Answers to Concept Checks
- 1. Restriction enzymes exist naturally in bacteria, which use them to prevent the entry of viral DNA.
- 2. c
- 3. Southern blotting is used to transfer DNA from a gel to a solid medium. Northern blotting is used to transfer RNA from a gel to a solid medium, and Western blotting is used to transfer protein from a gel to a solid medium.
- 4. The gene and plasmid are cut with the same restriction enzyme and mixed together. DNA ligase is used to seal nicks in the sugar–phosphate backbone.
- 5. A heat-stable DNA polymerase enzyme is important to the success of PCR because the first step of the reaction requires that the solution be heated to between 90° and 100°C to separate the two DNA strands. Most enzymes are denatured at this temperature. With the use of a heat-stable polymerase, the enzyme can be added at the beginning of the reaction and will function throughout multiple cycles.
- 6. With the use of the genetic code and the amino acid sequence of the protein, possible nucleotide sequences that cover a small region of the gene can be deduced. A mixture of all the possible nucleotide sequences that might encode the protein, taking into consideration synonymous codons, is used to probe the library. To minimize the number of sequences required, a region of the protein that has relatively little degeneracy in its codons is selected.
- 7. The expression pattern of the gene can be examined, and the coding region of copies of the gene from individuals with the mutant phenotype can be compared with the coding region of wild-type individuals.
- 8. d
- 9. By using PCR with primers that flank the region containing tandem repeats.
- 10. b
- 11. c
- 12. Possible concerns include: (a) ecological damage caused by introducing novel organisms into the environment; (b) negative effects of transgenic organisms on biodiversity; (c) possible spread of transgenes to native organisms by hybridization; and (d) health effects of eating genetically modified foods.
- 13. Somatic gene therapy modifies genes only in somatic tissue, and these modifications cannot be inherited. Germline gene therapy alters genes in germ-line cells and will be inherited.
WORKED PROBLEMS
Problem 1
A molecule of double-stranded DNA that is 5 million base pairs long has a base composition that is 62% G + C. How many times, on average, are the following restriction sites likely to be present in this DNA molecule?
- a. BamHI (recognition sequence is GGATCC)
- b. HindIII (recognition sequence is AAGCTT)
- c. HpaII (recognition sequence is CCGG)
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Solution Strategy
What information is required in your answer to the problem?
The number of restriction sites likely to be present in the DNA molecule for each of the specified restriction enzymes.
What information is provided to solve the problem?
The size of the DNA molecule.-
The G + C base composition of the DNA molecule.
-
The recognition sequences for each restriction enzyme.
For help with this problem, review:
Cutting and Joining DNA Fragments in Section 19.2.
Solution Steps
The percentages of G and C are equal in double-stranded DNA; so, if G + C = 62%, then %G = %C = 62%/2 = 31%. The percentage of A + T = (100% − G − C) = 38%, and %A = %T = 38%/2 = 19%. To determine the probability of finding a particular base sequence, we use the multiplication rule, multiplying together the probably of finding each base at a particular site.
Hint: If you know the percentage of any base in the DNA, you can determine the percentages of all the other bases because G = C and A = T.
Recall: The multiplication rule states that the probability of two or more independent events is calculated by multiplying their independent probabilities.
- a. The probability of finding the sequence GGATCC = 0.31 × 0.31 × 0.19 × 0.19 × 0.31 × 0.31 = 0.0003333. To determine the average number of recognition sequences in a 5-million-base-pair piece of DNA, we multiply 5,000,000 bp × 0.00033 = 1666.5 recognition sequences.
- b. The number of AAGCTT recognition sequences is 0.19 × 0.19 × 0.31 × 0.31 × 0.19 × 0.19 × 5,000,000 = 626 recognition sequences.
- c. The number of CCGG recognition sequences is 0.31 × 0.31 × 0.31 × 0.31 × 5,000,000 = 46,176 recognition sequences.
Problem 2
You are given the following DNA fragment to sequence: 5′-GCTTAGCATC-3′. You first clone the fragment in bacterial cells to produce sufficient DNA for sequencing. You isolate the DNA from the bacterial cells and carry out the dideoxy-sequencing method. You then separate the products of the polymerization reactions by gel electrophoresis. Draw the bands that should appear on the gel from the four sequencing reactions.
Solution Strategy
What information is required in your answer to the problem?
The positions of the bands on the sequencing gel.
What information is provided to solve the problem?
The base sequence of the DNA fragment to be sequenced.
For help with this problem, review:
DNA Sequencing in Section 19.4.
Solution Steps
The first task is to write out the sequence of the newly synthesized fragment, which will be complementary and antiparallel to the original fragment. The original sequence is 5′–GCTTAGCATC–3′, so the newly synthesized sequence will be:
Original (template) sequence: 5′–GCTTAGCATC–3′
Newly synthesized sequence: 3′–CGAATCGTAG–5′
Recall: In dideoxy sequencing, a new DNA strand is synthesized and that strand is what is sequenced. Thus, the bands that appear on the gel represent the complement of the original sequence.
Thus, the sequence of the newly synthesized strand, written 5′→3′ is: 5′-GATGCTAAGC-3′. Bands representing this sequence will appear on the gel, with the bands representing nucleotides near the 5' end of the molecule at the bottom of the gel.
Hint: Small fragments, those nearer the 5' end of the newly synthesized strand, will migrate faster and will appear near the bottom of the gel.
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