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How was Restriction Site of EcoRI sequenced?

How was Restriction Site of EcoRI sequenced?


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The sequence of restriction site of EcoRI - GAATTC was identified in the early 1970s, before Sanger Sequencing was invented.(1977)

How was the restriction site of EcoRI sequenced ?


The first determination of a recognition site for a restriction endonulease was reported in:

Kelly & Smith (1970) A restriction enzyme from Hemophilus influenzae II. Base sequence of the recognition site. J Mol. Biol. 51: 393-409

The enzyme was then called endonuclease R, but is now known as HindII (or HincII).

The method used was to cut DNA with the enzyme and then remove the exposed terminal phosphates with alkaline phosphatase. These ends could then be labelled with 32P using polynucleotide kinase and γ-labelled ATP.

The end-labelled DNA was then treated with nucleases in three different ways to release mononucleotides, dinucleotides, and a mixture enriched for trinucleotides. In each case the labelled products were identified by their behaviour in 2D thin-layer chromatography, and were further characterised for their total base composition after digestion to mononucleotides where appropriate.

The deduced site was/is

5'----GTY RAC----3' 3'----CAR YTG----5'

Thus the labelled mononucleotides were found to be A and G (purine=R), the dinucleotides were found to be GA and AA and so on.

The EcoRI sequence was determined by essentially the same methodology and was reported by

Hedgpeth et al (1972) DNA nucleotide sequence restricted by the RI endonuclease. Proc. Natl. Acad. Sci USA 69:3448 - 3452

Addendum:

Although Sanger sequencing (the plus-minus method) may have been developed through the 1970s, it really didn't catch on until the early 80s. So for example the first complete plasmid sequence to be determined was pBR322 in 1979:

Sutcliffe (1979) Complete nucleotide sequence of the Escherichia coli plasmid pBR322. CSH Symp. Quant. Biol. 43: 77 - 90

"I have determined the 4362-nucleotide-pair sequence of the plasmid cloning vector pBR322 using the DNA-sequencing technique of Maxam and Gilbert (1977). The DNA structure has several interesting features that lead to testable predictions."

The standard sequencing method in use at that time was chemical sequencing using the Maxam-Gilbert methodology. When I started a postdoc in the USA in 1979 M-G sequencing was the standard method. Sanger sequencing really only took off following the development of the M13 mp vectors (and later the pUC plasmids) by J Messing in the early 80s. And what a relief it was: M-G sequencing was extremely laborious, as anyone who ever did it will testify.


EcoRI

EcoRI (pronounced "eco R one") is a restriction endonuclease enzyme isolated from species E. coli. It is a restriction enzyme that cleaves DNA double helices into fragments at specific sites, and is also a part of the restriction modification system. The Eco part of the enzyme's name originates from the species from which it was isolated - "E" denotes generic name which is "Escherichia" and "co" denotes species name, "coli" - while the R represents the particular strain, in this case RY13, and the I denotes that it was the first enzyme isolated from this strain.

In molecular biology it is used as a restriction enzyme. EcoRI creates 4 nucleotide sticky ends with 5' end overhangs of AATT. The nucleic acid recognition sequence where the enzyme cuts is G↓AATTC, which has a palindromic, complementary sequence of CTTAA↓G. Other restriction enzymes, depending on their cut sites, can also leave 3' overhangs or blunt ends with no overhangs.


Restriction Enzymes

DNA can be cut by restriction endonucleases ( RE ). Endonucleases are enzymes that can hydrolyze the nucleic acid polymer by breaking the phosphodiester bond between the phosphate and the pentose on the nucleic acid backbone. This is a very strong covalent bond while the weaker hydrogen bonds maintain their interactions and double strandedness.

As the name implies, restriction endonucleases (or restriction enzymes) are “restricted” in their ability to cut or digest DNA. The restriction that is useful to biologists is usually palindromic DNA sequences. Palindromic sequences are the same sequence forwards and backwards. Some examples of palindromes: RACE CAR, CIVIC, A MAN A PLAN A CANAL PANAMA. With respect to DNA, there are 2 strands that run antiparallelel to each other. Therefore, the reverse complement of one strand is identical to the other. Molecular biologists also tend to use these special molecular scissors that recognize palindromes of 6 or 8. By using 6-cutters or 8-cutters, the sequences occur throughout large stretches rarely, but often enough to be of utility.

EcoRI generates sticky of cohesive ends SmaI generates blunt ends

Restriction enzymes hydrolyze covalent phosphodiester bonds of the DNA to leave either “sticky/cohesive” ends or “blunt” ends. This distinction in cutting is important because an EcoRI sticky end can be used to match up a piece of DNA cut with the same enzyme in order to glue or ligate them back together. While endonucleases cut DNA, ligases join them back together. DNA digested with EcoRI can be ligated back together with another piece of DNA digested with EcoRI, but not to a piece digested with SmaI. Another blunt cutter is EcoRV with a recognition sequence of GAT | ATC.


Application of HindIII and EcoRI Restriction Endonucleases in Identifying and Diagnosing Cystic Fibrosis caused by the CFTR ∆F508 Mutation

Cystic fibrosis (CF) is caused by a mutation on the CFTR protein preventing transport of salts across epithelial cell surfaces leading to mucus hyperproduction and eventually death. The purpose of this experiment was to determine if a 3-year old patient has cystic fibrosis. The hypothesis stated that Jeff would have cystic fibrosis caused by the CFTR ∆F508 mutation. It was predicted that Jeff’s DNA and the positive/negative controls would be cut by EcoRI two times producing three bands with the sizes 2,150 bp, 2,150 bp, and 4,700 bp and that Jeff’s DNA and the positive control would be cut by HindIII once producing two bands with the sizes 7,200 bp and 1,800 bp. The experiment used EcoRI and HindIII restriction endonucleases in RFLP analysis and visualized the results with electrophoresis on an agarose gel, the molecular size of these DNA fragments was calculated from an equation produced from a standard curve graph. The DNA of a patient with the CFTR ∆F508 mutation was included as the positive control and the DNA of a patient without the CFTR ∆F508 mutation was included as the negative control. The results of the experiment showed that all of the DNA samples cut by EcoRI produced the similar-sized DNA fragments, and Jeff’s DNA and the positive control were cut nearly identically by HindIII, while the negative control was cut differently. These results led to the acceptance of the hypothesis, meaning that Jeff was diagnosed with cystic fibrosis caused by the CFTR ∆F508 mutation.

Introduction

In healthy individuals, the body responds to respiratory infections by increasing mucus levels in the lungs and respiratory tract. However, this mucus can also trap bacteria and foreign material so the cystic fibrosis transmembrane conductance regulator (CFTR) protein transports salts including chlorine ions, bicarbonate ions, and anions across epithelial lung cells to hydrate the lungs and clear away excess mucus (Gentzsch, 2018 Southern et al. 2018).

However, there are over 2000 mutations of the CFTR gene that can result in the deadly disease Cystic Fibrosis (Southern et al. 2018). These CFTR mutations are grouped into five different classes based on how they affect the CFTR protein: Class I mutations produce a dysfunctional CFTR protein by adding an early stop codon to the genetic sequence, Class II mutations produce an abnormal CFTR protein most of which is degraded by the cell before it reaches the lung lining, Class III mutations produce CFTR proteins that cannot transport ions across epithelial lung cells, Class IV mutations produce impaired CFTR proteins that have a difficult time transporting ions across epithelial lung cells, Class V mutations produce a functional CFTR protein but in lower numbers than usual so this reduces its efficiency (Southern et al. 2018).

The most common of the CFTR mutations, responsible for 90% of Cystic Fibrosis cases (Cooney, 2018) is a Class II defect called the ∆F508 mutation, this specific mutation occurs when phenylalanine is deleted at position 508 of the CFTR gene (Suaud et al. 2011). This is a very common and lethal autosomal recessive disease that affects 1 in 2000 North Americans or 70,000 individuals globally (Cutting, 2015 Southern et al. 2018). This disease is lethal due to a dysfunctional CFTR protein which means that mucus builds up in the lungs and pancreatic ducts which traps bacteria, weakens the immune system, damages organs, causes inflammation, and often leads to diabetes and/or malnutrition, usually, patients die from respiratory failure (Cutting, 2015 Southern et al. 2018). Fortunately, there are several treatments to mitigate the effects of this disease and extend the life of affected individuals, and since cystic fibrosis is only caused by mutations on the CFTR gene it is relatively easy to test and diagnose patients (Cutting, 2015).

These single-gene autosomal recessive disorders (Cutting, 2015) create a type of genetic variation of the CFTR gene within a population called polymorphism (Pare, 2012). One of the most common ways for diagnosing single-mutation polymorphisms is with a technique called restriction fragment length polymorphism (RFLP) analysis, which uses restriction endonucleases to cut DNA sequences into fragments at specific sites to aid researchers in identifying genetic variations (Loenen et al. 2014 Sapienza, 2012). Restriction endonucleases are enzymes that are naturally produced in several species of prokaryotes but have many applications in laboratory genetic experiments (Pingoud et al. 2014). Restriction endonucleases or REases are grouped into four categories (Type I, Type II, Type III, and Type IV) the most commonly used in genetic testing are Type II REases which function by cleaving the phosphate bands on or near the recognition sequence in the DNA, this process produces consistent DNA fragments (Pingoud et al. 2014 Sapienza, 2012).

Two of the most understood of the Type II REases include EcoRI and HindIII: EcoRI which was discovered from Escherichia coli and HindIII was discovered from Haemophilus influenzae rd (Pingoud et al. 2014). Specifically, EcoRI and HindIII locate their specific staggered nucleotide recognition sequences and cleave the phosphate bonds between the nucleotides: EcoRI recognizes GAATTC and cleaves the phosphate bond between the G and A, additionally, it will recognize and cut any sequences that differ by one base-pair, HindIII recognizes AAGCTT and cleaves the phosphate bond between the two A’s (Loenen et al. 2014 Sapienza, 2012). Additionally, since these two REases produces symmetrical staggered cuts, the DNA fragments can anneal to their complementary strand this has been exploited for genetic cloning applications (Loenen et al. 2014). When EcoRI is added to the blood or saliva DNA sample of a patient either with or without the CF∆F508 mutation, the EcoRI cleaves the CFTR gene twice to produce three fragments that are 2,150 bp, 2,150 bp, and 4,700 bp. However, HindIII produces different results when added to the blood or salvia DNA sample of patients with and without the CF∆F508 mutation. In patients without the CF∆F508 mutation, HindIII cleaves the CFTR gene twice to produce three fragments that are 1,500 bp, 5,700 bp, and 1,800 bp. On the other hand, in patients with the CF∆F508 mutation, HindIII does not identify the recognition sequence at 1,500 bp because of the phenylalanine deletion at position 508 messes up the rest of the sequence, so instead HindIII cleaves the CFTR gene only to produce two fragments that are 7,200 bp and 1,800 bp. The reason that the HindIII produces different results in patients with and without the mutation, while EcoRI produces identical results is that HindIII is more specific and can only identify exact recognition sequences whereas EcoRI is slightly less specific and can identify slightly varied sequences. The results of RFLP analysis can be analyzed with electrophoresis on an agarose gel, a fluorescent dye is added to the DNA samples so that it fluoresces under a UV transilluminator to reveal the distance that the fragments have traveled. The distance that the samples travel is proportional to their length, therefore, the distances that marker DNA fragments travel can be used to create an equation from a standard curve that will use the distances that the other DNA samples traveled to determine their molecular size.

This experiment is going to be testing an adopted 3-year old named Jeff for Cystic Fibrosis caused by the CFTR ∆F508 mutation. To determine if Jeff has cystic fibrosis his DNA will undergo RFLP analysis with the restriction endonucleases EcoRI and HindIII. A positive control (patient with the ∆F508 mutation) and negative control (patient without the ∆F508 mutation) will also undergo RFLP analysis with EcoRI and HindIII so that Jeff’s results can be compared to something. The results of the RFLP analysis will be visualized by electrophoresis on an agarose gel and then the equation produced from a standard curve will be used to calculate the molecular sizes of the DNA fragments after they are cut by the REases. Determining if Jeff has cystic fibrosis and what type of mutation his cystic fibrosis is caused by is important so he can receive the best treatment so that he can live a pain-free and extended life.

It is hypothesized that Jeff has cystic fibrosis because he displays several of the symptoms associated with cystic fibrosis including wheezing, crackling, persistent cough, greasy stools, and a runny nose. Furthermore, because cystic fibrosis caused by a CFTR ∆F508 mutation is an autosomal recessive disorder it is possible that both his parents were carriers of the mutation and that he inherited two mutated alleles, this explains why his parents would not have a medical history of cystic fibrosis.

If this hypothesis is correct and Jeff does have cystic fibrosis caused by the CFTR ∆F508 mutation, then it is predicted that when his DNA will be cut by EcoRI two times to produce three bands that have molecular weights of 2,150 bp, 2,150 bp, and 4,700 bp and that his DNA will be cut by HindIII once to produce two bands that have molecular weights of 7,200 bp and 1,800 bp. Also, the controls (patients with and without the ∆F508 mutation) will also be cut by EcoRI two times to produce three bands that have molecular weights of 2,150 bp, 2,150 bp, and 4,700 bp. Therefore, Jeff’s DNA and the patients with and without the ∆F508 mutation will have bands with identical molecular weights when cut by EcoRI. Additionally, the positive control (patient with the ∆F508 mutation) will also be cut by HindIII once to produce two bands that have molecular weights of 7,200 bp and 1,800 bp. Whereas, the negative control (patient without the ∆F508 mutation) will be cut twice to produce three bands that have molecular weights of 5,700 bp, 1,800 bp, and 1,500 bp. This means that the bands that appear when Jeff’s DNA sample is cut with HindIII should be identical in weight to the patient with the ∆F508 mutation whose DNA was also cut with HindIII, and both of these samples should be different than when HindIII is used with the patient who does not have the ∆F508 mutation.

Materials

To begin, six DNA samples were obtained: two samples from the patient being tested for cystic fibrosis (Jeff), two samples from an individual with the ∆F508 mutation, and two samples from an individual without the ∆F508 mutation. The samples from the individual with the ∆F508 mutation had previously been cut with EcoRI and HindIII respectively, nothing else was added to these samples until the loading dye was added. Reaction buffer was combined with the other samples, and then EcoRI was added to one DNA sample of the individual without the ∆F508 mutation and to one of Jeff’s DNA sample. Next, HindIII was added to one DNA sample of the individual without the ∆F508 mutation and to one of Jeff’s DNA sample. These four samples were incubated at 37 degrees Celsius for thirty minutes. A solution of 0.8% agarose gel was prepared with agarose, 1x TAE buffer, and 10,000x Sybr Safe DNA gel stain. This solution was poured into a gel tray locked into a casting rack, a comb was added and then the gel was placed in a refrigerator for thirty minutes to solidify. The comb was removed and then the gel tray with solid gel was lowed into the electrophoresis chamber and submerged in 1x TAE buffer. A loading dye containing Ficol was added to each of the six DNA samples. A marker was loaded into the first well/lane to display bands of the following molecular sizes: 12,000 bp, 7,000 bp, 3,000 bp, 2,500 bp, 2,000 bp, 1,800 bp, and 1,500 bp. Then the six DNA samples that had been cut with either EcoRI or HindIII were loaded into separate wells. The cover was placed on the electrophoresis chamber so that the charge would run from the negative to the positive end, the apparatus was allowed to run for 45 minutes at 120V until the due was half-way through the gel. The gel was removed from the apparatus and viewed on the UV transilluminator next to a ruler so that the distance that each dye traveled could be measured. A standard curve graph was created with the data from the measurements of the marker. The equation that this graph generated was used to approximate the molecular size of the bands from the six DNA samples. (DeCicco-Skinner, 2019).

Results

Figure 1: Displaying the gel on top of the UV transilluminator next to a metric ruler. Lane 1 contained the marker and produced 7 visible bands, lane 2 contained -E (DNA of the patient without the ∆F508 mutation, cut with EcoRI), lane 3 contained +E (DNA of the patient with the ∆F508 mutation, cut with EcoRI), lane 4 contained JE (Jeff’s DNA cut with EcoRI), lane 5 contained -H (DNA of the patient without the ∆F508 mutation, cut with HindIII), lane 6 contained +H (DNA of the patient with the ∆F508 mutation, cut with HindIII), lane 7 contained JH (Jeff’s DNA cut with HindIII). Lanes 2, 3, and 4 each produced two bands that were nearly identical across the three wells. Lane 5 produced four bands. Lanes 6 and 7 produced two bands that were nearly identical.

