LEHNINGER PRINCIPLES OF BIOCHEMISTRY - VOL 1. THE FOUNDATIONS OF BIOCHEMISTRY: STRUCTURE AND CATALYSIS - 2011

PART I. STRUCTURE AND CATALYSIS

9. DNA-BASED INFORMATION TECHNOLOGY

Questions and Problems

1. Cloning.

When joining two or more DNA fragments, scientists can tailor the sequence at the junction in many clever ways, as explored in the following exercises.

a) Draw The Structure of each end of a linear DNA fragment generated by restriction with the enzyme EcoRI (including the sequences remaining from the EcoRI recognition site).

b) Draw the structure resulting from the reaction of this end sequence with DNA polymerase I and the four deoxynucleoside triphosphates (see Fig. 8-33).

c) Give the sequence resulting at the junction when two ends with the structure obtained in (b) are ligated (see Fig. 25-17).

d) Draw the structure that results if the structure from (a) is treated with a nuclease that degrades only single-stranded DNA.

e) Give the junction sequence when an end with structure (b) is ligated to an end with structure (g).

f) Draw the STRUCTURE OF THE end of a linear DNA fragment generated by restriction with the enzyme PvuII (consider the sequences remaining from the PvuII recognition sequence).

g) Give the junction sequence when an end with structure (b) is ligated to an end with structure (e).

h) "Synthesize" a short double-stranded DNA fragment with any sequence of your choice. Using this synthetic fragment and the Methods from parts (a) through (g), design a scheme to remove an EcoRI restriction site from a DNA molecule and insert a new BamHI restriction site at approximately the same Location (see Fig. 9-3).

i) Propose four different short synthetic double-stranded DNA fragments that would allow the ligation of structure (a) to a DNA fragment generated by restriction with the enzyme PstI. In one of these fragments, design the sequence so that the final joint contains recognition sequences for both EcoRI and PstI. In the second and third fragments, design the sequences so that the joint contains the recognition sequence for either EcoRI or PstI, respectively. In the fourth fragment, design the sequence so that the final joint contains neither an EcoRI nor a PstI restriction site.

2. Selection of recombinant Plasmids.

When cloning a foreign DNA fragment into a plasmid, it is often useful to insert the fragment at a site that disrupts a selectable marker (for example, the tetracycline resistance Gene of plasmid pBR322). The loss of function in the disrupted gene can be used to detect clones containing recombinant plasmids with foreign DNA. With a bacteriophage λ-based vector, this is not necessary because vectors carrying large foreign DNA fragments can be easily distinguished from those that do not. How are such recombinant vectors identified?

3. DNA Cloning.

The plasmid cloning vector pBR322 (see Fig. 9-4) is cut with the restriction endonuclease PstI. A DNA fragment taken from a eukaryotic genome (also obtained by PstI Digestion) is inserted into the prepared vector and ligated. The solution of DNAs ligated in this manner is then used to transform Bacteria, and bacteria containing plasmids are selected by growth in the presence of tetracycline.

a) In addition to the desired recombinant plasmid, what Other types of plasmids might be found among the tetracycline-resistant transformed bacteria? How can they be distinguished?

b) The cloned DNA fragment is 1,000 bp long, and an EcoRI site is located 250 bp from one end. Three different recombinant plasmids are digested with EcoRI and analyzed by gel Electrophoresis, yielding the given patterns. What can be concluded about the cloned DNA from each of the patterns? Keep in mind that in pBR322, the PstI and EcoRI restriction sites are 750 bp apart from each other. The entire plasmid without the cloned insert has a length of 4,361 bp. The size markers in lane 4 can be used to estimate the length of the nucleotide fragment.

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4. Identification of the gene for a protein with a known Amino Acid Sequence.

Using Figure 27-7 to translate METABOLISM/28.html">The Genetic Code, propose a DNA probe that would allow the identification of the gene for a protein with the following N-terminal amino acid sequence. The probe should be 18 to 20 Amino Acids long—a size that ensures proper Specificity, provided there is sufficient Homology between the gene and the probe.

