreign DNA inserted into the lacZ gene will be unable to make functional β-galactosidase. These colonies will be white.

2. What feature is commonly seen in the sequences recognized by type II restriction enzymes?

Solution: The recognition sequences are palindromic, and 4−8 base pairs long.

28. Will restriction sites for an enzyme that has 4 bp in its restriction site be closer together, farther apart, or similarly spaced, on average, compared with those of an enzyme that has 6 bp in its restriction site? Explain your reasoning.

Solution: The restriction sites for an enzyme with a 4-bp recognition sequence should be spaced closer together than the sites for an enzyme with a 6-bp recognition sequence. The 4- bp recognition sequence will occur with an average frequency of once every 44 = 256 bp, whereas the 6-bp recognition sequence will occur with an average frequency of once every 46 = 4096 bp.

42. You have discovered a gene in mice that is similar to a gene in yeast. How might you determine whether this gene is essential for development in mice?

Solution: This gene must first be cloned, possibly by using the yeast gene as a probe to screen a mouse genomic DNA library. The cloned gene is then used to create a knockout mouse strain. A neo gene is inserted into a cloned exon sequence, followed by the tk (thymidine kinase) gene. This construct is then introduced into mouse embryonic stem cells. Selection of G418 resistant and gancyclovir resistant cells will identify transformants where homologous recombination resulted in insertion of the neo gene into the target mouse gene exon. These knockout stem cells are injected into blastocyst stage embryos, which are then transferred to the uterus of a pseudopregnant mouse. The progeny are tested for the presence of the knockout allele. Progeny with the knockout allele are interbred. If the gene is essential for embryonic development, no homozygous knockout mice will be born. The arrested or spontaneously aborted fetuses can then be examined to determine how development has gone awry in fetuses that are homozygous for the knockout allele.

*29. About 60% of the base pairs in a human DNA molecule are AT. If the human genome has 3.2 billion base pairs of DNA, about how many times will the following restriction sites be present? a. BamHI (restriction site = 5'—GGATCC—3') b. EcoRI (restriction site = 5'—GAATTC—3') c. HaeIII (restriction site = 5'—GGCC—3')

Solution: We must first calculate the frequency of each base. Given that AT base pairs consist 60% of the DNA, we deduce that the frequency of A is 0.3 and frequency of T is 0.3. The GC base pairs must consist of 40% of the DNA; therefore, the frequency of G is 0.2 and the frequency of C is 0.2. a. BamH1 GGATCC is then (0.2)(0.2)(0.3)(0.3)(0.2)(0.2) = 0.000144 3,200,000,000(0.000144) = 460,800 times b. EcoRI GAATTC = (0.2)(0.3)(0.3)(0.3)(0.3)(0.2) = 0.000324 3,200,000,000(0.000324) = 1,036,800 times c. HaeIII GGCC = (0.2)(0.2)(0.2)(0.2) = 0.0016 3,200,000,000(0.0016) = 5,120,000 times

*look at #43 in solutions manual*

Yee

36. A protein has the following amino acid sequence: Met-Tyr-Asn-Val-Arg-Val-Tyr-Lys-Ala-Lys-Trp-Leu-Ile-His-Thr-Pro You wish to make a set of probes to screen a cDNA library for the sequence that encodes this protein. Your probes should be at least 18 nucleotides in length. a. Which amino acids in the protein should be used to construct the probes so that the least degeneracy results? (Consult the genetic code in Figure 15.10.) b. How many different probes must be synthesized to be certain that you will find the correct cDNA sequence that specifies the protein?

a. Solution: A probe of 18 nucleotides must be based on six amino acids. The six amino acid stretch with the least degeneracy is Val-Tyr-Lys-Ala-Lys-Trp. This sequence avoids the amino acids arg and leu, which have six codons each. b. Solution: Val and Ala have four codons each, Tyr and Lys have two codons each, and Trp has one codon. Therefore, there are 4 × 2 × 2 × 4 × 2 × 1 possible sequences, or 128.