The distance the bands traveled on the gel was measured from the distance between the well and the bottom of the band. The marker from lane 1 contained bands of several molecular weights: the 12,000 bp band traveled 21.5 mm, the 7,000 bp band traveled 26.5 mm, the 3,000 bp band traveled 29.2 mm, the 2,500 bp band traveled 34.4 mm, the 2,000 bp band traveled 37.9 mm, the 1,800 bp band traveled 40.5 mm, and the 1,500 bp band traveled 44.1 mm (Figure 1). Lane 2 was loaded with the DNA of the patient without the ∆F508 mutation that was cut with EcoRI, lane 2 displayed two bands: one traveled 28.1 mm and the other traveled 35.2 mm (Figure 1). Lane 3 was loaded with the DNA of the patient with the ∆F508 mutation that was cut with EcoRI, lane 3 had two bands as well: one traveled 27.8 mm and the other 36.1 mm (Figure 1). Lane 4 was loaded with Jeff’s DNA and had been cut with EcoRI, lane 4 also had two bands in nearly the same location as the previous two wells: one band traveled 28.0 mm and the other traveled 35.9 mm (Figure 1). Lane 5 was loaded with the DNA of the patient without the ∆F508 mutation that was cut with HindIII, lane 5 had four bands traveling 27.8 mm, 35.8 mm, 42.3 mm, and 44.1 mm (Figure 1). Lane 6 was loaded with the DNA of the patient with the ∆F508 mutation that was cut with HindIII, lane 6 had two bands traveling 26.1 mm and 43.5 mm (Figure 1). Lane 7 was loaded with Jeff’s DNA and had been cut with HindIII, lane 7 had two bands that were very similar to lane 6, traveling 26.0 mm and 43.5 mm (Figure 1).

Figure 2: Displaying the Standard Curve for CFTR ∆F508 mutation restriction digestion. The log of the molecular weight of each marker band was plotted on the y-axis, and the distance in millimeters that each band migrated from the well that the marker was loaded into was plotted on the x-axis. From this data a linear trendline was added, the equation of the line of best fit was calculated and displayed (y=-0.0391x + 4.8133), and the coefficient of determination was calculated and displayed (R^2=0.8987).

The measurements that were collected from Figure 1 were used to create the standard curve graph in Figure 2. Adding a linear trendline produced an equation of the line of best-fit (y=-0.0391x + 4.8133) and yielded a 0.8987 coefficient of determination. The distance that each band traveled for the six samples was plugged into this equation and then the antilog was taken to calculate the approximate band size.

Table 1: CFTR ∆F508 mutation restriction digestion by EcoRI and HindIII results for Jeff, patient with the CFTR ∆F508 mutation, and patient without the CFTR ∆F508 mutation.
Table 1: Displaying the expected vs observed molecular weight for the marker and six DNA samples cut with EcoRI or HindIII. The molecular weight for each sample was calculated from the standard curve equation (y=-0.0391x + 4.8133). Lanes 2, 3, and 4 had bands of very similar sizes and the values were close to what was expected. Lane 5 had more bands than were expected. Lane 6 and 7 had bands that were very similar and close to what was expected.

From the equation in Figure 2 the observed band sizes were calculated for the six DNA samples and recorded in Table 1. The observed band sizes were also calculated for the marker so that the amount of error could be quantified. Table 1 shows that the calculated molecular weight is different than the expected molecular weight, this difference is most notable when the band has more base pairs. This means that the observed band size for the DNA samples should not exactly match the expected band size and that the value will be most accurate for bands with fewer base pairs. For the three DNA samples cut with EcoRI (the patients with (+E) and without (-E) the CFTR ΔF508 mutation, and Jeff (JE)) it was expected that the EcoRI would cut the DNA strand twice to produce three bands: two bands that were 2,150 bp and one band that was 4,700 bp. However, observation of these samples revealed only two bands. Patient -E had calculated bands of 2,500 bp and 5,183 bp patient +E had calculated bands of 2,522 bp and 5,325 bp patient JE had calculated bands of 2,568 bp and 5,229 bp. These values are very close and are essentially identical. The patient without the CFTR ΔF508 mutation (-H) whose DNA was cut with HindIII was expected to be cut twice to produce three bands: 1,500 bp, 5,700 bp, and 1,800 bp. However, four bands were observed indicating the DNA was cut in four places. This produced bands that were calculated to be 1,227 bp, 1,443 bp, 2,591 bp, and 5,325 bp. The patient with the CFTR ΔF508 mutation (+H) and Jeff (JH) whose DNA was cut with HindIII was expected to be cut once to produce two bands that were 7,200 bp and 1,800 bp. In actuality, these two samples were cut once, in the +H patient the band sizes were 6,206 bp and 1,296 bp, whereas in the JH patient the band sizes were 6,262 bp and 1,296 bp. These values are so similar that the band sizes could be considered identical.

Discussion

With over 2000 polymorphisms of cystic fibrosis mutations, there is certainly a lot that can go wrong for the CFTR gene. In the example of the CFTR ∆F508 mutation, a single amino acid (phenylalanine) is deleted at position 508 of the gene (Suaud et al. 2011). The deletion of phenylalanine at this position is so severe that the CFTR protein becomes dysfunctional, resulting in hyperproduction of mucus across epithelial surfaces (Kreda et al. 2012). The CFTR protein has been identified through several experiments as a cyclic adenosine monophosphate-dependent phosphorylation (cAMP) activated anion channel that transports salts (chloride ions and bicarbonate ions) and other anions across epithelial cells (Gentzsch, 2018). Healthy, fully functional CFTR proteins transport these salts across the plasma membrane of epithelial cells that line the lungs and other organs to clear away excess mucus by hydrating the surface (Gentzch, 2018 Suaud et al. 2011). In the case of the CFTR ∆F508 mutation, an abnormal CFTR protein which cannot fold properly is produced the protein cannot fold because the section of CFTR that interacts with ATP to bind nucleotides called the nucleotide-binding domain (NBDI) and the fourth cytosolic loop within another section of CFTR that anchor other protein into the plasma membrane (MSD 2) form hydrogen bonds with the arginine amino acid at the 1070 codon (Cutting, 2015). The cell detects the misfolded CFTR proteins and degrades most of them within the endoplasmic reticulum (Suaud et al. 2011). This means that very little or none of the CFTR protein reaches the surface of the epithelial cells therefore, chloride and bicarbonate ions cannot be transported across the plasma membrane (Suaud et al. 2011). Since these salt ions cannot be transported across the plasma membrane then the salts are in higher concentration on the basal side of the epithelial cells, this draws water away from the apical surface and leads to dehydration of the lung, gastrointestinal, and pancreatic surfaces (Gentzsch, 2018). When there is no water on the apical surfaces of these organs excess mucus cannot be cleared away and then dense sticky mucus creates obstructions and leads to respiratory illnesses and inflammation that eventually cause death in patients (Gentzsch, 2018).

This experiment tested 3-year old Jeff for Cystic Fibrosis caused by the CFTR ∆F508 mutation using RFLP analysis with EcoRI and HindIII restriction endonucleases. His results were compared to a positive control (patient with the ∆F508 mutation) and a negative control (patient without the ∆F508 mutation) who also were included in the RFLP analysis with EcoRI and HindIII so that Jeff’s results can be compared to something. The results of the RFLP analysis were visualized with electrophoresis on agarose gel (Figure 1) and then an equation was produced from a standard curve (Figure 2) which was used to calculate the molecular sizes (Table 1) of the DNA fragments that the restriction endonucleases produced. This early diagnosis is critical to ensuring that Jeff receives the care he needs so he can live a painless and long life. Furthermore, RFLP analysis is a much more reliable method to test for autosomal recessive disorders than direct-to-consumers genetic tests such as 23andMeTM because there is less room for misunderstandings. Companies like 23andMeTM measure genetic variation by comparing the customer's genetic information to their database of mutations, this database can detect 715,000 single nucleotide mutations (Lu et al. 2017). This means that the database would detect the three-nucleotide deletion that occurs when phenylalanine is deleted at position 508 of the CFTR gene. RFLP is different than this because it does not compare genetic sequences, rather it uses restriction endonucleases to cut DNA into fragments at recognition sequences, in this case, cystic fibrosis could be diagnosed if the DNA of a patient was cut at the same locations as a positive control. One advantage to genetic tests like those provided by 23andMeTM is that they can also tell if a patient is a carrier of a recessive mutation, this is something that RFLP analysis cannot do (Lu et al. 2017). However, one of the largest concerns with direct-to-consumer genetic testing is that most people do not know how to interpret the results and may take drastic behaviors as a result of not receiving genetic counseling (Pare, 2012), this is another reason why it is important to test for cystic fibrosis with RFLP instead of using 23andMeTM.

It was hypothesized that Jeff has cystic fibrosis caused by the CFTR ∆F508 mutation because he displays several of the symptoms associated with the disease. This would be possible if both of his parents were carriers of the CFTR ∆F508 mutation because carriers do not have symptoms since the disease only occurs when both alleles on this location of the CFTR gene are mutated. It was predicted that when Jeff’s DNA would be cut by EcoRI two times producing three bands with the sizes 2,150 bp, 2,150 bp, and 4,700 bp and that his DNA would be cut by HindIII once producing two bands with the sizes 7,200 bp and 1,800 bp. Also, the EcoRI samples (patients with and without the ∆F508 mutation) would be cut by EcoRI two times in an identical manner to Jeff’s DNA producing three bands with the sizes 2,150 bp, 2,150 bp, and 4,700 bp. Meaning that Jeff’s DNA and the DNA of the patients with and without the ∆F508 mutation should have bands identical in size when cut by EcoRI. Additionally, the positive control (patient with the ∆F508 mutation) would also be cut by HindIII once, similar to Jeff’s DNA producing two bands with sizes 7,200 bp and 1,800 bp. This is in contrast to the negative control (patient without the ∆F508 mutation) whose DNA was predicted to be cut twice, producing three bands with sizes 5,700 bp, 1,800 bp, and 1,500 bp. If the hypothesis is correct then the bands that appear when Jeff’s DNA sample is cut with HindIII would be identical in weight to the patient with the ∆F508 mutation whose DNA was also cut with HindIII, and both of these samples should be different than when HindIII was used with the patient who does not have the ∆F508 mutation.

The purpose of including the positive control was so that the HindIII restriction digestion results of an individual with ∆F508 mutation can be visualized and then compared to how Jeff’s DNA was cut, if they were cut the same way then it means he also has the same CFTR polymorphism. If his DNA is cut differently than the positive control, then it would mean he does not have the ∆F508 mutation. The purpose of including the negative control was so that the HindIII restriction digestion results of an individual without ∆F508 mutation can be visualized and then compared to how Jeff’s DNA was cut. If the negative control was cut differently than both the positive control and Jeff’s DNA then it would mean that he does not have the ∆F508 mutation. However, if Jeff’s results did not match either the positive or negative it would mean that the results of the experiment are invalid. Also, if the positive and negative controls are identical when cut by HindIII then the experimental results would be invalid. Furthermore, even though the results of the RFLP analysis will be the same for Jeff and the controls when cutting with EcoRI, the EcoRI was still included because if this did not produce identical results for all of the samples then the results of the experiment would have to be called into question.

The marker that was loaded into lane 1 successfully moved across the gel (Figure 1) and produced bands at 12,000 bp, 7,000 bp, 3,000 bp, 2,500 bp, 2,000 bp, 1,800 bp, and 1,500 bp as was expected (Table 1). The distance that each band migrated was measured and this data was plotted against the log of each molecular weight to produce a standard curve that yielded a 0.8987 coefficient of determination and line of best fit with the equation y=-0.0391x +4.8133 (Figure 2). The distance that each band traveled for the six samples was plugged into this equation and then the antilog was taken to calculate the approximate band size. This equation was used to calculate the size of the markers as well to determine how much mathematical error is present in this model. Table 1 shows that the calculated molecular weight is slightly different than the actual molecular weight, this effect is amplified when the DNA has more base pairs. Meaning that the results of the other DNA samples could also have some variation in the expected vs observed molecular weight of the bands and that even with such variation the results would still be valid. This equation was also used to calculate the molecular sizes of the bands shown on the gel in Figure 1. If the hypothesis is correct then Jeff’s DNA, the patient with the ∆F508 mutation, and the patient without the ∆F508 mutation would all be cut twice by EcoRI to produce bands with molecular weights of 2,150 bp, 2,150 bp, and 4,700 bp. Since two of these bands are the same molecular weight then only two bands would appear on the gel: 2,150 bp and 4,700 bp. Observation of these DNA samples showed that they all have two bands (Figure 1 – lanes 2,3,4) and calculations from the standard curve equation revealed that the molecular weight of these two bands in the three samples were very similar: in JE the sizes of the DNA fragments were calculated as 2,568 bp and 5,229 bp in -E the sizes of the bands were calculated as 2,500 bp and 5,183 bp in +E the sizes of the bands were calculated as 2,522 bp and 5,325 bp (Table 1). Since the EcoRI cut all of the samples in approximately the same place then it can be concluded that the RFLP analysis worked correctly and that the other results of the experiment should have also worked correctly. This also means that the CFTR gene was actually present in all of the DNA samples because if any of the samples lacked the CFTR gene the samples would not have produced identical DNA fragments. However, these results alone cannot provide a diagnosis for cystic fibrosis because the EcoRI enzyme cuts the positive and negative control the same way. If EcoRI and the positive control had been the only things used to diagnose Jeff, then the hypothesis would have been accepted because the CFTR gene is cut into the same DNA fragments for the positive control and Jeff. Paradoxically, if EcoRI and the negative control had been the only things used to diagnose Jeff then the hypothesis would have been rejected because the CFTR gene is cut into the same DNA fragments for the negative control and Jeff. Obviously, this is problematic because the hypothesis cannot both be accepted and rejected, therefore, to accurately diagnosis Jeff with cystic fibrosis a restriction endonuclease that cuts the positive and negative control differently must be used, this is why HindIII was used as well.

If the hypothesis was correct then Jeff’s DNA and the (positive control) patient with the ∆F508 mutation would be cut once by HindIII to produce bands with molecular weights of 7,200 bp and 1,800 bp. Observation of these DNA samples showed that they both were cut once and had two bands (Figure 1 – lanes 6 and 7) and calculations from the standard curve equation revealed that the molecular weight of these two bands in both samples was similar: in JH the sizes of the bands were calculated as 6,262 bp and 1,296 bp and in +H the sizes of the bands were calculated as 6,206 bp and 1,296 bp (Table 1). Since the HindIII cut both Jeff’s DNA and the DNA of the patient who has the ∆F508 mutation it is likely that Jeff also has the ∆F508 mutation which causes cystic fibrosis. To confirm that these results are valid both these samples must be compared to the negative control – the patient without the CFTR ∆F508 mutation. Observation revealed that the DNA of the patient (-H) without the ∆F508 mutation was cut three times by HindIII to produce four bands (Figure 1) and these bands were calculated to have the molecular weights of 1,227 bp, 1,443 bp, 2,591 bp, and 5,325 bp (Table 1). Obviously, these results do not match the prediction which stated that this patient’s DNA would be cut twice by HindIII to produce three bands with the molecular weights of 5,700 bp, 1,800 bp, and 1,500 bp. However, the top two bands appear to align with the bands from the samples cut with EcoRI. Meaning that this sample was likely contaminated with some EcoRI. This interferes with the analysis of the results however, some conclusions can still be drawn due to the placement of these bands in relation to the other HindIII samples. First, the results are different enough that it can be concluded that JH and +H were identical whereas -H was different. Second, the -H sample has a band that is below the lowest band in +H and JH (Figure 1– lanes 5, 6, 7). This would make sense if the hypothesis was correct because -H should have its smallest band be 1,500 bp and +H/JH should have their smallest bands at 1,800 BP (Table 1). Third, the only way that Jeff’s DNA sample could be cut by HindIII at 7,200 bp would be if he had the CFTR ∆F508 mutation (Figure 1 and Table 1). Therefore, the hypothesis is accepted, Jeff has cystic fibrosis caused by the CFTR ∆F508 mutation.

If the positive control had been omitted then the diagnosis would be very difficult to make and the hypothesis could not be confirmed, because the calculated molecular size of Jeff’s DNA fragments is different than the expected molecular sizes. Observing that the calculated molecular size of the also differed from the expected size for the positive control’s DNA fragments and that the molecular size of these fragments and Jeff’s were nearly identical when cut by HindIII was the ultimate factor that led to accepting the hypothesis. If the negative control had been omitted, then the hypothesis would have still been accepted because Jeff’s DNA would have still matched the positive control. However, in the alternative scenario where he did not have cystic fibrosis, missing a negative control would mean that it would be impossible to confirm that Jeff did not have cystic fibrosis because his DNA fragments would not have matched anything when cut by HindIII.

