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 K 12 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.

Short Answer

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Answer: The slow-migrating band may have emerged due to incorrect re-ligation of the cleaved plasmid DNA during the control experiment. This could result in the plasmid forming a larger molecule containing extra or incorrectly re-ligated DNA sequences. Larger molecules migrate slower on the gel compared to smaller molecules.

Step by step solution

01

Understand the control experiment

In this control experiment, a plasmid containing a HindIII recognition sequence within a kanamycin resistance gene is cut with HindIII, which cleaves the DNA at the recognition site. The DNA is then re-ligated and used to transform \(E. coli\) K 12 cells. The kanamycin-resistant colonies are selected, and plasmid DNA from these colonies is subjected to electrophoresis.
02

Analyze the electrophoresis results

Most of the colonies contain plasmids that produce single bands that migrate at the same rate as the original intact plasmid. On the other hand, a few colonies produce two bands, one of original size and one that migrates much less further down the gel (slow-migrating band).
03

Explain the presence of the slow band

The slow-migrating band could have emerged due to incorrect re-ligation of the cleaved plasmid DNA during the control experiment. This could result in the plasmid forming a larger molecule containing extra or incorrectly re-ligated DNA sequences. Larger molecules migrate slower on the gel compared to smaller molecules.
04

Diagram the origin of the slow band as a product of ligation

To illustrate the origin of the slow band, we can draw the following diagram: 1. Start with the original plasmid containing a HindIII recognition site within a kanamycin resistance gene. (A circle with the HindIII site and kanamycin resistance gene marked) 2. Cut the plasmid with HindIII, resulting in a linear DNA molecule with two overhangs where it was cut. (A straight line with the two ends marked as HindIII overhangs, and the kanamycin resistance gene portion separated) 3. During re-ligation, instead of correctly joining the two ends back together, the DNA molecule may form a larger loop or concatemer with other plasmid molecules or extra DNA sequences. (An enlarged circle with the HindIII site and kanamycin resistance gene marked, plus any extra or misaligned DNA sequences) 4. On gel electrophoresis, this larger looped or concatenated molecule would migrate slower on the gel, resulting in the observed slow-migrating band. (Electrophoresis gel image showing two bands: one for the original intact plasmid and the other for the larger ligation product)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

HindIII Restriction Enzyme
Restriction enzymes like HindIII are vital tools in molecular biology. They act as molecular scissors, cutting DNA at specific recognition sites. HindIII, in particular, recognizes a specific 6-base pair sequence and typically creates a staggered cut, producing sticky ends. These sticky ends are crucial for the ligation process that allows the insertion of new genetic material into plasmids.

When performing plasmid ligation experiments, the choice of restriction enzyme is essential. An enzyme like HindIII allows precise cutting, ensuring that the gene of interest, such as the kanamycin resistance gene in this case, can be easily inserted or removed. This specificity facilitates the study of how genes can affect bacterial cells, such as E. coli, and enables the selection of successfully transformed cells using antibiotic resistance markers.
Kanamycin Resistance Gene
Antibiotic resistance genes like the kanamycin resistance gene are often used as markers in genetic engineering. These genes provide bacteria with resistance to certain antibiotics, which can be used to select for those cells that have successfully incorporated a plasmid containing the gene.

In the context of the experiment, the kanamycin resistance gene is used to ensure that only the bacteria which have taken up the plasmid with the gene survive on media containing kanamycin. This creates a strong selection pressure, making it easier to identify the E. coli that have undergone successful transformation. This selection process is an important tool in molecular biology for verifying the presence of the plasmid within the bacterial colonies.
E. coli Transformation
Transformation is a genetic engineering technique used to introduce foreign DNA into a bacterial cell. E. coli transformation involves taking up a plasmid, which is a small, circular DNA molecule, and incorporating it into the bacteria's own genetic material.

There are several methods to make E. coli cells competent to accept foreign DNA, such as chemically treating the cells or using electroporation. Once the plasmid enters the E. coli cell, it can replicate independently and express the genes it carries. In the experiment mentioned, the E. coli K 12 strain is used, which is a common laboratory strain known for its efficiency in transformation. After the cells take up the plasmid containing the kanamycin resistance gene, they're exposed to kanamycin to select for successful transformations.
DNA Gel Electrophoresis
DNA gel electrophoresis is a technique used to separate DNA fragments based on size. During an electrophoresis experiment, DNA samples are loaded into a gel, and an electric current is applied. The DNA fragments migrate through the gel towards the positive electrode since DNA is negatively charged.

The rate at which DNA moves through the gel is inversely proportional to its size: smaller fragments travel faster, while larger fragments move slower. This process allows researchers to visualize the size and purity of DNA samples, making it a standard method for analyzing DNA after a restriction digest and ligation. In this experiment, DNA gel electrophoresis is used to differentiate between successfully ligated plasmid DNA and larger, possibly concatenated DNA that forms the slow-migrating bands observed in some colonies.

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Most popular questions from this chapter

Most of the techniques described in this chapter (blotting, cloning. \(\mathrm{PCR},\) etc.) are dependent on hybridization (annealing) between different populations of nucleic acids. Length of the strands, temperature, and percentage of GC nucleotides weigh considerably on hybridization. Two other components commonly used in hybridization protocols are monovalent ions and formamide. A formula that takes monovalent \(\mathrm{Na}^{+}\) ions \(\left(\mathrm{M}\left|\mathrm{Na}^{+}\right|\right)\) and formamide concentrations into consideration to compute a \(T_{m}\) (temperature of melting is as follows: $$T_{m}=81.5+16.6\left(\log \mathrm{M}\left[\mathrm{Na}^{+}\right]\right)+0.41(96 \mathrm{GC})-0.72(\% \text { formamide })$$ (a) For the following concentrations of Na' and formamide, calculate the \(T_{m \cdot}\) Assume \(45 \%\) GC content. (b) Given that formamide competes for hydrogen bond locations on nucleic acid bases and monovalent cations are attracted to the negative charges on nucleic acids, explain why the \(T_{m}\) varies as described in part (a).

A widely used method for calculating the annealing temperature for a primer used in PCR is 5 degrees below the melting temperature, \(T_{m}\left(^{\circ} \mathrm{C}\right),\) which is computed by the equation \(81.5+0.41 \times(\% \mathrm{GC})-(675 / N),\) where \(96 \mathrm{GC}\) is the percentage of GC nucleotides in the oligonucleotide and \(N\) is the length of the oligonucleotide. Notice from the formula that both the GC content and the length of the oligonucleotide are variables. Assuming you have the following oligonucleotide as a primer, $$5'-TTGAAAATATTTCCCATTGCC-3'$$ compute the annealing temperature for PCR. What is the relationship between \(T_{m}\left(^{\circ} \mathrm{C}\right)\) and \(\% \mathrm{GC} ?\) Why? (Note: In reality, this computation provides only a starting point for empirical determination of the most useful annealing temperature.)

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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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