5. Development of diagnostic tests for Genetic Disorders.

Huntington's disease (HD) is an inherited neurodegenerative disorder characterized by a gradual, irreversible decline in mental, motor, and cognitive abilities. Symptoms typically appear in middle age, although onset can occur at virtually any age. The disease generally lasts 15 to 20 years. The Molecular Basis of the condition is now becoming better understood. The genetic mutation underlying HD was discovered in a gene encoding a protein ($M_r$ = 350,000) of unknown function. In individuals unaffected by HD, the region of the gene encoding the N-terminus of the protein contains a sequence of CAG codons (corresponding to glutamine) repeated 6 to 39 times in tandem. In individuals with adult-onset HD, this codon is typically repeated 40 to 55 times, while in those with juvenile-onset HD, it is repeated more than 70 times. Based on the length of this simple trinucleotide repeat, it is possible to determine whether a person will develop HD and approximately when the first symptoms will appear.

Shown below is a small portion of the N-terminal coding sequence of the HD gene, which consists of 3,143 codons. The DNA nucleotide sequence is shown in black, The amino acid sequence encoded by the gene is shown in blue, and the CAG repeat is shaded. Using Fig. 27-7 to translate the genetic code, outline a PCR-based test for HD that could be performed on a Blood sample. Assume that the PCR primer must be 25 NUCLEOTIDES long. By convention, unless otherwise specified, the protein-coding DNA sequence is depicted with the coding strand (the sequence identical to the mRNA transcribed from the gene) on top, so that it reads from the 5' end to the 3' end from left to right.

6. Using PCR to detect circular DNA molecules.

In ciliated single-celled organisms, a segment of genomic DNA is sometimes excised. This deletion is a genetically programmed response associated with Cell mating. Scientists hypothesize that the DNA is removed via Site-Specific Recombination, where the DNA strands at both ends of the segment are joined together, and the excised DNA appears as a circular molecule as a product of the reaction.

Consider how the Polymerase Chain Reaction (PCR) could be used to detect the circularized form of the excised DNA in an extract of single-celled organisms.

7. Glowing plants.

Plants carrying the green fluorescent protein gene (see Fig. 9-15a) glow in the dark when grown in standard garden soil under normal watering conditions, whereas plants carrying the firefly luciferase gene (see Fig. 9-29) do not. How can this be explained?

8. Using RFLP analysis for paternity testing.

DNA fingerprinting and RFLP analysis are very commonly used in paternity testing. A child inherits Chromosomes from both the mother and the father; therefore, the child's DNA contains distinct restriction fragments of genetic material derived from each parent. Based on the electrophoresis data presented here, determine which child, if any, is not the biological offspring of the putative father. Explain your reasoning. Lane M corresponds to the sample taken from the mother, F to the putative father, and C₁, C₂, and C₃ to the children.

9. Mapping a chromosomal segment.

A set of overlapping clones, designated A through F, was obtained from the same region of a chromosome. Each clone was digested individually with a restriction endonuclease, and the resulting fragments were separated by agarose gel electrophoresis, as shown in the figure below. Restriction mapping of this chromosomal segment yielded nine distinct fragments, with each clone containing a specific subset of them. Based on this information, determine the order of the restriction fragments along the chromosome.

10. Plant cloning.

The technique schematically illustrated in Fig. 9-28 utilizes Cells of the bacterium *Agrobacterium* species containing two separate plasmids. Consider why the sequences of the two plasmids do not combine into one.

11. DNA fingerprinting and RFLP analysis.

DNA is extracted from Blood Cells of two different individuals (1 and 2). In separate experiments, the DNA from each individual is digested with Restriction Endonucleases A, B, and C, and the fragments are separated by electrophoresis. The figure depicts a hypothetical restriction map of a 10,000 bp human chromosomal segment. Individual 2 carries point Mutations that eliminate the restriction recognition sites B* and C*. You probe the gel with a radioactive oligonucleotide complementary to the specified sequence and expose it to X-ray film. Indicate where you would expect to see bands on the film. The gel lanes are labeled in the accompanying diagram.

12. Using photolithography to prepare DNA Microarrays.

Figure 9-21 shows the initial steps in the preparation of a DNA microarray or DNA chip using photolithography. Describe the remaining steps required to synthesize the desired sequences (a different four-nucleotide sequence in each of the four features) shown in the first boxed illustration in the figure. After each step, show the resulting nucleotide sequence attached to each feature.