This experiment only used the segment of Jeff’s genome that included the CFTR gene. If we had used his entire genome of a patient with or without the mutation then EcoRI and HindIII would have located many more recognition sites (GAATTC and AAGCTT respectively) and cleaved phosphate groups at these sites, this would produce many more DNA fragments. In fact, so many DNA fragments that it would be difficult to tell what you were looking at, Instead of seeing a few bands on Figure 1 instead there might be hundreds of bands. This would make it impossible to identify whether or not Jeff had a mutation in CFTR. The lab protocol would need to be modified if entire genomes were used so that a specific region of DNA could be examined. One way to achieve this is through southern blotting: where the DNA fragments from electrophoresis are transferred to a membrane by upward capillary transfer and then immobilized, allowing for the bands matching the CFTR sequence to be identified with a probe (Brown, 2001).

In conclusion, these findings are significant because it reveals that HindIII is a very useful restriction endonuclease for diagnosing cystic fibrosis and Jeff’s cystic fibrosis diagnosis (caused by the CFTR ∆F508 mutation) means that he can receive personalized-treatment, lumacaftor, for example, is a drug that might benefit Jeff by preventing the degradation of the misfolded CFTR proteins caused by the CFTR ∆F508 mutation, this helps the proteins reach the apical epithelial cell surface so that function can be partially restored (Kreda et al. 2012). This experiment could be improved doubling the number of samples to ensure that the samples have not been contaminated by the incorrect restriction endonuclease and by using a different trendline that produces an equation that calculates molecular sizes with improved accuracy. Future studies can explore the application of other restriction endonucleases such as BamHI, EcoRII, Hand HaeIII to see how they cut the CFTR in the positive and negative controls.

References

Brown, T (2001) Southern Blotting. Curr Protoc Immunol 10(6a).

Cooney, A.L., P.B. McCray Jr, and P.L. Sinn (2018) Cystic Fibrosis Gene Therapy: Looking Back, Looking Forward. Genes (Basel) 9(11).

Cutting, G. R. (2015) Cystic fibrosis genetics: from molecular understanding to clinical application. Nat Rev Genet 16(1): 45-56.


Important Questions for CBSE Class 12 Biology Principles of Biotechnology and Tools of Recombination DNA Technology

1. Biotechnology can be defined as the use of microorganisms, plants or animal cells or their components to produce products and processes useful to humans. According to the European Federation of Biotechnology (EFB), biotechnology is the integration of natural science and organisms, cells, parts thereof and molecular analogues for products and services. The term ‘Biotechnology’ was coined by Karl Ereky in 1919.
2. Principles of biotechnology are based on the concept of the following techniques:
(i) Genetic engineering is the technique to alter the chemistry of genetic material (DNA/RNA), to introduce these into another organisms and thus, change the phenotype of the host organism.
(ii) Adequate maintenance of sterile conditions to support growth of only the desired microbes/eukaryotic cells in large quantities for the manufacture of biotechnological products like antibiotics, vaccines, enzymes, etc.

3. The techniques of genetic engineering include the following:

  • Creation of recombinant DNA by combining desired genes.
  • Gene transfer.
  • Maintenance of DNA in host and gene cloning.
  • The basic steps in genetic engineering can be summarised as:
  • Identification of DNA with desirable genes.
  • Introduction of the identified DNA into the host.
  • Maintenance of introduced DNA in the host and transfer of the DNA to its progeny.

4. Construction of First Artificial Recombinant DNA
(i)It was achieved by linking a gene encoding antibiotic resistance with a native plasmid (an autonomously replicating circular extrachromosomal DNA) of Salmonella typhimurium.
(ii) Stanley Cohen and Herbert Boyer accomplished this in 1972.
(iii) They isolated the antibiotic resistance gene by cutting out a piece of DNA from a plasmid.
(iv) The cutting of DNA at specific locations was carried out by molecular scissors, i.e. restriction enzymes.
(v) The cut piece of DNA was then linked to the plasmid DNA with the enzyme DNA ligase. The plasmid DNA acts as vectors to transfer the piece of DNA attached to it.
(vi) When this DNA is transferred into coli, it could replicate using the new host’s DNA polymerase enzyme and make multiple copies.
(vii) This ability to multiply copies of antibiotic resistance gene in E. coli was called cloning of antibiotic resistance gene in E. coli.

5. Tools of recombinant DNA technology are as follow:
(i) Restriction enzymes (ii) Polymerase enzymes
(iii) Ligases (iv) Vectors
(v) Competent host organism.

6. Restriction enzymes or ‘molecular scissors’ are used for cutting DNA.
(i) Two enzymes from E. coli that were responsible for restricting the growth of bacteriophage were isolated in 1963, one of them added methyl group to DNA and the other cut DNA into segments. The later was called restriction endonuclease.
(ii) The first restriction endonuclease Hind II was isolated by Smith Wilcox and Kelley (1968). They found that it always cut DNA molecules at a particular point by recognising a specific sequence of six base pairs known as recognition sequence.
(iii) Besides Hind II, more than 900 restriction enzymes have been isolated now, from over 230 strains of bacteria, each of which recognise different recognition sequences.
(iv) Naming of Restriction Enzymes
(a) The first letter is derived from the genus name and the next two letters from the species name of the prokaryotic cell from which enzymes are extracted.
(b) The Roman numbers after name show the order in which the enzymes were isolated from the bacterial strain.
For example, Eco RI comes from Escherichia coli RY13 and Eco RII comes from E. coli R 245, etc.
(v) Restriction enzymes belong to a class of enzymes called Nucleases.
Nucleasesare of two types:
Exonucleases They remove nucleotides from the ends.
Endonucleases They cut at specific positions within the DNA.
(a) Each restriction endonuclease recognises a specific palindromic nucleotide sequences in the DNA.
(b) Palindrome in DNA is a. sequence of base pairs that reads same on the two strands when orientation of reading is kept same.
For example, the following sequences reads the same on the two strands in 5′ –> 3′ direction as well as 3′–> 5′ direction.
5′ — GAATTC — 3′
3′ — CTTAAG — 5′
(vi) Mechanism of Action of Restriction Enzymes
(a) Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands.
(b) This leaves single stranded portions at the ends.
(c) There are overhanging stretches called sticky ends on each strands as given in above figure. These are named so, because they form hydrogen bonds with their complementary cut counterparts.
(d) The stickiness of the ends facilitates the action of the enzyme DNA ligase.
(e) Restriction endonucleases are used in genetic engineering to form recombinant molecules of DNA, which are composed of DNA from different sources/genomes.
(f) These sticky ends are complementary to each other when cut by same restriction enzyme, therefore can be joined together (end-to-end) using DNA ligases.

7. Separation and Isolation of DNA Fragments
(i) The cutting of DNA by restriction’endonucleases results in the fragments of DNA.
(ii) The technique, which separates DNA fragments based on their size is called
gel electrophoresis.
(iii) DNA fragments are negatively charged molecules. They can be separated by forcing them to move towards the anode under an electric field through a medium/matrix.
(iv) The most common medium used is agarose, a natural polymer extracted from sea weeds.
(v) The DNA fragments separate (resolve) according to their size through sieving effect provided by the agarose gel. The smaller the fragment size, the farther it moves.
(vi) The separated DNA fragments can be visualised only after staining the DNA with a compound known as ethidium bromide followed by exposure to UV radiation.
(vii) The DNA fragments can be seen as bright orange coloured bands. These separated bands are cut out from the agarose gel and extracted from the gel piece. This is called elution.
(viii) The purified DNA fragments can be used in constructing recombinant DNA by joining them with cloning vectors.

8. Cloning vectors are the DNA molecules that can carry a foreign DNA segment into the host cell.
(i) The vectors used in recombinant DNA technology can be:
(a) Plasmids Autonomously replicating circular extra-chromosomal DNA.
(b) Bacteriophages Viruses infecting bacteria.
(c) Cosmids Hybrid vectors derived from plasmids which contain cos site of X
(ii) Copy number can be defined as the number of copies of vectors present in a cell.
(iii) Bacteriophages have high number per cell, so their copy number is also high in genome.
(iv) Plasmids have only one or two copies per cell.
(v) Copy number can vary from 1-100 or more than 100 copies per cell.
(vi) If an alien piece of DNA is linked with bacteriophage or plasmid DNA, its number can be multiplied equal to the copy number of the plasmid or bacteriophage.
(vii) Features Required to Facilitate Cloning into Vector
(a) Origin of replication (Ori) (b) Selectable marker
(c) Cloning sites (d) Vectors for cloning genes in plants and animals.
(a) Origin of replication (Ori) is a sequence from where replication starts.

  • Any piece of DNA when linked to this sequence can be made to replicate within the host cells.
  • The sequence is also responsible for controlling the copy number of the linked DNA.

(b) Selectable marker helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants.

  • Transformation is a process through which a piece of DNA is introduced in a host bacterium.
  • The genes encoding resistance to antibiotics such as ampicillin, chloramphenicol, tetracycline or kanamycin, etc, are some useful selectable markers for coli.


E.coli cloning vector pBR322 showing restriction sites (Hind III, Eco Rl
Bam HI, Sal I, Pvu II, Pst I, Cla I), ori and antibiotic resistance genes (amp R and tet R ).
rop codes for the proteins involved in the replication of the plasmid.

  • Ligation of alien DNA is carried out at a restriction site present in one of the twoantibiotic resistance genes. Example is ligating a foreign DNA at the Bam HI site of tetracycline resistance gene in the vector pBR322.
    –>>The recombinant plasmids will lose tetracycline resistance due to insertion of foreign DNA. But, it still can be selected out from non-recombinant ones by plating the transformants on ampicillin containing medium.
    –>> The transformants growing on ampicillin containing medium are then transferred on a medium containing tetracycline.
    –>>The recombinants will grow in ampicillin containing medium but not on that containing tetracycline.
    –>>The non-recombinants will grow on the medium containing both the antibiotics.
    In this example, one antibiotic resistance gene helps in selecting the transformants whereas, the other antibiotic resistance gene gets ‘inactivated due to insertion’ of alien DNA and helps in selection of recombinants.
  • Selection of recombinants due to inactivation of antibiotics is a cumbersome procedure, so alternative selectable markers are developed which differentiate recombinants from non-recombinants on the basis of their ability to produce colour in the presence of a chromogenic substrate.

–>> In this method, a recombinant DNA is inserted within the coding sequence of an enzyme J3-galactosidase.
–>> This results into inactivation of the enzyme, P-galactosidase (insertional inactivation).
–>> The bacterial colonies whose plasmids do not have an insert, produce blue colour, but others do not produce any colour, when grown on a chromogenic substrate.
(c) Cloning sites are required to link the alien DNA with the vector.

  • The vector requires very few or single recognition sites for the commonly used restriction enzymes.
  • The presence of more than one recognition sites within the vector will generate several fragments leading to complication in gene cloning.

(d) Vectors for cloning genes in plants and animals are many which are used to clone genes in plants and animals.
In plants, the Tumour inducing (Ti) plasmid of Agrobacterium tumefaciens is used as a cloning vector.
–>>Agrobacterium tumefaciens is a pathogen of several dicot plants.
–>> It delivers a piece of DNA known as T-DNA in the Ti plasmid which transforms normal plant cells into tumour cells to produce chemicals required by pathogens.Retrovirus, adenovirus, papillomavirus are also now used as cloning vectors in animals because of their ability to transform normal cells into cancerous cells.

9. Competent host organism (for transformation with recombinant DNA) is required because DNA being a hydrophilic molecule, cannot pass through cell membranes,
Hence, the bacteria should be made competent to accept the DNA molecules,
(i) Competency is the ability of a cell to take up foreign DNA.
(ii) Methods to make a cell competent are as follow.
(a) Chemical method In this method, the cell is treated with a specific concentration of a divalent cation such as calcium to increase pore size in cell wall.
The cells are then incubated with recombinant DNA on ice, followed by placing
them briefly at 42°C and then putting it back on ice. This is called heat shock treatment. This enables the bacteria to take up the recombinant DNA.
(b) Physical methods In this method, a recombinant DNA is directly injected into the nucleus of an animal cell by microinjection method.
In plants, cells are bombarded with high velocity microparticles of gold or
tungsten coated with DNA called as biolistics or gene gun method.
(c) Disarmed pathogen vectors when allowed to infect the cell, transfer the recombinant DNA into the host.

Previous Years Examination Questions

1 Mark Questions
1. Why is it not possible for an alien DNA to become part of a chromosome anywhere along its length and replicate normally?
[All India 2014]
Ans.The integration of alien DNA is required to become a part of chromosome. As the DNA itself cannot multiply and replicate but rather requires a specific sequence for initiating its replication called origin of replication. Therefore, the alien DNA needs to be joined with the host DNA with the help of enzymes and linked to Ori, so as to be part of chromosome and replicate normally to produce its copies.

2. Mention the type of host cells suitable for the gene guns to introduce an alien DNA. [Delhi 2014]
Ans. The host cells suitable for the gene guns to introduce an alien DNA are plant cells.

3. Write the two components of first artificial recombinant DNA molecule constructed by Cohen and Boyer. [Foreign 2014]
Ans.The two components of first artificial recombinant DNA molecule constructed by Gohen and Boyer are:
(i) antibiotic resistance gene
(ii) plasmid of Salmonella typhimurium

4. Name the host cells in which microinjection technique is used to introduce an alien DNA. [Foreign 2014]
Ans. The microinjection technique to introduce alien DNA is usually carried out in animal cell, i.e. directly into the nucleus.

5. Name the material used as matrix in gel electrophoresis and mention its role. [All India 2014C]
Ans. The material used as matrix in gel electrophoresis is agarose.
This agarose gel acts as a sieve to separate the DNA fragments according to their size.

6. Write any four ways used to introduce a desired DNA segment into a bacterial cell in recombinant technology experiments. [All India 2013]
Ans. Ways to introduce desired DNA into bacterial cell are:
(i) microinjection (ii) disarmed pathogen vectors
(iii) portion by bivalent cation such as calcium
(iv) bidlistic or gene gun

7. State what happens when an alien gene is ligated at Sal I site of pBR322 plasmid. [Delhi 2013c]
Ans. If an alien gene is ligated at Sal I site of tetracycline resistance gene in the vector pBR322, the recombinant plasmid will lose its tetracycline resistance.

8. Mention the uses of cloning vector in biotechnology ? [Delhi 2011]
Ans. Uses of cloning vector in biotechnology.
(i) Helps in linking the foreign/alien DNA with that of host’s DNA.
(ii) Help in the selection erf recombinants from the non-recombinants.

9. Biotechnologists refer to Agrobacterium tumefaciens as natural genetic engineer of plants. Give reasons to support the statement.
[ HOTS All India 2011]
Ans. Agrobacterium tumefacieps is a pathogen of several dicot plants. It is used as a natural genetic engineer because it is able to deliver a piece of its DNA (called T-DNA) to transform normal plant cells into tumour cells and direct the tumour cells to synthesise the chemicals required by the pathogen.

10. Why is it essential to have a selectable marker in a cloning vector? [All India 2011]
Ans.Selectable marker in cloning vector helps in identifying and selecting the recombinants and eliminating the non-recombinants.

11. Why do DNA fragments move towards the anode during gel electrophoresis? [hots Delhi 2011c]
Ans. DNA fragments are negatively charged molecules and hence, moves toward the anode during gel electrophoresis.

12. In the year 1963, two enzymes responsible for restricting the growth of bacteriophage in E. coli were isolated. How did enzymes act to restrict the growth of the bacteriophage? [All India 2011c]
Ans.Two enzymes responsible for restricting the growth of bacteriophage in coli are: Exonucleases Add methyl group to DNA. Endonucleases Cut DNA at specific points.

13. How is the action of exonuclease different from that of endonuclease.
[All India 2010]
Ans.Exonuclease removes nucleotides from theends of DNA, while endonuclease cuts the DNA at specific positions.

14. Mention the role of molecular scissors in recombinant DNA technology. [All India 2009]
Ans. Molecular scissors or restriction enzymes cut DNA at specific site, thus allowing to extract desired gene and like it with DNA of host.

15. Name the technique used for separating DNA fragments in the laboratory. [Delhi 2008]
Ans. Gel electrophoresis is used for separating DNA fragments in the lab.

16. What is the role of ethidium bromide during agarose gel electrophoresis of DNA fragments? [All India 2008C]
Ans. The separated DNA fragments during agarose gel electrophoresis are visualised after staining the DNA with ethidium bromide, in UV light. This staining imparts DNA a bright orange colour.
2 Marks Questions
17. How does a restriction nuclease function? Explain. [All India 2014]
Ans. Restriction nucleases function by inspecting the length of DNA sequence, and then binding to specific recognition sequences and cutting the strands at sugar phosphate backbones.
These nucleases are of two types depending on their mode of action.
(i) Restriction exonucleases cut sequences at terminal ends of DNA.
(ii) Restriction endonucleases cut between the two bases of recognition sequence.

