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Ch. 20 - Recombinant DNA Technology

Chapter 19, Problem 15

You have recovered a cloned DNA segment from a vector and determine that the insert is 1300 bp in length. To characterize this cloned segment, you isolate the insert and decide to construct a restriction map. Using enzyme I and enzyme II, followed by gel electrophoresis, you determine the number and size of the fragments produced by enzymes I and II alone and in combination, as recorded in the following table. Construct a restriction map from these data, showing the positions of the restriction-enzyme cutting sites relative to one another and the distance between them in units of base pairs.

Enzyme   Restriction Fragment Sizes (bp)
     I           350, 950
    II           200, 1100
I and II      150, 200, 950

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Welcome back. Here's our next problem. It says a 1500 base base pair, clone DNA segment was recovered from a vector and its restriction map must be constructed in order to characterize this clone segment. The clone segment was digested with two restriction enzymes. And then electrophoresis on Aguero's gel. The band size, observed band sizes observed in the gel are as follows, enzyme A 450 base pairs and 10 50 base pairs, Enzyme B, 200 base pairs and 1300 base pairs, enzymes A. And B. 200 base pairs, 250 base pairs, 10 50 base pairs. Using this information, create a restriction map and then calculate the distance between the cutting sites of enzymes A and B. In base pairs on the cologne DNA segment. Well, I've put up the start of a little diagram to help us think through this and calculate these things. So we know that our entire segment is 1500 base pairs long. So I've drawn out that segment three different times. And we'll think about when we cut with enzyme A. Enzyme B. And then A. And B. Together, remembering that we want to look for the distance between the cutting sites and make a little map of where they are. So we know that enzyme a. We get uh two bands of 450 base pairs and 1050 base pairs. So we know that this um cutting site. If we start at zero will be 450 base pairs From the end, giving us 450 base pairs and 1050 base pairs. So let's go down to answer or I'm sorry, enzyme b. Enzyme B. We have are fragments of 200 base pairs and 1300 base pairs. So we know our cutting site here. It's going to be about 200 base pairs from the end. And that would give us those two fragment links. So we cut with enzymes A. And B. We know that we cut in this site here at 200 base pairs. Enzyme B. is going to cut there and then Enzyme A. is going to cut at its 450 base parasite. And let's make sure that gives us the fragments we observe. So we would get Our 200 base pair fragment here at the beginning. So that will give us our 200 base pair. And then this distance here between the cutting site at 200 from enzyme B. And the cutting site at 450 from enzyme A. Will be 250 base pairs apart. And then finally this last part here we know will be 1050 base pairs. And these are indeed the three band length that we expect when we cut. So here's our restriction map here with A. And B. Cutting and it says, calculate the distance between The cutting sites of enzymes a and b. In base pairs. So that would be this distance here between these two cutting sites, which is 250 base pairs. So come over here and we see the choice B is 250 base pairs. And again, that is the distance between the cutting sites of enzymes a and b. Given the different lengths were given. When the two enzymes cut this 1500 base pair fragment. See you in the next video.
Related Practice
Textbook Question

In a control experiment, a plasmid containing a HindIII recognition sequence within a kanamycin resistance gene is cut with HindIII, re-ligated, and used to transform E. coli K12 cells. Kanamycin-resistant colonies are selected, and plasmid DNA from these colonies is subjected to electrophoresis. Most of the colonies contain plasmids that produce single bands that migrate at the same rate as the original intact plasmid. A few colonies, however, produce two bands, one of original size and one that migrates much less far down the gel. Diagram the origin of this slow band as a product of ligation.

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Textbook Question

Describe how the team from the J. Craig Venter Institute created a synthetic genome. How did the team demonstrate that the genome converted the recipient strain of bacteria into a different strain?

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What advantages do cDNA libraries provide over genomic DNA libraries? Describe cloning applications where the use of a genomic library is necessary to provide information that a cDNA library cannot.

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Textbook Question

To create a cDNA library, cDNA can be inserted into vectors and cloned. In the analysis of cDNA clones, it is often difficult to find clones that are full length—that is, many clones are shorter than the mature mRNA molecules from which they are derived. Why is this so?

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Textbook Question

Although the capture and trading of great apes has been banned in 112 countries since 1973, it is estimated that about 1000 chimpanzees are removed annually from Africa and smuggled into Europe, the United States, and Japan. This illegal trade is often disguised by simulating births in captivity. Until recently, genetic identity tests to uncover these illegal activities were not used because of the lack of highly polymorphic markers (markers that vary from one individual to the next) and the difficulties of obtaining chimpanzee blood samples. A study was reported in which DNA samples were extracted from freshly plucked chimpanzee hair roots and used as templates for PCR. The primers used in these studies flank highly polymorphic sites in human DNA that result from variable numbers of tandem nucleotide repeats. Several offspring and their putative parents were tested to determine whether the offspring were 'legitimate' or the product of illegal trading. The data are shown in the following Southern blot.

Examine the data carefully and choose the best conclusion.

a. None of the offspring are legitimate.
b. Offspring B and C are not the products of these parents and were probably purchased on the illegal market. The data are consistent with offspring A being legitimate.
c. Offspring A and B are products of the parents shown, but C is not and was therefore probably purchased on the illegal market.
d. There are not enough data to draw any conclusions. Additional polymorphic sites should be examined.
e. No conclusion can be drawn because 'human' primers were used. <>

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Textbook Question
The Age of Genetics was created by remarkable advances in the use of biotechnology to manipulate plant and animal genomes. Given that the world population reached 7.5 billion people in 2017 and is expected to reach 9.7 billion in 2050, some scientists have proposed that only the worldwide introduction of genetically modified (GM) foods will increase crop yields enough to meet future nutritional demands. Pest resistance, herbicide, cold, drought, and salinity tolerance, along with increased nutrition, are seen as positive attributes of GM foods. However, others caution that unintended harm to other organisms, reduced effectiveness to pesticides, gene transfer to nontarget species, allergenicity, and as yet unknown effects on human health are potential concerns regarding GM foods. If you were in a position to control the introduction of a GM primary food product (rice, for example), what criteria would you establish before allowing such introduction?
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