13. Mammalian cloning.

The retroviral vectors shown in Fig. 9-32 enable the efficient integration of introduced DNA into the mammalian genome. Explain how these vectors, which lack viral Replication and packaging genes (gag, pol, env), are incorporated into infectious Viral Particles. Consider why it is crucial that these vectors lack replication and packaging genes.

Analysis of experimental data

14. Hindi: The first restriction endonuclease.

The discovery of a restriction endonuclease that could be used in practice was first reported in two papers published in 1970. In the first study, Smith and Wilcox described the isolation of an enzyme that cleaved double-stranded DNA. The nuclease activity of the enzyme was measured by the decrease in viscosity of a DNA solution upon Treatment with the enzyme.

a) Why does the viscosity of the DNA solution decrease after enzyme treatment?

To determine whether the enzyme is an endonuclease or an exonuclease, DNA was labeled with ³²P and then treated with the enzyme, followed by The addition of trichloroacetic acid (TCA). Under these conditions, single nucleotides should remain in solution, whereas oligonucleotides should precipitate.

b) Treatment of the labeled DNA with the enzyme yielded no TCA-soluble fragments. Based on this result, determine whether the enzyme was an endonuclease or an exonuclease. Explain your reasoning.

During the Cleavage of a polynucleotide chain, the phosphate group is typically not cleaved off, but remains attached to either the 5'- or 3'-end of the resulting DNA fragment. Smith and Wilcox proposed the following protocol to determine THE POSITION OF the phosphate group.

1. Treatment of unlabeled DNA with the nuclease.

2. Addition to an aliquot of the product obtained in step 1 of radioactively labeled ATP (y-[³²P] ATP) and polynucleotide kinase (an enzyme that transfers the y-phosphate of ATP to a 5'-hydroxyl group, but not to a 5'-phosphate, 3'-hydroxyl, or 3'-phosphate group). Measurement of The amount of ³²P incorporated into the DNA (Sample A).

3. Addition to the product obtained in step 1 of alkaline phosphatase (an enzyme that removes phosphate groups from both 5'- and 3'-ends), followed by polynucleotide kinase and y-[³²P] ATP. Measurement of the amount of ³²P incorporated into the DNA (Sample B).

c) Sample A was found to have a radioactivity of 136 cpm, whereas Sample B had 3,740 cpm. Based on these findings, did the phosphate group remain on the 5'- or the 3'-end of the DNA fragment after enzyme treatment? Explain your answer.

d) Treatment of bacteriophage T7 DNA with the nuclease produced about 40 DNA fragments of varying lengths. How does this finding support the Conclusion that the restriction enzyme recognizes specific DNA sequences rather than introducing random breaks into The Double Helix?

At this stage of the research, two hypotheses existed: 1) the enzyme introduces breaks within the recognition site itself; 2) the enzyme recognizes a specific sequence but cleaves the DNA at an adjacent site. To resolve this issue, Kelly and Smith determined the 5'-terminal sequences of the DNA fragments generated by nuclease digestion. The following protocol was used for this purpose.

1. Treatment of phage T7 DNA with the restriction enzyme.

2. Treatment of the resulting fragments with alkaline phosphatase to remove 5'-phosphate groups.

3. Treatment of the fragments obtained in step 2 with polynucleotide kinase in the presence of y-[³²P]ATP to radiolabel the 5'-ends of the DNA fragments.

4. Treatment of the labeled fragments with DNases to convert the DNA fragments into a mixture of mono-, di-, and trinucleotides.

5. Determination of the sequences of the labeled mono-, di-, and trinucleotides by comparing them with oligonucleotides of known sequence using Thin-Layer Chromatography.

Among the labeled products, the following were detected: mononucleotides A and G, dinucleotides 5'-pApA-3' and 5'-pGpA-3', and trinucleotides 5'-pApApC-3' and 5'-pGpApC-3'.

e) Which hypothesis regarding the mode of DNA cleavage by the nuclease is consistent with these results? Explain your reasoning.

Kelly and Smith then proceeded to determine the 3'-terminal sequences of the fragments generated by nuclease digestion. They detected a mixture of 5'-pTpC-3' and 5'-pTpT-3'. They were unable to determine The sequence of any trinucleotide at the 3'-ends of the fragments.

f) Based on these data, determine the sequence recognized by the nuclease and indicate the exact cleavage site within this fragment. Use Table 9-2 as a model for your answer.