18. Write the role of Ori and restriction site in a cloning vector pBR322.
[Delhi 2014]
or
How do Ori and cloning sites facilitate cloning into a vector?
[All India 2008C]
Ans.Ori is a sequence of DNA from where replication starts. Any piece of DNA that needs to replicate in the host cell has to be linked to it.
Cloning sites refers to the site/sequence of DNA where the alien DNA is linked.

19. Explain with the help of a suitable example the naming of a restriction endonuclease. [Delhi 2014]
Ans. Naming of restriction endonuclease are:
(i) The first letter of the name comes from the genus and next two letters from species of the prokaryotic cell from where enzymes are extracted.
(ii) The Roman numbers following the name show the order in which the enzymes, were isolated from the bacterial strain. For example, Eco Rl is derived from Escherichia coli RY 13, Hind II from Haemophilus influenzae Rd, etc.

20. How are sticky ends formed on a DNA strand? Why are they so called? [Delhi 2014]
Ans.Sticky ends on DNA are formed by action of enzymes restriction endonucleases. These enzymes cut the strand of DNA a little away from the centre of the palindrome sequence between the same two bases on both the strands. This results in single stranded stretches on both the complementary strands at their ends.
These overhanging stretches are called sticky ends as they form hydrogen bonds with the complementary base pair sequences.

21.How is insertional inactivation of an enzyme used as a selectable marker to differentiate recombinants from non-recombinants?
[Foreign 2014]
Ans. The insertional activation of J3-galactosidase enzyme, i.e. by inserting the desired gene in the coding region of enzyme, results in inactivation of (3-galactosidase gene in recombinants. The recombinant on transformed hosts are unable to produce any colour when grown on chromogenic substrate, thus acting as a selectable marker to differentiate recombinants from non-recombinants.

22. Explain palindromic nucleotide sequence with the help of a suitable example. [Foreign 2014]
Ans.The palindromic nucleotide sequence is the sequence of base pairs in DNA that reads the same on both the complementary strands of DNA, with same orientation of reading.
For example
5′-GAATTC-3′
3′-CTTAAG-5′

23. Why are molecular scissors so called? Write their use in biotechnology? [Foreign 2014]
Ans. Molecular scissors are so called because they cut DNA at specific sequences between base pairs.
Since, molecular scissors on restriction enzymes cut DNA at desired sequences and generate sticky ends that facilitate to join with host genome or vector DNA, they play an important role in genetic engineering or biotechnology. It is because with the help of these enzymes we can cut the desired gene and introduce into vectors for expression.

24. Why is making cells competent essential for biotechnology experiments? List any two ways by which this can be achieved.
[Delhi 2014C]
Ans. Since, DNA molecules are hydrophillic, they cannot pass through cell membranes. For recombinant DNA to be integrated into vector or host genome it is necessary for the DNA to be inserted in the cell. Therefore, making the host cells competent is necessary in biotechnology experiments.
The two ways by which cells can be made competent to take up DNA are:
(i) Chemical action By increasing concentration of divalent cation, calcium, thereby increasing the efficiency of DNA entering through pores in cell, wall.
(ii) Heat shock treatment Incubating the cells with recombinant DNA on ice, followed by brief treatment of heat at 42 °C and again putting them back on ice.

25. How is an exonuclease functionally different from an endonuclease? Give an example of any two endonucleases other than Sal I. [Delhi 2013C]
Ans. Exonucleases are the enzymes which cleaves base pairs of DNA at their terminal ends and act on single strand of DNA or gaps in double stranded DNA. While, endonucleases cleaves DNA at any point except the terminal ends and can make cut on one strand or on both strands of double stranded DNA, e.g. Eco Rl and Hind II.

26. Explain the work carried out by Cohen and Boyer that contributed immensely in biotechnology. [Delhi 2012]
Ans. (i) Stanley Cohen and Herbert Boyer constructed the first artificial recombinant DNA (rDNA) molecule.
(ii) They isolated the antibiotic-resistance gene by cutting out a piece of DNA from a plasmid with the help of restriction enzyme and linked it to a native plasmid of Salmonella typhimurium with the help of DNAIigase.

27. (i) A recombinant vector with a gene of interest inserted within the gene of a-galactosidase enzyme is introduced into a’ bacterium. Explain the method that would help in selection of recombinant colonies from non-recombinant colonies.
(ii) Why is this method (.of selection referred to as insertional inactivation? [HOTS All India 2012]
Ans. (i) The recombinant colonies can be differentiated from non-recombinant colonies by their inability to produce colour in the presence of a chromogenic substrate.
The recombinants do not produce any colour while, the non-recombinants produce a blue colour with chromogenic substrate in the medium.
(ii) The enzyme a-galactosidase become inactivated on insertion of recombinant DNA, within the coding sequence of enzyme. Thus, the method is called insertional inactivation.

28. Explain giving reasons why an alien piece of DNA needs to be integrated to a specific sequence of host DNA for its cloning?
[All India 2011]
Ans.The replication of DNA is initiated from the specific DNA sequence called origin of replication. For multiplication of alien DNA in the host, it has to be integrated to the origin of replication (ori).

29. List the key tools used in recombinant DNA technology. [Delhi 2011]
Ans.Key tools used in recombinant technology are restriction enzymes, polymerases, ligases, cloning vectors.and competent host organism or cells.

30. Explain the role of Ti plasmids in biotechnology. [Delhi 2011]
Ans.The Ti plasmid of Agrobacterium is responsible for the natural transformation of plant cells into tumours. So, it is modified into a non-pathogen ic vector but still is able to deliver the DNA. This disarmed plasmid of Agrobacterium is used as a vector for the transformation of plant cells, thus proved to play an important role in biotechnology.

31. How are recombinant vectors created? Why is only one type of restriction endonuclease required for creating one recombinant vector? [Foreign 2011]
Ans.Creation of recombinant vectors Vector DNA is cut at a particular restriction site by a restriction enzyme, the same that was used to cut the desired DNA segment. The alien DNA is then linked with the vector DNA using enzyme ligase’ to form the recombinant vector.
Since, a restriction enzyme Recognises and cuts the DNA at a particular sequence
called recognition site, the same restriction enzyme is used for cutting the DNA segment from both the vector and the other source, sp as to produce same sticky ends in both DNA molecules to facilitate their joining.

32. Study the diagram given below and answer the following questions
(i) Why have DNA fragments in bank D moved farther away in comparision to those in band C ?
(ii) Identify the anode end in digram.
(iii) How are these DNA fragments visualised ? [Foreign 2011]
Ans. (i) In band D, DNA fragments are smaller than those on band C. The fragments separate according to their size through the sieving effect provided by the gel So, the smaller fragments move farther away than the larger ones.
(ii) B is anode end in the diagram.
(iii) Gel containing DNA fragments is stained with ethidium bromide and exposed to UV radiation. Orange colour bands of DNA becomes visible.

33. A recombinant DNA is formed when sticky ends of vector DNA and foreign DNA join. Explain how the sticky ends are formed and get joined ? [All India 2010]
Ans.Sticky ends are formed when a restriction enzyme cuts the strands of DNA a little away from the centre of the palindromic sequence.
The sticky ends are joined via complementary factor of two polynucleotide strands bases and by the action of enzyme, DNA ligase.

34. Explain the action of the restriction endonuclease Eco RI.
[Foreign 2010]
Ans.Restriction endonuclease Eco Rl cuts the DNA strands a little away from the centre of the palindromic sequence, but between the same two bases on the two strands, i.e. G and A.
5′-GAATTC-3′
3′- C T T A A G- 5′
(i) Due to this, single stranded portions called sticky ends, overhang at the end of each strand.
(ii) Because of the stickiness, they easily form hydrogen bonds with their complementary counterparts.

35. How are the DNA fragments separated by gel electrophoresis visualised and separated for use in constructing recombinant DNA ?
[ Foreign 201 All India 2008]
Ans. The separated DNA fragments are stained with ethidium bromide.
(i) By the exposure to UV radiation, the separated DNA fragments become visible as orange-coloured bands.
(ii) The separated bands of DNA are cut out from the agarose gel and DNA is extracted from these gel pieces, this process is called elution.

36. (i) Illustrate the recognition sequence of Eco RI and mention what such sequences are called?
(ii) How does restriction endonuclease act on a DNA molecule?
[All India 2010 C]
Ans.(/’) Recognition sequence of Eco Rl
5′- G A A T T C -3′
3′- C T T A A G -3′
These sequences are called palindromic nucleotide sequences.
(ii) Restriction endonuclease acts on specified length of a DNA and binds to the DNA at the specific recognition sequence. It cuts both the double strands of DNA, at the sugar phosphate backbones, a little away from the centre of palindromic sites in between the specific sequence or points.

37. Name the source organism from which Ti plasmid is isolated. Explain the use of this plasmid in biotechnology. [Foreign 2009]
Ans. Ti plasmid is isolated from Agrobacterium tumefaciens bacteria.
The Ti plasmid of Agrobacterium is responsible for the natural transformation of plant cells into tumours. So, it is modified into a non-pathogen ic vector but still is able to deliver the DNA. This disarmed plasmid of Agrobacterium is used as a vector for the transformation of plant cells, thus proved to play an important role in biotechnology.

38. Name the natural source of agarose. Mention one role of agarose in biotechnology. [Delhi 2009c]
Ans.The natural source of agarose is sea weed. Role of agarose in biotechnology
(i) It is used to develop the matrix for gel electrophoresis.
(ii) It helps in the separation of fragments on the basis of their size.

39. Study the linking of DNA fragments shown below:

E.coli cloning vector pBR322 showing restriction sites (Hind III, Eco Rl
Bam HI, Sal I, Pvu II, Pst I, Cla I), ori and antibiotic resistance genes (amp R and tet R ).
rop codes for the proteins involved in the replication of the plasmid.

  • Name A DNA and B DNA.
  • Name the restriction enzyme that recognises this palindrome.

Name the enzyme that can link these two DNA fragments.
[Delhi 2008]
Ans. (i) A – Vector/plasmid DNA
B – Foreign DNA
(ii) Eco Rl
(iii) DNA ligase40. The following illustrates the linking of DNA fragments.

(i) Name A and B.
(ii) Complete the palindrome, which is recognised by Eco RI.
(iii) Name the enzyme that can link the two DNA fragments.
[Foreign 2008]
Ans.(i) A – Vector DNA, B-Foreign DNA
(ii) 5′- GAATTC-3′
3′- CTTAAG-5′
(iii) DNA ligase
3 Marks Questions

41. Name and describe the technique that helps in separating the DNA fragments formed by the use of restriction endonuclease. [All India 2014]
Ans. DNA fragments formed by the use of restriction endonucleases are separated by gel electrophoresis.
(i) DNA fragments are negatively charged molecules. Thus, they move towards the anode under electric field through the medium.
(ii) DNA fragments separate according to their size due to seiving effect of agarose gel.
(iii) The separated DNA fragments can be viewed by staining the DNA with ethidium bromide followed by exposure to UV radiation.
(iv) The separated bands of DNA are cut and extracted from gel piece. This is known as elution.

42. Draw a schematic diagram of the colicloning vector pBR322 and mark the following in it.

  • ori
  • rop
  • ampicillin resistance gene
  • tetracycline resistance gene
  • restriction site Bam HI
  • restriction site EcoRI

(i) Draw schematic diagrams of segments of a vector and a foreign DNA with the sequence of nucleotides recognised by Eco
(ii) Draw the vector DNA segment arid foreign DNA segments after the action of EcoRI and label the sticky ends produced. [Delhi 2014C]
or
Draw a schematic sketch of pBR322 plasmid and label the following in it

  • Any two restriction sites.
  • Ori and rop
  • An antibiotic resistant gene. [Delhi 2012].

Ans. coli cloning vector pBR322.
E.coli cloning vector pBR322 showing restriction sites (Hind III, Eco Rl
Bam HI, Sal I, Pvu II, Pst I, Cla I), ori and antibiotic resistance genes (amp R and tet R ).
rop codes for the proteins involved in the replication of the plasmid.
or
(a) Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands.
(b) This leaves single stranded portions at the ends.
(c) There are overhanging stretches called sticky ends on each strands as given in above figure. These are named so, because they form hydrogen bonds with their complementary cut counterparts.
(d) The stickiness of the ends facilitates the action of the enzyme DNA ligase.
(e) Restriction endonucleases are used in genetic engineering to form recombinant molecules of DNA, which are composed of DNA from different sources/genomes.
(f) These sticky ends are complementary to each other when cut by same restriction enzyme, therefore can be joined together (end-to-end) using DNA ligases.43. What are ‘cloning sites’ in a cloning vector? Explain their role. Name any two such sites in pBR322. [All India 2014C]
Ans.The cloning sites are actually the specific unique recognition sequence for a particular restriction enzyme, so as to link the foreign DNA with the vector DNA to create a recombinant DNA molecules, (l) These sites are important for joining of DNA fragments of vector and alien DNA. And also multiple recognition sequences for a particular restriction enzyme within a DNA or vector will complicate the process of gene cloning.
The two cloning sites in pBR322 are Bam HI of tetracycline resistant gene and Pvu I of ampiciMies resistant genes.

44. How are the DNA fragments separated and isolated for DNA fingerprinting? Explain. [Foreign 2012]
Ans. DNA fragments formed by the use of restriction endonucleases are separated by gel electrophoresis.
(i) DNA fragments are negatively charged molecules. Thus, they move towards the anode under electric field through the medium.
(ii) DNA fragments separate according to their size due to seiving effect of agarose gel.
(iii) The separated DNA fragments can be viewed by staining the DNA with ethidium bromide followed by exposure to UV radiation.
(iv) The separated bands of DNA are cut and extracted from gel piece. This is known as elution

45. (i) Why are restriction endonucleases, so called?
(ii) What is a palindromic , nucleotide sequence? How do restriction endonucleases act on palindromic sites, to create sticky ends?
[Delhi 2011C]
Ans. (i) Restriction endonucleases are so called because they recognise and make a cut at specific positions within the DNA and restrict the growth of bacteriophage.
(ii) (a) The palindrome in DNA is a sequence of base pairs that reads same on the two strands of DNA when orientation of reading is kept the same.
For example,
5’—GAATTG—3′
3’—CTTAAG—5′
(b) Restriction enzymes cut the strand of DNA a little away from the centre of palindrome site, but between the same two bases in both the strands. This creates single stranded stretches, overhanging at the ends of the palindrome, called sticky ends.

46. How are the following used in biotechnology?

  • Plasmid DNA
  • Recognition sequence
  • Gel electrophoresis [All India 2011c]

Ans. (i) Plasmid DNA It is used for constructing recombinant DNA, by ligating the alien piece of DNA with it. It is used as a cloning vector and helps in the selection of recombinants from non-recombinants.

(ii) Recognition sequences These are specific base sequences of DNA, where restriction enzyme cuts the DNA. They are utilised to extract the desired gene or fragments from DNA molecules.
(iii) Gel electrophoresis It is a technique, used to separate the DNA fragments according to their size through seiving effect of the gel.47. EcoRI is used to cut a segment of foreign DNA and that of a vector DNA to form a recombinant DNA. Show with the help of schematic diagrams.

  • The set of palindromic nucleotide sequence of base pairs the EcoRI will recognise in both the DNA segments. Mark the site at which EcoRI will act and cut both the segments.
  • Sticky ends formed on both the segments, where the two DNA segments will join later to form a recombinant DNA. [Delhi 2010]

Ans. Palindromic sequence of Eco Rl is

5′ G AATTC 3′
3’C TTAA G 5′ t
(the arrows indicate the site, where it cuts the strands.)
(ii) Sticky ends formed on both the segments.
(a) Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands.
(b) This leaves single stranded portions at the ends.
(c) There are overhanging stretches called sticky ends on each strands as given in above figure. These are named so, because they form hydrogen bonds with their complementary cut counterparts.
(d) The stickiness of the ends facilitates the action of the enzyme DNA ligase.
(e) Restriction endonucleases are used in genetic engineering to form recombinant molecules of DNA, which are composed of DNA from different sources/genomes.
(f) These sticky ends are complementary to each other when cut by same restriction enzyme, therefore can be joined together (end-to-end) using DNA ligases.48. (i) Name the organism in which the vector shown is inserted to get the copies of the desired gene.

  • Mention the area labelled in the vector responsible for controlling the copy number of the inserted gene.
  • Name and explain the role of aselectable marker in the vector Shown. [All India 2010]

Ans. (i) Escherichia coli (E. coli)
(ii) Ori in the vector is responsible for controlling the copy number of inserted gene.
(iii) The genes encoding resistance to antibiotics like tet R resistant to tetracycline, amp R resistant to ampicillin are used as selectable markers. If a foreign DNA is ligated at the Bam HI site of tetracycline resistance gene, the recombinant plasmids will lose the tetracycline resistance.
The selectable markers help in identifying and eliminating non-transformants and selectively permitting the growth of transformants.

