Molecular Genetics and Biotechnology Practice Problems
Review the following concepts if needed:
- History of Molecular Genetics and Biotechnology for Genetics
- Recombinant DNA Technology for Genetics
- DNA Sequencing for Genetics
- Production of Recombinant Gene Products in Industry for Genetics
- Bioinformatics for Genetics
Molecular Genetics and Biotechnology Practice Problems
In 1953,Watson and Crick proposed that DNA replicates semiconservatively; i.e., both strands of the double helix become templates against which new complementary strands are made so that a replicated molecule would contain one original strand and one newly synthesized strand. A different hypothesis proposes that DNA replicates conservatively; i.e., the original double helix remains intact so that a replicated molecule would contain two newly synthesized strands. Bacterial DNA can be "labeled" with a heavy isotope of nitrogen (15N) by growing cells for several generations in a medium that has 15NH4Cl as its only nitrogen source. The common "light" form of nitrogen is 14N. Light and heavy DNA molecules can be separated by high-speed centrifugation (50,000 rpm = 105 × gravity) in a 6M (molar) CsCl (cesium chloride) solution, the density of which is 1.7 g/cm3 (very close to that of DNA). After several hours of spinning, the CsCl forms a density gradient, being heavier at the bottom and lighter at the top. In 1957, Matthew Meselson and Franklin W. Stahl performed a density-gradient experiment to clarify which of the two replication hypotheses was correct. How could this be done, and what results are expected after the first, second, and third generations of bacterial replication according to each of these hypotheses?
Solution 1Bacteria from 15N-labeled culture are transferred into medium containing 14N as the only source of nitrogen. A sample is immediately taken and its DNA is extracted and subjected to density-gradient equilibrium centrifugation. The DNA forms a single band relatively low in the tube where its density matches that of the CsCl in that region of the gradient.
After each generation of growth and replication of all DNA molecules, DNA is again extracted and measured for its density. According to the semiconservative theory, the first generation of DNA progeny molecules should all be "hybrid" (one strand containing only 14N and the other strand only 15N). Hybrid molecules would form a band at a density intermediate between fully heavy and fully light molecules. The second generations of DNA molecules should be 50% hybrids and 50% totally light, the latter forming a band relatively high in the tube where the density is lighter. After three generations, the ratio of light: hybrid molecules should be 3 : 1, respectively. The amount of hybrid molecules should be decreased by 50% in each subsequent generation.
According to the conservative replication scheme, the first generation of DNA molecules should be 50% heavy : 50% light. The second generation should be 25% heavy : 75% light. The third generation should be 12.5% heavy : 87.5% light. No hybrid molecules should be detected. The results of the Meselson and Stahl experiment supported the semiconservative theory of DNA replication.
Suppose that a circular plasmid contains 1000 bp. It is cut by three different restriction endonucleases, both singly and in pairs, with the results as shown below.
Reconstruct the plasmid, indicating where each enzyme cuts and the distances between all cuts.
Enzyme A cuts the circular plasmid at only one position, producing a linear molecule that is 1000 bp in length.
Enzyme B cuts the plasmid at three positions, producing fragments that are 100, 300, and 600 bp in length. Let us label these fragments B1, B2, and B3, respectively.
Enzyme C cuts the plasmid at two places, giving fragments that are 200 and 800 bp in length. Let us label them C1 and C2, respectively.
Digestion by both enzymes A and B produced four fragments of lengths 50, 100, 300, and 550 bp. Since the 100 and 300-bp fragments generated by enzyme B are still intact, enzyme A must have cut fragment B3 (600 bp) into two fragments of 50 and 550 bp. The location of this cut could be either closer to or farther from B2.
Digestion with both enzymes A and C produced three fragments of lengths 200, 375, and 425 bp. Obviously, the single cut by enzyme A must have been in fragment C2 (800 bp). The 425-bp fragment could be either to the left or right of A in the linear map.
Double digestion with enzymes B and C yields five pieces: 75, 100, 125, 225, and 475 bp in length. Fragment B1 (100 bp) is still intact, butB2 (300 bp) and B3 (600 bp) have been degraded. Therefore, one cut by enzyme C occurs in B2 and the other cut occurs in B3. The sum of the 75- and 225-bp fragments is 300 bp, corresponding to the length of B2. Thus, one cut by enzyme C is 75 bp from one end of B2. Likewise, the other cut by enzyme C is 125 bp from one end of B3. Recall that the single digest with enzyme C produced fragments of 200 bp (C1) and 800 bp (C2). Therefore, the only way the double digest of B and C can make sense is to have the 75-bp and 125-bp fragments adjacent to one another so that they total 200 bp as in C1.
The results of all three double digests can only be combined in one meaningful way. Since the cut made by enzyme A is in fragment B3, 50 bp from a B cut, and far from either C cut, A must map as follows:
A researcher desires to clone the gene encoding a particular biosynthesis enzyme (E1) from the yeast Saccharomyces cerevisiae. Its genome size is 1.4 × 104 kb and the average size of the library fragments is 5 kb. The genomic library was created in vectors that were transformed into bacterial cells. With 95% probability, how many recombinant bacterial colonies will have to be screened in order to find this particular gene fragment?
First, solve for n (ratio of the organism's genome size relative to the average fragment size in the gene library).
Then, N (the number of recombinant cells of a genomic library that need to be screened in order to find at least one containing the desired gene) can be calculated.
Thus, nearly 8400 bacterial colonies obtained from transformation will have to be screened in order to find the DNA fragment of interest with 95% confidence. If the average gene library insert size were larger, one can see that the number of clones to be screened would decrease. In contrast, if the genome size were larger, many more colonies would have to be screened in order to achieve success. This formula can act as a guide for estimating the number of clones to screen.
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