5′-GAAAGTCCGCGTTATAGGCATG-3′

3′-ACGTCTTTCAGGCGCAATATCCGTACTTAA-5′

A YAC vector is not stably maintained as a yeast chromosome during mitosis unless it carries an insert of more than 100,000 bp.

(a) Some original pBR322 plasmids will be present, regenerated without insertion of a foreign DNA fragment. Also, two or more pBR322 plasmids might be ligated, with or without insertion of a segment of foreign DNA. All of these would retain resistance to ampicillin. (b) The clones in lanes 1 and 2 each have one DNA fragment, inserted in different orientations. The clone in lane 3 has incorporated two of the DNA fragments, ligated such that the ends closest to the EcoRI sites are joined.

The sequence will appear about once every 4⁸ = 65,536 bp. If the G + C content is greater than the A + T content (or vice versa), the frequency of occurrence of the restriction site will decrease.

In a large DNA molecule with a random sequence, a BamHI site will occur, on average, every 4,096 bp (assuming all four nucleotides are present in equal proportions). Cleavage of all BamHI sites in the DNA would produce fragments much smaller than the 100,000 to 300,000 bp needed for a BAC library.

Primer 1: CCTCGAGTCAATCGATGCTG

Primer 2: CGCGCACATCAGACGAACCA

Recall that all DNA sequences are written in the 5′→3′ direction, left to right; that the two strands of a DNA molecule are antiparallel; and that both PCR primers must target the end sequences so that their 3′ ends are oriented toward the segment to be amplified.

Primer 1: GAATTCCCTCGAGTCAATCGATGCTG

Primer 2: GAATTCCGCGCACATCAGACGAACCA

The test requires DNA primers, a heat-stable DNA polymerase, deoxynucleoside triphosphates, and a PCR machine. The primers are designed to amplify a DNA segment encompassing the CAG repeat. The DNA strand shown in the problem is the coding strand, oriented 5′ → 3′, left to right. The primer targeted to the DNA to the left of the CAG repeat should be identical to any 25-nucleotide sequence in the region to the left of the repeat. The primer on the right side must be complementary and antiparallel to a 25-nucleotide sequence to the right of the CAG repeat. With these primers, an investigator would use PCR to amplify the DNA including the CAG repeat, and then determine its size by comparison with size markers on electrophoresis. The length of the DNA reflects the length of the CAG repeat, providing a simple test for the disease.

The researcher could design PCR primers complementary to DNA in the deleted segment that will direct DNA synthesis away from each other. No PCR product will be generated unless the ends of the deleted segment are joined to create a circle.

No. The orientation of the cloned gene is very important, because the information specifying the protein is contained in only one of the two DNA strands. The promoter specifies not only where RNA polymerase binds the DNA but also the direction in which it travels and the DNA strand that it uses as template for RNA synthesis. When the correct DNA strand is used as template, a functional protein results. If the gene is inverted, the opposite DNA strand will become the template for synthesis of RNA, with a much different nucleotide sequence. The resulting protein will be completely different, unrelated to the normal gene product, and most likely nonfunctional.

S-8

(a) The bacterial recA gene is readily cloned by using the bacterial plasmid. (b) For the mammalian DNA polymerase, the baculovirus system may have a better chance of generating an active protein.

The production of labeled antibodies is difficult and expensive. The labeling of every antibody to every protein target would be impractical. By labeling one antibody preparation for binding to all antibodies of a particular class, the same labeled antibody preparation can be used in many different Western blot experiments.

Express the protein in yeast strain 1 as a fusion protein with one of the domains of Gal4p, such as the DNA- binding domain. Using yeast strain 2, make a library in which essentially every protein of the fungus is expressed as a fusion protein with the interaction domain of Gal4p. Mate strain 1 with the strain 2 library, and look for colonies that are colored due to expression of the reporter gene. These colonies will generally arise from mated cells containing fusion protein that interacts with your target protein.

Cover spot 4, add solution containing activated T, irradiate, wash. Result: 1. A–T; 2. G–T; 3. A–T; 4. G–C. Cover spots 2 and 4, add solution containing activated G, irradiate, wash. Result: 1. A–T–G; 2. G–T; 3. A–T–G; 4. G–C. Cover spot 3, add solution containing activated C, irradiate, wash. Result: 1. A–T–G–C; 2. G–T–C; 3. A–T–G; 4. G–C–C. Cover spots 1, 3, and 4, add solution containing activated C, irradiate, wash. Result: 1. A–T–G–C; 2. G–T–C–C; 3. A–T–G; 4. G–C–C. Cover spots 1 and 2, add solution containing activated G, irradiate, wash. Result: 1. A–T–G–C; 2. G–T–C–C; 3. A–T–G–G; 4. G–C–C–G.

In long repetitive DNA sequences, the fragments can be overlapped (see Figure 7-15), but there are many ways to overlap two fragments that contain such repeats. It is impossible to determine exactly how long the repeated region is unless all the repeats are contained in one continuous sequenced fragment. The read lengths for many next-generation sequencing technologies are insufficient to capture an entire region of this kind in one continuous sequence.

The sgRNA is required to pair the CRISPR/Cas9 with its intended target site in the DNA and to activate the CRISPR nuclease domains for cleavage.

(a) R6-5, at least 11; pSC101, 1; pSC102, 3. (b) Each of the observed bands in a given lane represents a DNA fragment, and each fragment is present in the same concentration (total molecules). However, the fragments to the left are longer than the ones to the right and thus take up more of the fluorescent stain. (c) Two EcoRI fragments derived from R6-5 are very nearly the same size, and they migrate together at this position. Thus, there are 12 fragments, derived from cleavage of R6-5 at its 12 EcoRI recognition sites. (d) The plasmid pSC102 is made up of these three EcoRI fragments from R6-5. (e) The larger fragment on the left, which comigrates with pSC101, is the only possible source of a tetracycline-resistance gene in the parent plasmids. (f) The smaller fragment on the right, which comigrates with one of the fragments of pSC102, is the only possible source of a kanamycin-resistance gene in the parent plasmids. (g) 7,000 bp. (h) Four phosphodiester bonds; two fragments were ligated, with two new phosphodiester bonds created at each of the two ligation sites. (i) The original ligation mixture included a wide range of combinations of DNA fragments. When the mixture was used to transform E. coli cells, only cells that took up a combination of fragments that allowed survival on the selection media would grow. Evidently, these two fragments of pSC101 did not include a gene for resistance to tetracycline or kanamycin, the antibiotics used for selection. The pSC101 plasmid that included the tetracycline-resistance gene also had a replication origin, so no new replication origin was required. The joining of three fragments into a circle by ligation is considerably less probable than the ligation of two fragments. In effect, the selection generated the simplest possible recombinant plasmid from the available fragments.