49. (i) EcoRI is a restriction endonuclease. How is it named so? Explain.
(ii) Write the sequence of DNA bases that the enzyme recognises. Mention the point at which the enzyme makes a cut in the DNA segment. [Delhi 2010c]
Ans.(I) Naming of restriction endonuclease are:
(i) The first letter of the name comes from the genus and next two letters from species of the prokaryotic cell from where enzymes are extracted.
(ii) The Roman numbers following the name show the order in which the enzymes, were isolated from the bacterial strain. For example, Eco Rl is derived from Escherichia coli RY 13, Hind II from Haemophilus influenzae Rd, etc.
(II) The recognition sequence is a palindrome, where the sequence of the base pair reads the same on both the DNA strands, when orientation of reading is kept same,
e.g., 5′- GAATTC -3′
3′- CTTAAG -5′
The enzyme Eco Rl it cuts the DNA segment at bases G and A on both the strands only when the sequence GAATTC

50. (i) Name the technique used for separation of DNA fragments.

  • Write the type of matrix used in this technique.
  • How is the separated DNA visualised and extracted for use in recombinant technology? [All India 2010]

Ans. (i) Gel electrophoresis.

(ii)The material used as matrix in gel electrophoresis is agarose.
This agarose gel acts as a sieve to separate the DNA fragments according to their size.
(iii) The separated DNA fragments are stained with ethidium bromide.
(a) By the exposure to UV radiation, the separated DNA fragments become visible as orange-coloured bands.
(b) The separated bands of DNA are cut out from the agarose gel and DNA is extracted from these gel pieces, this process is called elution.

51. (i) Identify the selectable markets in the diagram of E. coli vector shown below.
Ans. (i) A – Ampicillin resistanceD – Tetracycline resistance are used as selectable markers in E.coli cloning vector.
(ii) Coding sequence of a-galactosidase is a better marker, as the recombinants and non-recombinants are differentiated on the basis of their ability to produce colour in the presence of a chromogenic substrate, while the selection of recombinants due to inactivation of antibiotic resistant gene is a tedious and time taking process to grow them simultaneously on two antibiotics separately.
(a) Introduction of rDNA into the coding sequence of a-galactosidase leads to insertional inactivation.
(ii) The recombinants do not produce a blue colour, while the non-recomninants produces a blue colour.

52. Name and explain the techniques used in the separation and isolation of DNA fragments to be used in recombinant DNA technology.
[All India 2009]
Ans.DNA fragments formed by the use of restriction endonucleases are separated by gel electrophoresis.
(i) DNA fragments are negatively charged molecules. Thus, they move towards the anode under electric field through the medium.
(ii) DNA fragments separate according to their size due to seiving effect of agarose gel.
(iii) The separated DNA fragments can be viewed by staining the DNA with ethidium bromide followed by exposure to UV radiation.
(iv) The separated bands of DNA are cut and extracted from gel piece. This is known as elution.

53. Why are genes encoding resistance of antibiotics considered useful selectable markers for coli cloning vector? Explain with the help of one example. [Delhi 2009c]
Ans. The genes encoding resistance to antibiotics are cosidered useful selectable
markers because the normal £. coli cells do not carry resistance against any of these antibiotics.
Example A foreign DNA is ligated at the Bam HI site of tetracycline resistance gene in the vector pBR322. The recombinant plasmids will lose tetracycline resistance due to the insertion of foreign DNA but can still be selected from non-recombinants by plating the transformants on ampicillin containing medium.
The transformants growing on ampicillin containing medium are then transferred on medium containing tetracycline. The recombinants will not grow, while non-recombinants will grow on the medium containing both the antibiotics. Antibiotic resistance gene, thus helps in selecting the transformants.

54. (i) Write the palindromic nucleotidesequence for the following DNA segment
5′- GAATTC- 3′

  • Name the restriction endonuclease that recognises this sequence.
  • How are sticky-ends produced? Mention their role. [All India 2009]

Ans. (i) Palindromic sequence for

5′- GAATTC- 3′
3′- CTTAAG -5′.
(ii) Restriction endonuclease Eco Rl recognises the above palindromic sequence.
(iii)Sticky ends on DNA are formed by action of enzymes restriction endonucleases. These enzymes cut the strand of DNA a little away from the centre of the palindrome sequence between the same two bases on both the strands. This results in single stranded stretches on both the complementary strands at their ends.
These overhanging stretches are called sticky ends as they form hydrogen bonds with the complementary base pair sequences.
Role of the sticky ends These sticky ends produced form hydrogen bonds with their complementary cut counterparts. This stickiness of the ends faciIitates the action of the enzyme DNA ligase.

55. (i) What are molecular scissors?
Give one example.
(ii) Explain their role in recombinant DNA technology.
[Delhi 2008 Foreign 2008]
or
(i) Name the category of enzymes that cut at a specific site within the DNA molecule. Give an example.
(ii) Explain, how do these enzymes function? Mention their use in genetic engineering. [All India 2008C]
Ans. (i) Molecular scissors are the restriction endonucleases because they cut the DNA segments at particular locations or specific sequence, e.g. Eco Rl.
(ii) The restriction enzymes cut the DNA strand a little away from the centre of the palindromic sites, but between the same two bases on the opposite strands. This leaves single stranded portions at the ends with overhanging stretches called sticky ends on each strand. These ends form hydrogen bonds with their complementary cut counterparts. This stickiness of the ends facilitates the action of the enzyme DNA ligase.

56. Why is Agrobacterium tumefaciens a good cloning vector? Explain.
[All India 2008]
Ans. Agrobacterium tumefaciens is a soil bacterium, which causes disease in many dicot plants. It is a natural vector as it is able to deliver a piece of DNA known as T-DNA to transform the normal cells into tumour cells and direct these tumour cellsto produce the chemicals required by the pathogen.
The Tumour inducing (Ti) plasmid of A. tumefaciens has now been modified into a cloning vector, which is no more pathogenic to the plants but still deliver the gene of interest, that gets incorporated into genome of plant.

57. Explain the importance of
(i) ori (ii) amp R and (iii) rop in the E.coli vector shown below.
[All India 2008]
Ans.(i) Ori is the sequence of DNA from where replication starts and any piece of DNA, when linked to this sequence can be made to replicate within the host cells.
It is also responsible for controlling the copy number of linked DNA.
(ii) amp R is an antibiotic resistance gene in which ligation of alien DNA is carried out at a restriction site.
(iii) rop codes for the proteins involved in the replication of the plasmid.

58. A vector is engineered with three features, which facilitate its clonincj within the host cell. List the three features and explain each of them.
[Foreign 2008]
Ans. Features which facilitate cloning of vector are:
(i) Origin of replication (Ori)

  • This is the sequence of DNA from where replication starts.
  • Any piece of alien/foreign DNA linked to it is made to replicate within host cell. It also decides the copy number of the linked[ DNA.
    (ii) Selectable marker is a marker gene, which helps in selecting the host cells, which are transformants/recombinants from the non-recombinant ones.
    For example, ampicillin and tetracycline resistant genes in, £. coli.
    (iii) Cloning site is a unique recognition site in a vector to link the foreign DNA. The presence of a particular cloning/recognition site helps the partjcular restriction enzyme to cut the vector DNA. Single recognition site is commonly preferred.
    (iv) Small size of the vector The small size facilitates the introduction of the DNA into the host easily.

59. Why a cell must be made competent to take up DNA?Explain the steps by which a bacterial cells made competent to take up plasmid/rDNA. [Delhi 2008C]

Ans. DNA being a hydrophilic molecule cannot pass through cell membrane. Hence, the bacterial cell is made competent to accept the DNA molecule. Thus, competency is the ability of a cell to take up foreign DNA.
Following methods are used to make a bacterial cell competent:
the bacteria should be made competent to accept the DNA molecules,
(i) Competency is the ability of a cell to take up foreign DNA.
(ii) Methods to make a cell competent are as follow.
(a) Chemical method In this method, the cell is treated with a specific concentration of a divalent cation such as calcium to increase pore size in cell wall.

  • The cells are then incubated with recombinant DNA on ice, followed by placing
  • them briefly at 42°C and then putting it back on ice. This is called heat shock treatment.
    • This enables the bacteria to take up the recombinant DNA.
    (b) Physical methods In this method, a recombinant DNA is directly injected into the nucleus of an animal cell by microinjection method.
    • In plants, cells are bombarded with high velocity microparticles of gold or
    tungsten coated with DNA called as biolistics or gene gun method.
    (c) Disarmed pathogen vectors when allowed to infect the cell, transfer the recombinant DNA into the host.

60. Read the following base sequence of a certain DNA strand and answer the questions that follow:

  • What is called a palindromic sequence in a DNA?
  • Write the palindromic nucleotide sequence shown in the DNA strand given and mention the enzyme that will recognise such a sequence.
  • State the significance of enzymes that identify palindromic nucleotide sequences. [All India 2008C]

Ans. (i) Palindromic sequence in a DNA are base sequences that reads same, both in forward and backward direction.
(ii) Palindromic sequence in DNA 5′- GAATTC- 3′
3 – CTTAAG- 5′
The sequence reads the same on the two strands in 5′-> 3′ direction. This is also same if read in the 3′->5′ direction. Restriction endonuclease recognises such sequence.
DNA being a hydrophilic molecule cannot pass through cell membrane. Hence, the bacterial cell is made competent to accept the DNA molecule. Thus, competency is the ability of a cell to take up foreign DNA.
Following methods are used to make a bacterial cell competent:
the bacteria should be made competent to accept the DNA molecules,
(i) Competency is the ability of a cell to take up foreign DNA.
(ii) Methods to make a cell competent are as follow.
(a) Chemical method In this method, the cell is treated with a specific concentration of a divalent cation such as calcium to increase pore size in cell wall.

them briefly at 42°C and then putting it back on ice. This is called heat shock treatment.
• This enables the bacteria to take up the recombinant DNA.
(b) Physical methods In this method, a recombinant DNA is directly injected into the nucleus of an animal cell by microinjection method.
• In plants, cells are bombarded with high velocity microparticles of gold or
tungsten coated with DNA called as biolistics or gene gun method.
(c) Disarmed pathogen vectors when allowed to infect the cell, transfer the recombinant DNA into the host.
Significance of restriction endonucleases They are used in genetic engineering to form recombinant molecules of DNA,
which are composed of DNA from different sources/genomes and allow isolating the desired gene fragment and joining it with host on vector DNA due to sticky ends generated.
5 Marks Questions

61. (i) Describe the characteristics a cloning vector must possess.
(ii) Why DNA cannot pass through the cell membrane? Explain. How is a bacterial cell made competent to take up recombinant DNA from the medium? [All India 2011]
Ans.(I) The following features are required tofacilitate cloning into a vector
Features which facilitate cloning of vector are:
(i) Origin of replication (Ori)

  • This is the sequence of DNA from where replication starts.
  • Any piece of alien/foreign DNA linked to it is made to replicate within host cell. It also decides the copy number of the linked[ DNA.

(ii) Selectable marker is a marker gene, which helps in selecting the host cells, which are transformants/recombinants from the non-recombinant ones.
For example, ampicillin and tetracycline resistant genes in, £. coli.
(iii) Cloning site is a unique recognition site in a vector to link the foreign DNA. The presence of a particular cloning/recognition site helps the partjcular restriction enzyme to cut the vector DNA. Single recognition site is commonly preferred.
(iv) Small size of the vector The small size facilitates the introduction of the DNA into the host easily.
(II) DNA being a hydrophilic molecule cannot pass through cell membrane. Hence, the bacterial cell is made competent to accept the DNA molecule. Thus, competency is the ability of a cell to take up foreign DNA.
Following methods are used to make a bacterial cell competent:
the bacteria should be made competent to accept the DNA molecules,
(i) Competency is the ability of a cell to take up foreign DNA.
(ii) Methods to make a cell competent are as follow.
(a) Chemical method In this method, the cell is treated with a specific concentration of a divalent cation such as calcium to increase pore size in cell wall.

them briefly at 42°C and then putting it back on ice. This is called heat shock treatment.
• This enables the bacteria to take up the recombinant DNA.
(b) Physical methods In this method, a recombinant DNA is directly injected into the nucleus of an animal cell by microinjection method.
• In plants, cells are bombarded with high velocity microparticles of gold or
tungsten coated with DNA called as biolistics or gene gun method.
(c) Disarmed pathogen vectors when allowed to infect the cell, transfer the recombinant DNA into the host.62. (i) With the help of diagrams show the different steps in the formation of recombinant DNA by action of restriction endonuclease enzyme Eco RI.
(ii) Name the technique that is used for separating the fragments of DNA cut by restriction endonucleases. [All India 2011]
Ans. (i) Formation of recombinant DNA by the action of restriction endonuclease enzyme Eco Rl.
The restriction enzyme cuts both DNA strands at the same site, producing sticky ends.
(a) Restriction enzymes cut the strand of DNA a little away from the centre of the palindrome sites, but between the same two bases on the opposite strands.
(b) This leaves single stranded portions at the ends.
(c) There are overhanging stretches called sticky ends on each strands as given in above figure. These are named so, because they form hydrogen bonds with their complementary cut counterparts.
(d) The stickiness of the ends facilitates the action of the enzyme DNA ligase.
(e) Restriction endonucleases are used in genetic engineering to form recombinant molecules of DNA, which are composed of DNA from different sources/genomes.
(f) These sticky ends are complementary to each other when cut by same restriction enzyme, therefore can be joined together (end-to-end) using DNA ligases.
(ii) Gel electrophoresis.

63. (i) Why are engineered vectors preferred by biotechnologists for transferring the desired genes into another organism?(ii) Explain, how do ori, selectable marker and cloning sites facilitate cloning into a vector? [Foreign 2009]
Ans. (I) Engineered vectors are preferred by biotechnologists because they help in easy linking of foreign DNA and selection of recombinants from non-recombinants.
(II) Features which facilitate cloning of vector are:
(i) Origin of replication (Ori)

  • This is the sequence of DNA from where replication starts.
  • Any piece of alien/foreign DNA linked to it is made to replicate within host cell. It also decides the copy number of the linked[ DNA.

(ii) Selectable marker is a marker gene, which helps in selecting the host cells, which are transformants/recombinants from the non-recombinant ones.
For example, ampicillin and tetracycline resistant genes in, £. coli.
(iii) Cloning site is a unique recognition site in a vector to link the foreign DNA. The presence of a particular cloning/recognition site helps the partjcular restriction enzyme to cut the vector DNA. Single recognition site is commonly preferred.
(iv) Small size of the vector The small size facilitates the introduction of the DNA into the host easily.


How was Restriction Site of EcoRI sequenced? - Biology

Restriction sites, or restriction recognition sites, are specific sequences of nucleotides that are recognized by restriction enzymes. The sites are generally palindromic, and a particular restriction enzyme may cut the sequence between two nucleotides within its.
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A restriction enzyme is an enzyme that cuts double-stranded or single stranded DNA at specific recognition nucleotide sequences known as restriction sites. Such enzymes, found in bacteria and archaea, are thought to have.
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How Can I Add a Site to Internet Explorer's Restricted Sites Zone? . Because we are talking about the Restricted Sites zone you will likely want to .
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Finds restriction endonuclease cleavage sites in DNA sequences. Supports linear and circular DNA and provides several ways to sort and filter output.
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CLC bio offers a wide range of world class bioinformatics solutions: From highly advanced sequence analysis . interactive restriction site analysis .
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CLC bio offers a wide range of world class bioinformatics solutions: From highly advanced sequence analysis . Restriction site detection. Reverse .
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. zones in Microsoft Internet Explorer, and how to configure different levels of security for Web sites that you visit. . site to the Restricted Sites zone, .
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Indicates restriction sites of Fermentas enzymes that are sensitive (cleavage . There are no restriction sites in PhiX174 DNA for the following enzymes: .
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Indicates restriction sites of Fermentas enzymes that are sensitive (cleavage . There are no restriction sites in M13mp18 DNA for the following enzymes: .
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restriction site synonyms, restriction site antonyms. Information about restriction site . restriction-site polymorphism. Restrictionary. restrictionism .
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Restricted Site. This site is for the use of the staff of the Northeast Georgia Council, Boy . confidential or private information contained on this web site. .
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This article provides details on blocking Unwanted Ads, Banners, pop-ups, Parasites, Hijackers, and Search Engines, by placing them in the Restricted Zone
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Degenerate sequences Analyze restriction maps of sequences . Please indicate how you would like the restriction sites displayed. Map of restriction sites .
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You can see our list of restricted sites is very large. . Let's add a site to the Restricted Sites list. Note the format of the entry. .
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How surfers find their way in times of war to restricted or censored sites . Both have restricted free versions. These sites hide your location and protect .
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Sites placed in the high security zone will then be restricted in the access . You are protected if a site is in your restricted zone. .
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The length of restriction recognition sites varies: The enzymes EcoRI, SacI and . Different restriction enzymes can have the same recognition site - such enzymes .
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Regulators of G-Protein Signaling, Part B

Patrizia Tosetti , Kathleen Dunlap , in Methods in Enzymology , 2004

Production of Replication-Competent Retroviruses.

A 5′ Xba I restriction site (TCT AGA) followed by a translation/transcription accessory sequence (CCA CGT TGG CCA GGC GGT AGC TGG GAC GTG CAG CCG ACC ACC) and a 3′ HA epitope (TAC GAC GTG CCC GAC TAC GCC) followed by a stop codon (TAG) and a HindIII restriction site (AAG CTT) are engineered in-frame into RGS3 variant cDNAs by PCR, using primers carrying the appropriate sequences. The PCR products are then digested and subcloned in frame into the XbaI/HindIII sites of the adaptor plasmid Cla12 NCO. The sequences of interest are then excised from the adaptor plasmid by ClaI digestion and are subcloned nondirectionally into the ClaI site of RCASBP retroviral variants of the B subgroup (both plasmids were a generous gift of Dr. Donna Fekete, Purdue University, West Lafayette, IN). All ligations are carried out using the rapid DNA ligation kit from Boehringer Mannheim (Indianapolis, IN), according to the manufacturer's instructions. The restriction enzymes are from New England Biolabs (Beverly, MA). Properly oriented constructs are identified by PCR and sequencing.


You can also filter enzymes by where or how many times they cut on the sequence. For example, you can filter to see only enzymes that are single or double cutters. You can also see enzymes that cut anywhere outside of your current selection.

Click the scissors icon again to close the Digests panel. Double-click on a cut site on the sequence map, and notice that the panel automatically opens to show more information about the enzyme.


Bacterial Autoimmunity Due to a Restriction-Modification System

Restriction-modification (RM) systems represent a minimal and ubiquitous biological system of self/non-self discrimination in prokaryotes [1], which protects hosts from exogenous DNA [2]. The mechanism is based on the balance between methyltransferase (M) and cognate restriction endonuclease (R). M tags endogenous DNA as self by methylating short specific DNA sequences called restriction sites, whereas R recognizes unmethylated restriction sites as non-self and introduces a double-stranded DNA break [3]. Restriction sites are significantly underrepresented in prokaryotic genomes [4-7], suggesting that the discrimination mechanism is imperfect and occasionally leads to autoimmunity due to self-DNA cleavage (self-restriction) [8]. Furthermore, RM systems can promote DNA recombination [9] and contribute to genetic variation in microbial populations, thus facilitating adaptive evolution [10]. However, cleavage of self-DNA by RM systems as elements shaping prokaryotic genomes has not been directly detected, and its cause, frequency, and outcome are unknown. We quantify self-restriction caused by two RM systems of Escherichia coli and find that, in agreement with levels of restriction site avoidance, EcoRI, but not EcoRV, cleaves self-DNA at a measurable rate. Self-restriction is a stochastic process, which temporarily induces the SOS response, and is followed by DNA repair, maintaining cell viability. We find that RM systems with higher restriction efficiency against bacteriophage infections exhibit a higher rate of self-restriction, and that this rate can be further increased by stochastic imbalance between R and M. Our results identify molecular noise in RM systems as a factor shaping prokaryotic genomes.


Frequencies of Restriction Sites

The table below summarizes the frequencies with which restriction endonuclease sites occur in commonly used DNA molecules. Detailed restriction maps can be found on DNA sequences and maps. The sites listed in these tables were identified by computer analysis of published sequences. Although we have tried to ensure their accuracy, the sites have not necessarily been confirmed by experimentation. When the same specificity is displayed by several enzymes, the site is listed by the name of the enzyme that is available from New England Biolabs. Other enzymes with the same specificity are listed in the table of isoschizomers. Recognition sequences are written 5´ to 3´.

DNASU is a central repository for plasmid clones and collections that may also be helpful.

Enzyme Recognition Sequence Adeno-2 Lambda** M13mp18* pBR322 pKLAC2 pMAL-p5x pSNAPf pTXB1 pTYB21 pUC19 T7
AatII GACGT/C 3 10 0 1 0 0 5 1 0 1 1
AbaSI CNNNNNNNNNNN/NNNNNNNNNG
AccI GT/MKAC 17 9 1 2 5 1 2 5 3 1 33
Acc65I G/GTACC 8 2 1 0 0 0 2 0 1 1 5
AciI CCGC(-3/-1) 582 516 42 67 81 80 75 102 92 34 199
AclI AA/CGTT 3 7 2 4 2 5 3 12 9 2 19
AcuI CTGAAG(16/14) 23 40 0 2 8 4 5 2 2 2 1
AfeI AGC/GCT 13 2 1 4 0 2 0 1 2 0 0
AflII C/TTAAG 4 3 0 0 0 0 0 0 0 0 19
AflIII A/CRYGT 25 20 3 1 4 2 5 3 4 1 23
AgeI § A/CCGGT 5 13 0 0 1 0 1 1 0 0 2
AgeI-HF® A/CCGGT 5 13 0 0 1 0 1 1 0 0 2
AhdI GACNNN/NNGTC 9 9 0 1 1 2 1 2 2 1 14
AleI-v2 CACNN/NNGTG 10 20 1 0 1 0 1 0 2 0 8
AluI AG/CT 158 143 27 17 38 28 27 30 34 16 140
AlwI GGATC(4/5) 35 58 3 12 18 12 20 15 16 10 1
AlwNI CAGNNN/CTG 25 41 1 1 5 2 4 1 2 1 15
ApaI GGGCC/C 12 1 0 0 0 1 1 1 1 0 0
ApaLI G/TGCAC 7 4 0 3 3 6 4 4 4 3 1
ApeKI G/CWGC 179 199 10 21 27 25 20 27 29 12 116
ApoI § R/AATTY 29 58 11 0 19 5 3 5 7 1 13
ApoI-HF R/AATTY 29 58 11 0 19 5 3 5 7 1 13
AscI GG/CGCGCC 2 2 0 0 0 0 1 0 0 0 0
AseI AT/TAAT 3 17 7 1 5 4 7 10 4 3 12
AsiSI GCGAT/CGC 1 0 0 0 0 0 0 0 0 0 0
AvaI C/YCGRG 40 8 2 1 2 1 2 3 1 1 4
AvaII G/GWCC 73 35 1 8 7 9 6 6 8 2 54
AvrII C/CTAGG 2 2 0 0 0 0 1 0 0 0 3
BaeGI GKGCM/C 45 10 1 3 8 8 8 6 5 3 16
BaeI (10/15)ACNNNNGTAYC(12/7) 5 10 3 0 1 0 0 0 1 0 3
BamHI § G/GATCC 3 5 1 1 1 1 1 1 1 1 0
BamHI-HF® G/GATCC 3 5 1 1 1 1 1 1 1 1 0
BanI G/GYRCC 57 25 7 9 7 4 8 8 12 4 33
BanII GRGCY/C 57 7 2 2 5 2 5 3 6 1 1
BbsI § GAAGAC(2/6) 27 24 0 3 3 3 2 4 4 0 38
BbsI-HF ® GAAGAC(2/6) 27 24 0 3 3 3 2 4 4 0 38
BbvCI CCTCAGC(-5/-2) 9 7 2 0 0 0 0 0 0 0 10
BbvI GCAGC(8/12) 179 199 10 21 27 25 20 27 29 12 116
BccI CCATC(4/5) 62 145 14 9 22 16 8 14 25 3 121
BceAI ACGGC(12/14) 80 115 7 3 13 11 8 13 10 2 47
BcgI (10/12)CGANNNNNNTGC(12/10) 10 28 0 3 6 4 1 4 6 1 19
BciVI GTATCC(6/5) 9 26 0 2 3 4 3 4 4 2 23
BclI § T/GATCA 5 8 0 0 2 2 1 1 2 0 1
BclI-HF T/GATCA 5 8 0 0 2 2 1 1 2 0 1
BcoDI GTCTC(1/5) 60 37 5 3 11 8 4 8 9 4 95
BfaI C/TAG 54 13 5 5 19 3 14 8 8 4 60
BfuAI ACCTGC(4/8) 39 41 3 1 5 4 4 2 3 1 18
BglI GCCNNNN/NGGC 20 29 1 3 3 1 7 2 3 2 2
BglII A/GATCT 11 6 1 0 1 1 2 0 2 0 1
BlpI GC/TNAGC 8 6 0 0 0 1 0 1 1 0 20
BmgBI CACGTC(-3/-3) 15 17 0 0 1 1 2 0 0 0 8
BmrI ACTGGG(5/4) 22 4 1 5 2 5 11 8 2 6
BmtI § GCTAG/C 4 1 0 1 4 0 1 1 1 0 1
BmtI-HF® GCTAG/C 4 1 0 1 4 0 1 1 1 0 1
BpmI CTGGAG(16/14) 32 25 2 4 3 4 1 5 6 1 23
BpuEI CTTGAG(16/14) 19 13 4 6 7 5 9 7 9 4 56
Bpu10I CCTNAGC(-5/-2) 23 19 4 1 0 1 2 0 2 0 39
BsaAI YAC/GTR 22 14 5 1 4 0 3 2 4 0 35
BsaBI GATNN/NNATC 2 21 2 1 2 2 1 1 3 0 7
BsaHI GR/CGYC 44 40 1 6 6 5 8 12 7 3 8
BsaI-HF®v2 GGTCTC(1/5) 18 2 0 1 3 2 2 2 2 1 29
BsaJI C/CNNGG 234 105 9 8 18 10 17 15 15 5 85
BsaWI W/CCGGW 28 81 6 5 8 7 5 8 6 3 32
BsaXI (9/12)ACNNNNNCTCC(10/7) 29 19 4 0 3 1 1 2 3 1 12
BseRI GAGGAG(10/8) 63 19 1 0 2 0 3 0 0 0 13
BseYI CCCAGC(-5/-1) 31 32 3 2 3 4 5 4 4 1 29
BsgI GTGCAG(16/14) 34 41 0 1 4 6 3 5 4 0 21
BsiEI CGRY/CG 50 22 3 7 11 8 6 9 8 5 17
BsiHKAI GWGCW/C 38 28 3 8 8 9 9 7 7 5 24
BsiWI § C/GTACG 4 1 0 0 0 1 0 1 0 0 0
BsiWI-HF ® C/GTACG 4 1 0 0 0 1 0 1 0 0 0
BslI CCNNNNN/NNGG 216 176 17 20 18 16 26 31 25 6 90
BsmAI GTCTC(1/5) 60 37 5 3 11 8 8 10 4 95
BsmBI-v2 CGTCTC 21 14 1 1 2 2 0 2 2 2 16
BsmFI GGGAC(10/14) 59 38 2 4 4 1 5 4 6 0 46
BsmI GAATGC(1/-1) 10 46 1 1 3 1 5 1 0 0 15
BsoBI C/YCGRG 40 8 2 1 2 1 2 3 1 1 4
BspCNI CTCAG(9/7) 75 80 24 7 10 10 9 20 16 5 142
BspDI AT/CGAT 2 15 2 1 2 0 0 0 0 0 3
BspEI T/CCGGA 8 24 0 1 1 2 0 1 1 0 0
BspHI T/CATGA 3 8 1 4 2 1 2 2 3 3 13
Bsp1286I GDGCH/C 105 38 5 10 16 11 15 11 12 5 40
BspMI ACCTGC(4/8) 39 41 3 1 5 4 4 2 3 1 18
BspQI GCTCTTC(1/4) 7 10 0 1 2 1 3 1 1 1 4
BsrBI CCGCTC(-3/-3) 28 17 4 2 6 4 6 9 5 3 17
BsrDI GCAATG(2/0) 14 44 3 2 7 4 4 4 4 2 18
BsrFI-v2 R/CCGGY 40 61 1 7 9 2 6 11 9 1 3
BsrGI § T/GTACA 5 5 1 0 1 0 1 1 2 0 13
BsrGI-HF® T/GTACA 5 5 1 0 1 0 1 1 2 0 13
BsrI ACTGG(1/-1) 86 110 19 18 23 26 19 32 28 11 118
BssHII G/CGCGC 52 6 0 0 1 2 2 1 1 0 1
BssSI-v2 CACGAG(-5/-1) 11 8 0 3 5 3 4 2 4 3 31
BstAPI GCANNNN/NTGC 20 34 0 2 3 2 0 3 2 1 12
BstBI TT/CGAA 1 7 0 0 2 0 0 0 1 0 7
BstEII § G/GTNACC 10 13 0 0 1 1 0 1 1 0 1
BstEII-HF® G/GTNACC 10 13 0 0 1 1 2 1 1 0 1
BstNI CC/WGG 136 71 7 6 15 14 19 19 14 5 2
BstUI CG/CG 303 157 17 23 26 31 19 41 35 10 65
BstXI CCANNNNN/NTGG 10 13 0 0 2 4 1 4 3 0 11
BstYI R/GATCY 22 21 2 8 12 9 11 10 12 7 2
BstZ17I-HF ® GTATAC 3 3 0 1 3 0 1 1 1 0 8
Bsu36I CC/TNAGG 7 2 1 0 1 1 1 0 0 0 30
BtgI C/CRYGG 82 46 2 2 4 3 6 1 4 0 26
BtgZI GCGATG(10/14) 23 45 4 3 4 6 6 4 3 0 24
BtsCI GGATG(2/0) 78 150 4 12 20 17 7 12 18 5 97
BtsIMutI CAGTG(2/0) 20
BtsI-v2 GCAGTG(2/0) 22 34 1 2 7 5 5 4 4 3 20
Cac8I GCN/NGC 285 238 28 31 33 32 41 49 42 14 104
ClaI AT/CGAT 2 15 2 1 2 0 0 0 0 0 3
CspCI (11/13)CAANNNNNGTGG(12/10) 6 7 1 0 1 0 1 0 0 0 9
CviAII C/ATG 183 181 14 26 38 23 21 23 31 11 148
CviKI-1 RG/CY 680 692 103 73 131 86 119 112 121 45 562
CviQI G/TAC 83 113 19 3 15 7 14 6 11 3 168
DdeI C/TNAG 97 104 30 8 17 11 11 20 19 6 282
DpnI GA/TC 87 116 6 22 35 23 31 24 29 15 6
DpnII /GATC 87 116 6 22 35 23 31 24 29 15 6
DraI TTT/AAA 12 13 5 3 5 1 6 3 2 3 9
DraIII-HF® CACNNN/GTG 10 10 1 0 2 0 6 1 1 0 16
DrdI GACNNNN/NNGTC 6 3 1 2 5 2 2 4 4 2 11
EaeI Y/GGCCR 70 39 3 6 10 5 15 4 8 3 2
EagI-HF® C/GGCCG 19 2 0 1 4 1 2 2 2 0 0
EarI CTCTTC(1/4) 29 34 2 2 11 6 4 3 5 3 46
EciI GGCGGA(11/9) 29 32 2 4 6 6 8 9 8 3 2
Eco53kI GAG/CTC 16 2 1 0 2 1 2 0 1 1 0
EcoNI CCTNN/NNNAGG 10 9 0 1 3 0 0 2 1 0 1
EcoO109I RG/GNCCY 44 3 0 4 1 2 5 1 3 1 22
EcoP15I CAGCAG(25/27) 50 72 4 7 7 10 6 6 5 3 36
EcoRI § G/AATTC 5 5 1 1 1 1 1 1 1 1 0
EcoRI-HF® G/AATTC 5 5 1 1 1 1 1 1 1 1 0
EcoRV § GAT/ATC 9 21 0 1 1 1 1 1 3 0 0
EcoRV-HF® GAT/ATC 9 21 0 1 1 1 1 1 3 0 0
Esp3I CGTCTC(1/5) 21 14 1 1 2 2 0 2 2 2 16
FatI /CATG 183 181 14 26 38 23 21 23 31 11 148
FauI CCCGC(4/6) 147 90 10 10 14 17 11 28 22 5 24
Fnu4HI GC/NGC 411 380 17 42 52 42 47 49 52 19 156
FokI GGATG(9/13) 78 150 4 12 20 17 7 12 18 5 97
FseI GGCCGG/CC 3 0 0 0 0 0 0 0 0 0 0
FspEI CC(12/16)
FspI TGC/GCA 17 15 1 4 3 2 2 1 2 2 7
HaeII RGCGC/Y 76 48 6 11 6 9 3 7 12 3 26
HaeIII GG/CC 216 149 15 22 31 23 36 34 36 11 68
HgaI GACGC(5/10) 87 102 7 11 10 12 7 20 15 4 70
HhaI GCG/C 375 215 26 31 36 39 41 46 17 103
HincII GTY/RAC 25 35 1 2 9 7 4 7 6 1 61
HindIII § A/AGCTT 12 6 1 1 1 1 4 0 2 1 0
HindIII-HF® A/AGCTT 12 6 1 1 1 1 4 0 2 1 0
HinfI G/ANTC 72 148 26 10 31 9 11 16 20 6 218
HinP1I G/CGC 375 215 26 31 36 39 27 41 46 17 103
HpaI GTT/AAC 6 14 0 0 3 1 1 1 2 0 18
HpaII C/CGG 171 328 18 26 32 25 24 50 35 13 58
HphI GGTGA(8/7) 99 168 18 12 15 19 14 21 20 7 102
HpyAV CCTTC(6/5) 84 106 14 10 24 14 11 16 20 6 110
HpyCH4III ACN/GT 122 187 31 14 25 20 15 18 22 8 174
HpyCH4IV A/CGT 83 143 22 10 21 10 19 23 22 5 170
HpyCH4V TG/CA 207 273 18 21 39 28 30 26 27 13 116
Hpy188I TCN/GA 80 170 31 15 24 19 17 19 28 10 153
Hpy99I CGWCG/ 61 102 8 9 14 9 13 18 16 5 29
Hpy166II GTN/NAC 116 125 10 8 29 20 13 27 24 5 199
Hpy188III TC/NNGA 103 185 28 19 32 22 25 27 28 13 173
I-CeuI TAACTATAACGGTCCTAAGGTAGCGAA(-9/-13) 0 0 0 0 0 0
I-SceI TAGGGATAACAGGGTAAT(-9/-13) 0 0 0 0 0 0
KasI G/GCGCC 20 1 1 4 1 1 1 1 4 1 2
KpnI § GGTAC/C 8 2 1 0 0 0 2 0 1 1 5
KpnI-HF® GGTAC/C 8 2 1 0 0 0 2 0 1 1 5
LpnPI CCDG(10/14)
MboI /GATC 87 116 6 22 35 23 31 24 29 15 6
MboII GAAGA(8/7) 113 130 10 11 38 15 14 14 18 8 140
MfeI § C/AATTG 4 8 0 0 2 1 2 1 1 0 8
MfeI-HF® C/AATTG 4 8 0 0 2 1 2 1 1 0 8
MluCI /AATT 87 189 62 8 43 22 19 31 44 7 79
MluI § A/CGCGT 5 7 0 0 0 1 2 2 1 0 1
MluI-HF® A/CGCGT 5 7 0 0 0 1 2 2 1 0 1
MlyI GAGTC(5/5) 40 61 8 4 17 5 6 11 10 4 115
MmeI TCCRAC(20/18) 25 18 3 4 8 3 5 4 4 2 33
MnlI CCTC(7/6) 397 262 62 26 56 24 41 35 39 13 342
MscI TGG/CCA 17 18 1 1 0 1 2 0 1 0 2
MseI T/TAA 115 195 63 15 41 24 23 32 39 13 207
MslI CAYNN/NNRTG 35 62 3 7 10 10 6 7 9 3 38
MspA1I CMG/CKG 95 75 4 6 14 11 8 10 12 6 35
MspI C/CGG 171 328 18 26 32 25 24 50 35 13 58
MspJI CNNR(9/13)
MwoI GCNNNNN/NNGC 391 347 19 34 33 30 42 41 43 13 170
NaeI GCC/GGC 13 1 1 4 3 0 2 5 5 0 0
NarI GG/CGCC 20 1 1 4 1 1 1 1 4 1 2
Nb.BbvCI CCTCAGC 9 7 2 0 0 0 0 0 0 10
Nb.BsmI GAATGC 10 46 1 1 3 1 1 0 0 15
Nb.BsrDI GCAATG 14 44 3 2 7 4 4 4 2 18
Nb.BssSI CACGAG 11 8 0 3 5 3 2 4 3 31
Nb.BtsI GCAGTG 22 34 1 2 7 5 4 4 3 20
NciI CC/SGG 97 114 4 10 10 13 9 27 15 7 9
NcoI § C/CATGG 20 4 0 0 1 1 3 0 1 0 1
NcoI-HF® C/CATGG 20 4 0 0 1 1 3 0 1 0 1
NdeI CA/TATG 2 7 3 1 1 1 1 1 1 1 7
NgoMIV G/CCGGC 13 1 1 4 3 0 2 5 5 0 0
NheI-HF® G/CTAGC 4 1 0 1 4 0 1 1 0 1
NlaIII CATG/ 183 181 14 26 38 23 21 23 31 11 148
NlaIV GGN/NCC 178 82 18 24 22 14 20 22 30 11 99
NmeAIII GCCGAG(21/19) 17 8 0 3 2 3 2 2 4 1 14
NotI § GC/GGCCGC 7 0 0 0 1 1 1 1 1 0 0
NotI-HF® GC/GGCCGC 7 0 0 0 1 1 1 1 1 0 0
NruI § TCG/CGA 5 5 0 1 1 0 1 1 0 0 3
NruI-HF® TCG/CGA 5 5 0 1 1 0 1 1 0 0 3
NsiI § ATGCA/T 9 14 0 0 6 0 1 0 0 0 8
NsiI-HF® ATGCA/T 9 14 0 0 6 0 1 0 0 0 8
NspI RCATG/Y 41 32 6 4 9 3 5 5 5 3 24
Nt.AlwI GGATC(4/-5) 35 58 3 12 18 12 15 16 10 1
Nt.BbvCI CCTCAGC(-5/-7) 9 7 2 0 0 0 0 0 0 10
Nt.BsmAI GTCTC(1/-5) 60 37 5 3 11 8 8 10 4 95
Nt.BspQI GCTCTTC(1/-7) 7 10 0 1 2 1 1 1 1 4
Nt.BstNBI GAGTC(4/-5) 40 61 8 4 17 5 6 11 10 4 115
Nt.CviPII (0/-1)CCD 4148 4641 570 457 806 570 609 716 729 251 3575
PacI TTAAT/TAA 1 0 1 0 0 0 1 0 0 0 1
PaeR7I C/TCGAG 6 1 0 0 1 0 1 1 0 0 0
PaqCI CACCTGC(4/8) 9 12 0 0 0 0 1 0 0 0 5
PciI A/CATGT 9 2 3 1 3 1 2 1 1 1 6
PflFI GACN/NNGTC 12 2 0 1 1 1 1 1 2 0 1
PflMI CCANNNN/NTGG 18 14 0 2 3 1 5 2 2 0 8
PI-PspI TGGCAAACAGCTATTATGGGTATTATGGGT(-13/-17) 0 0 0 0 0 0
PI-SceI ATCTATGTCGGGTGCGGAGAAAGAGGTAAT(-15/-19) 0 0 0 0 0 0
PleI GAGTC(4/5) 40 61 8 4 17 5 6 11 10 4 115
PluTI GGCGC/C
PmeI GTTT/AAAC 1 2 0 0 0 0 1 1 0 0 2
PmlI CAC/GTG 10 3 0 0 0 0 1 0 1 0 1
PpuMI RG/GWCCY 23 3 0 2 1 2 1 0 0 0 12
PshAI GACNN/NNGTC 2 7 0 1 1 0 0 1 2 0 6
PsiI-v2 TTA/TAA 4 12 2 0 2 1 1 1 1 0 5
PspGI /CCWGG 136 71 7 6 15 14 19 19 14 5 2
PspOMI G/GGCCC 12 1 0 0 0 1 1 1 1 0 0
PspXI VC/TCGAGB 3 1 0 0 0 0 1 1 0 0 0
PstI § CTGCA/G 30 28 1 1 3 1 4 1 1 1 0
PstI-HF® CTGCA/G 30 28 1 1 3 1 4 1 1 1 0
PvuI § CGAT/CG 7 3 1 1 2 1 1 2 2 0
PvuI-HF® CGAT/CG 7 3 1 1 3 2 1 1 2 2 0
PvuII § CAG/CTG 24 15 3 1 3 3 3 3 4 2 3
PvuII-HF® CAG/CTG 24 15 3 1 3 3 3 3 4 2 3
RsaI GT/AC 83 113 19 3 15 7 14 6 11 3 168
RsrII CG/GWCCG 2 5 0 0 0 1 1 0 0 0 1
SacI § GAGCT/C 16 2 1 0 2 1 2 0 1 1 0
SacI-HF® GAGCT/C 16 2 1 0 2 1 2 0 1 1 0
SacII CCGC/GG 33 4 0 0 2 0 1 1 1 0 0
SalI § G/TCGAC 3 2 1 1 1 1 1 4 1 1 0
SalI-HF® G/TCGAC 3 2 1 1 1 1 1 4 1 1 0
SapI GCTCTTC(1/4) 7 10 0 1 2 1 3 1 1 1 4
Sau3AI /GATC 87 116 6 22 35 23 31 24 29 15 6
Sau96I G/GNCC 164 74 4 15 14 20 21 26 24 6 79
SbfI § CCTGCA/GG 3 5 1 0 1 1 1 0 1 1 0
SbfI-HF® CCTGCA/GG 3 5 1 0 1 1 1 0 1 1 0
ScaI-HF® AGT/ACT 5 5 0 1 2 1 1 2 1 4
ScrFI CC/NGG 233 185 11 16 25 27 28 46 29 12 11
SexAI A/CCWGGT 9 5 0 0 3 0 0 0 0 0 0
SfaNI GCATC(5/9) 85 169 7 22 18 20 17 23 19 8 96
SfcI C/TRYAG 47 38 7 4 9 4 10 6 8 4 48
SfiI GGCCNNNN/NGGCC 3 0 0 0 0 0 1 0 0 0 1
SfoI GGC/GCC 20 1 1 4 1 1 1 1 4 1 2
SgrAI CR/CCGGYG 6 6 0 1 0 0 0 1 1 0 0
SmaI CCC/GGG 12 3 1 0 1 0 1 0 0 1 0
SmlI C/TYRAG 29 17 4 6 8 5 10 8 9 4 75
SnaBI TAC/GTA 0 1 1 0 1 0 1 0 0 0 13
SpeI § A/CTAGT 3 0 0 0 0 0 1 1 1 0 2
SpeI-HF® A/CTAGT 3 0 0 0 0 0 1 1 1 0 2
SphI § GCATG/C 8 6 1 1 2 0 2 2 1 1 0
SphI-HF® GCATG/C 8 6 1 1 2 0 2 2 1 1 0
SrfI GCCC/GGGC
SspI § AAT/ATT 5 20 6 1 6 2 1 3 5 1 6
SspI-HF® AAT/ATT 5 20 6 1 6 2 1 3 5 1 6
StuI AGG/CCT 11 6 0 0 1 0 0 1 1 0 1
StyD4I /CCNGG 233 185 11 16 25 27 28 46 29 12 11
StyI-HF® C/CWWGG 44 10 0 1 4 1 4 2 4 0 36
SwaI ATTT/AAAT 1 0 1 0 0 0 1 1 1 0 1
TaqI-v2 T/CGA 50 121 12 7 32 16 15 22 17 4 111
TfiI G/AWTC 32 87 18 6 14 4 5 5 10 2 103
TseI G/CWGC 179 199 10 21 27 25 20 27 29 12 116
Tsp45I /GTSAC 73 81 9 9 9 7 5 12 13 4 108
TspMI C/CCGGG 12 3 1 0 1 0 1 0 0 1 0
TspRI NNCASTGNN/ 83 119 9 11 22 14 16 16 15 10 94
Tth111I GACN/NNGTC 12 2 0 1 1 1 1 1 2 0 1
XbaI T/CTAGA 5 1 1 0 1 0 1 1 1 1 3
XcmI CCANNNNN/NNNNTGG 14 12 0 0 1 3 0 3 4 0 8
XhoI C/TCGAG 6 1 0 0 1 0 1 1 0 0 0
XmaI C/CCGGG 12 3 1 0 1 0 1 0 0 1 0
XmnI GAANN/NNTTC 5 24 2 2 3 1 3 7 2 1 12
ZraI GAC/GTC 3 10 0 1 0 0 5 1 0 1 1

§ An HF version of this enzyme is available.

* For M13mp18, only double-stranded regions will be cut.
** Refers to the wild-type DNA substrate Hind III has 6 restriction sites on the wild-type lambda phage DNA, while NEB’s lambda phage mutant (Lambda DNA, NEB #N3011) has 7 Hind III sites.


Frequencies of Restriction Sites

The table below summarizes the frequencies with which restriction endonuclease sites occur in commonly used DNA molecules. Detailed restriction maps can be found on DNA sequences and maps. The sites listed in these tables were identified by computer analysis of published sequences. Although we have tried to ensure their accuracy, the sites have not necessarily been confirmed by experimentation. When the same specificity is displayed by several enzymes, the site is listed by the name of the enzyme that is available from New England Biolabs. Other enzymes with the same specificity are listed in the table of isoschizomers. Recognition sequences are written 5´ to 3´.

DNASU is a central repository for plasmid clones and collections that may also be helpful.

Enzyme Recognition Sequence Adeno-2 Lambda** M13mp18* pBR322 pKLAC2 pMAL-p5x pSNAPf pTXB1 pTYB21 pUC19 T7
AatII GACGT/C 3 10 0 1 0 0 5 1 0 1 1
AbaSI CNNNNNNNNNNN/NNNNNNNNNG
AccI GT/MKAC 17 9 1 2 5 1 2 5 3 1 33
Acc65I G/GTACC 8 2 1 0 0 0 2 0 1 1 5
AciI CCGC(-3/-1) 582 516 42 67 81 80 75 102 92 34 199
AclI AA/CGTT 3 7 2 4 2 5 3 12 9 2 19
AcuI CTGAAG(16/14) 23 40 0 2 8 4 5 2 2 2 1
AfeI AGC/GCT 13 2 1 4 0 2 0 1 2 0 0
AflII C/TTAAG 4 3 0 0 0 0 0 0 0 0 19
AflIII A/CRYGT 25 20 3 1 4 2 5 3 4 1 23
AgeI § A/CCGGT 5 13 0 0 1 0 1 1 0 0 2
AgeI-HF® A/CCGGT 5 13 0 0 1 0 1 1 0 0 2
AhdI GACNNN/NNGTC 9 9 0 1 1 2 1 2 2 1 14
AleI-v2 CACNN/NNGTG 10 20 1 0 1 0 1 0 2 0 8
AluI AG/CT 158 143 27 17 38 28 27 30 34 16 140
AlwI GGATC(4/5) 35 58 3 12 18 12 20 15 16 10 1
AlwNI CAGNNN/CTG 25 41 1 1 5 2 4 1 2 1 15
ApaI GGGCC/C 12 1 0 0 0 1 1 1 1 0 0
ApaLI G/TGCAC 7 4 0 3 3 6 4 4 4 3 1
ApeKI G/CWGC 179 199 10 21 27 25 20 27 29 12 116
ApoI § R/AATTY 29 58 11 0 19 5 3 5 7 1 13
ApoI-HF R/AATTY 29 58 11 0 19 5 3 5 7 1 13
AscI GG/CGCGCC 2 2 0 0 0 0 1 0 0 0 0
AseI AT/TAAT 3 17 7 1 5 4 7 10 4 3 12
AsiSI GCGAT/CGC 1 0 0 0 0 0 0 0 0 0 0
AvaI C/YCGRG 40 8 2 1 2 1 2 3 1 1 4
AvaII G/GWCC 73 35 1 8 7 9 6 6 8 2 54
AvrII C/CTAGG 2 2 0 0 0 0 1 0 0 0 3
BaeGI GKGCM/C 45 10 1 3 8 8 8 6 5 3 16
BaeI (10/15)ACNNNNGTAYC(12/7) 5 10 3 0 1 0 0 0 1 0 3
BamHI § G/GATCC 3 5 1 1 1 1 1 1 1 1 0
BamHI-HF® G/GATCC 3 5 1 1 1 1 1 1 1 1 0
BanI G/GYRCC 57 25 7 9 7 4 8 8 12 4 33
BanII GRGCY/C 57 7 2 2 5 2 5 3 6 1 1
BbsI § GAAGAC(2/6) 27 24 0 3 3 3 2 4 4 0 38
BbsI-HF ® GAAGAC(2/6) 27 24 0 3 3 3 2 4 4 0 38
BbvCI CCTCAGC(-5/-2) 9 7 2 0 0 0 0 0 0 0 10
BbvI GCAGC(8/12) 179 199 10 21 27 25 20 27 29 12 116
BccI CCATC(4/5) 62 145 14 9 22 16 8 14 25 3 121
BceAI ACGGC(12/14) 80 115 7 3 13 11 8 13 10 2 47
BcgI (10/12)CGANNNNNNTGC(12/10) 10 28 0 3 6 4 1 4 6 1 19
BciVI GTATCC(6/5) 9 26 0 2 3 4 3 4 4 2 23
BclI § T/GATCA 5 8 0 0 2 2 1 1 2 0 1
BclI-HF T/GATCA 5 8 0 0 2 2 1 1 2 0 1
BcoDI GTCTC(1/5) 60 37 5 3 11 8 4 8 9 4 95
BfaI C/TAG 54 13 5 5 19 3 14 8 8 4 60
BfuAI ACCTGC(4/8) 39 41 3 1 5 4 4 2 3 1 18
BglI GCCNNNN/NGGC 20 29 1 3 3 1 7 2 3 2 2
BglII A/GATCT 11 6 1 0 1 1 2 0 2 0 1
BlpI GC/TNAGC 8 6 0 0 0 1 0 1 1 0 20
BmgBI CACGTC(-3/-3) 15 17 0 0 1 1 2 0 0 0 8
BmrI ACTGGG(5/4) 22 4 1 5 2 5 11 8 2 6
BmtI § GCTAG/C 4 1 0 1 4 0 1 1 1 0 1
BmtI-HF® GCTAG/C 4 1 0 1 4 0 1 1 1 0 1
BpmI CTGGAG(16/14) 32 25 2 4 3 4 1 5 6 1 23
BpuEI CTTGAG(16/14) 19 13 4 6 7 5 9 7 9 4 56
Bpu10I CCTNAGC(-5/-2) 23 19 4 1 0 1 2 0 2 0 39
BsaAI YAC/GTR 22 14 5 1 4 0 3 2 4 0 35
BsaBI GATNN/NNATC 2 21 2 1 2 2 1 1 3 0 7
BsaHI GR/CGYC 44 40 1 6 6 5 8 12 7 3 8
BsaI-HF®v2 GGTCTC(1/5) 18 2 0 1 3 2 2 2 2 1 29
BsaJI C/CNNGG 234 105 9 8 18 10 17 15 15 5 85
BsaWI W/CCGGW 28 81 6 5 8 7 5 8 6 3 32
BsaXI (9/12)ACNNNNNCTCC(10/7) 29 19 4 0 3 1 1 2 3 1 12
BseRI GAGGAG(10/8) 63 19 1 0 2 0 3 0 0 0 13
BseYI CCCAGC(-5/-1) 31 32 3 2 3 4 5 4 4 1 29
BsgI GTGCAG(16/14) 34 41 0 1 4 6 3 5 4 0 21
BsiEI CGRY/CG 50 22 3 7 11 8 6 9 8 5 17
BsiHKAI GWGCW/C 38 28 3 8 8 9 9 7 7 5 24
BsiWI § C/GTACG 4 1 0 0 0 1 0 1 0 0 0
BsiWI-HF ® C/GTACG 4 1 0 0 0 1 0 1 0 0 0
BslI CCNNNNN/NNGG 216 176 17 20 18 16 26 31 25 6 90
BsmAI GTCTC(1/5) 60 37 5 3 11 8 8 10 4 95
BsmBI-v2 CGTCTC 21 14 1 1 2 2 0 2 2 2 16
BsmFI GGGAC(10/14) 59 38 2 4 4 1 5 4 6 0 46
BsmI GAATGC(1/-1) 10 46 1 1 3 1 5 1 0 0 15
BsoBI C/YCGRG 40 8 2 1 2 1 2 3 1 1 4
BspCNI CTCAG(9/7) 75 80 24 7 10 10 9 20 16 5 142
BspDI AT/CGAT 2 15 2 1 2 0 0 0 0 0 3
BspEI T/CCGGA 8 24 0 1 1 2 0 1 1 0 0
BspHI T/CATGA 3 8 1 4 2 1 2 2 3 3 13
Bsp1286I GDGCH/C 105 38 5 10 16 11 15 11 12 5 40
BspMI ACCTGC(4/8) 39 41 3 1 5 4 4 2 3 1 18
BspQI GCTCTTC(1/4) 7 10 0 1 2 1 3 1 1 1 4
BsrBI CCGCTC(-3/-3) 28 17 4 2 6 4 6 9 5 3 17
BsrDI GCAATG(2/0) 14 44 3 2 7 4 4 4 4 2 18
BsrFI-v2 R/CCGGY 40 61 1 7 9 2 6 11 9 1 3
BsrGI § T/GTACA 5 5 1 0 1 0 1 1 2 0 13
BsrGI-HF® T/GTACA 5 5 1 0 1 0 1 1 2 0 13
BsrI ACTGG(1/-1) 86 110 19 18 23 26 19 32 28 11 118
BssHII G/CGCGC 52 6 0 0 1 2 2 1 1 0 1
BssSI-v2 CACGAG(-5/-1) 11 8 0 3 5 3 4 2 4 3 31
BstAPI GCANNNN/NTGC 20 34 0 2 3 2 0 3 2 1 12
BstBI TT/CGAA 1 7 0 0 2 0 0 0 1 0 7
BstEII § G/GTNACC 10 13 0 0 1 1 0 1 1 0 1
BstEII-HF® G/GTNACC 10 13 0 0 1 1 2 1 1 0 1
BstNI CC/WGG 136 71 7 6 15 14 19 19 14 5 2
BstUI CG/CG 303 157 17 23 26 31 19 41 35 10 65
BstXI CCANNNNN/NTGG 10 13 0 0 2 4 1 4 3 0 11
BstYI R/GATCY 22 21 2 8 12 9 11 10 12 7 2
BstZ17I-HF ® GTATAC 3 3 0 1 3 0 1 1 1 0 8
Bsu36I CC/TNAGG 7 2 1 0 1 1 1 0 0 0 30
BtgI C/CRYGG 82 46 2 2 4 3 6 1 4 0 26
BtgZI GCGATG(10/14) 23 45 4 3 4 6 6 4 3 0 24
BtsCI GGATG(2/0) 78 150 4 12 20 17 7 12 18 5 97
BtsIMutI CAGTG(2/0) 20
BtsI-v2 GCAGTG(2/0) 22 34 1 2 7 5 5 4 4 3 20
Cac8I GCN/NGC 285 238 28 31 33 32 41 49 42 14 104
ClaI AT/CGAT 2 15 2 1 2 0 0 0 0 0 3
CspCI (11/13)CAANNNNNGTGG(12/10) 6 7 1 0 1 0 1 0 0 0 9
CviAII C/ATG 183 181 14 26 38 23 21 23 31 11 148
CviKI-1 RG/CY 680 692 103 73 131 86 119 112 121 45 562
CviQI G/TAC 83 113 19 3 15 7 14 6 11 3 168
DdeI C/TNAG 97 104 30 8 17 11 11 20 19 6 282
DpnI GA/TC 87 116 6 22 35 23 31 24 29 15 6
DpnII /GATC 87 116 6 22 35 23 31 24 29 15 6
DraI TTT/AAA 12 13 5 3 5 1 6 3 2 3 9
DraIII-HF® CACNNN/GTG 10 10 1 0 2 0 6 1 1 0 16
DrdI GACNNNN/NNGTC 6 3 1 2 5 2 2 4 4 2 11
EaeI Y/GGCCR 70 39 3 6 10 5 15 4 8 3 2
EagI-HF® C/GGCCG 19 2 0 1 4 1 2 2 2 0 0
EarI CTCTTC(1/4) 29 34 2 2 11 6 4 3 5 3 46
EciI GGCGGA(11/9) 29 32 2 4 6 6 8 9 8 3 2
Eco53kI GAG/CTC 16 2 1 0 2 1 2 0 1 1 0
EcoNI CCTNN/NNNAGG 10 9 0 1 3 0 0 2 1 0 1
EcoO109I RG/GNCCY 44 3 0 4 1 2 5 1 3 1 22
EcoP15I CAGCAG(25/27) 50 72 4 7 7 10 6 6 5 3 36
EcoRI § G/AATTC 5 5 1 1 1 1 1 1 1 1 0
EcoRI-HF® G/AATTC 5 5 1 1 1 1 1 1 1 1 0
EcoRV § GAT/ATC 9 21 0 1 1 1 1 1 3 0 0
EcoRV-HF® GAT/ATC 9 21 0 1 1 1 1 1 3 0 0
Esp3I CGTCTC(1/5) 21 14 1 1 2 2 0 2 2 2 16
FatI /CATG 183 181 14 26 38 23 21 23 31 11 148
FauI CCCGC(4/6) 147 90 10 10 14 17 11 28 22 5 24
Fnu4HI GC/NGC 411 380 17 42 52 42 47 49 52 19 156
FokI GGATG(9/13) 78 150 4 12 20 17 7 12 18 5 97
FseI GGCCGG/CC 3 0 0 0 0 0 0 0 0 0 0
FspEI CC(12/16)
FspI TGC/GCA 17 15 1 4 3 2 2 1 2 2 7
HaeII RGCGC/Y 76 48 6 11 6 9 3 7 12 3 26
HaeIII GG/CC 216 149 15 22 31 23 36 34 36 11 68
HgaI GACGC(5/10) 87 102 7 11 10 12 7 20 15 4 70
HhaI GCG/C 375 215 26 31 36 39 41 46 17 103
HincII GTY/RAC 25 35 1 2 9 7 4 7 6 1 61
HindIII § A/AGCTT 12 6 1 1 1 1 4 0 2 1 0
HindIII-HF® A/AGCTT 12 6 1 1 1 1 4 0 2 1 0
HinfI G/ANTC 72 148 26 10 31 9 11 16 20 6 218
HinP1I G/CGC 375 215 26 31 36 39 27 41 46 17 103
HpaI GTT/AAC 6 14 0 0 3 1 1 1 2 0 18
HpaII C/CGG 171 328 18 26 32 25 24 50 35 13 58
HphI GGTGA(8/7) 99 168 18 12 15 19 14 21 20 7 102
HpyAV CCTTC(6/5) 84 106 14 10 24 14 11 16 20 6 110
HpyCH4III ACN/GT 122 187 31 14 25 20 15 18 22 8 174
HpyCH4IV A/CGT 83 143 22 10 21 10 19 23 22 5 170
HpyCH4V TG/CA 207 273 18 21 39 28 30 26 27 13 116
Hpy188I TCN/GA 80 170 31 15 24 19 17 19 28 10 153
Hpy99I CGWCG/ 61 102 8 9 14 9 13 18 16 5 29
Hpy166II GTN/NAC 116 125 10 8 29 20 13 27 24 5 199
Hpy188III TC/NNGA 103 185 28 19 32 22 25 27 28 13 173
I-CeuI TAACTATAACGGTCCTAAGGTAGCGAA(-9/-13) 0 0 0 0 0 0
I-SceI TAGGGATAACAGGGTAAT(-9/-13) 0 0 0 0 0 0
KasI G/GCGCC 20 1 1 4 1 1 1 1 4 1 2
KpnI § GGTAC/C 8 2 1 0 0 0 2 0 1 1 5
KpnI-HF® GGTAC/C 8 2 1 0 0 0 2 0 1 1 5
LpnPI CCDG(10/14)
MboI /GATC 87 116 6 22 35 23 31 24 29 15 6
MboII GAAGA(8/7) 113 130 10 11 38 15 14 14 18 8 140
MfeI § C/AATTG 4 8 0 0 2 1 2 1 1 0 8
MfeI-HF® C/AATTG 4 8 0 0 2 1 2 1 1 0 8
MluCI /AATT 87 189 62 8 43 22 19 31 44 7 79
MluI § A/CGCGT 5 7 0 0 0 1 2 2 1 0 1
MluI-HF® A/CGCGT 5 7 0 0 0 1 2 2 1 0 1
MlyI GAGTC(5/5) 40 61 8 4 17 5 6 11 10 4 115
MmeI TCCRAC(20/18) 25 18 3 4 8 3 5 4 4 2 33
MnlI CCTC(7/6) 397 262 62 26 56 24 41 35 39 13 342
MscI TGG/CCA 17 18 1 1 0 1 2 0 1 0 2
MseI T/TAA 115 195 63 15 41 24 23 32 39 13 207
MslI CAYNN/NNRTG 35 62 3 7 10 10 6 7 9 3 38
MspA1I CMG/CKG 95 75 4 6 14 11 8 10 12 6 35
MspI C/CGG 171 328 18 26 32 25 24 50 35 13 58
MspJI CNNR(9/13)
MwoI GCNNNNN/NNGC 391 347 19 34 33 30 42 41 43 13 170
NaeI GCC/GGC 13 1 1 4 3 0 2 5 5 0 0
NarI GG/CGCC 20 1 1 4 1 1 1 1 4 1 2
Nb.BbvCI CCTCAGC 9 7 2 0 0 0 0 0 0 10
Nb.BsmI GAATGC 10 46 1 1 3 1 1 0 0 15
Nb.BsrDI GCAATG 14 44 3 2 7 4 4 4 2 18
Nb.BssSI CACGAG 11 8 0 3 5 3 2 4 3 31
Nb.BtsI GCAGTG 22 34 1 2 7 5 4 4 3 20
NciI CC/SGG 97 114 4 10 10 13 9 27 15 7 9
NcoI § C/CATGG 20 4 0 0 1 1 3 0 1 0 1
NcoI-HF® C/CATGG 20 4 0 0 1 1 3 0 1 0 1
NdeI CA/TATG 2 7 3 1 1 1 1 1 1 1 7
NgoMIV G/CCGGC 13 1 1 4 3 0 2 5 5 0 0
NheI-HF® G/CTAGC 4 1 0 1 4 0 1 1 0 1
NlaIII CATG/ 183 181 14 26 38 23 21 23 31 11 148
NlaIV GGN/NCC 178 82 18 24 22 14 20 22 30 11 99
NmeAIII GCCGAG(21/19) 17 8 0 3 2 3 2 2 4 1 14
NotI § GC/GGCCGC 7 0 0 0 1 1 1 1 1 0 0
NotI-HF® GC/GGCCGC 7 0 0 0 1 1 1 1 1 0 0
NruI § TCG/CGA 5 5 0 1 1 0 1 1 0 0 3
NruI-HF® TCG/CGA 5 5 0 1 1 0 1 1 0 0 3
NsiI § ATGCA/T 9 14 0 0 6 0 1 0 0 0 8
NsiI-HF® ATGCA/T 9 14 0 0 6 0 1 0 0 0 8
NspI RCATG/Y 41 32 6 4 9 3 5 5 5 3 24
Nt.AlwI GGATC(4/-5) 35 58 3 12 18 12 15 16 10 1
Nt.BbvCI CCTCAGC(-5/-7) 9 7 2 0 0 0 0 0 0 10
Nt.BsmAI GTCTC(1/-5) 60 37 5 3 11 8 8 10 4 95
Nt.BspQI GCTCTTC(1/-7) 7 10 0 1 2 1 1 1 1 4
Nt.BstNBI GAGTC(4/-5) 40 61 8 4 17 5 6 11 10 4 115
Nt.CviPII (0/-1)CCD 4148 4641 570 457 806 570 609 716 729 251 3575
PacI TTAAT/TAA 1 0 1 0 0 0 1 0 0 0 1
PaeR7I C/TCGAG 6 1 0 0 1 0 1 1 0 0 0
PaqCI CACCTGC(4/8) 9 12 0 0 0 0 1 0 0 0 5
PciI A/CATGT 9 2 3 1 3 1 2 1 1 1 6
PflFI GACN/NNGTC 12 2 0 1 1 1 1 1 2 0 1
PflMI CCANNNN/NTGG 18 14 0 2 3 1 5 2 2 0 8
PI-PspI TGGCAAACAGCTATTATGGGTATTATGGGT(-13/-17) 0 0 0 0 0 0
PI-SceI ATCTATGTCGGGTGCGGAGAAAGAGGTAAT(-15/-19) 0 0 0 0 0 0
PleI GAGTC(4/5) 40 61 8 4 17 5 6 11 10 4 115
PluTI GGCGC/C
PmeI GTTT/AAAC 1 2 0 0 0 0 1 1 0 0 2
PmlI CAC/GTG 10 3 0 0 0 0 1 0 1 0 1
PpuMI RG/GWCCY 23 3 0 2 1 2 1 0 0 0 12
PshAI GACNN/NNGTC 2 7 0 1 1 0 0 1 2 0 6
PsiI-v2 TTA/TAA 4 12 2 0 2 1 1 1 1 0 5
PspGI /CCWGG 136 71 7 6 15 14 19 19 14 5 2
PspOMI G/GGCCC 12 1 0 0 0 1 1 1 1 0 0
PspXI VC/TCGAGB 3 1 0 0 0 0 1 1 0 0 0
PstI § CTGCA/G 30 28 1 1 3 1 4 1 1 1 0
PstI-HF® CTGCA/G 30 28 1 1 3 1 4 1 1 1 0
PvuI § CGAT/CG 7 3 1 1 2 1 1 2 2 0
PvuI-HF® CGAT/CG 7 3 1 1 3 2 1 1 2 2 0
PvuII § CAG/CTG 24 15 3 1 3 3 3 3 4 2 3
PvuII-HF® CAG/CTG 24 15 3 1 3 3 3 3 4 2 3
RsaI GT/AC 83 113 19 3 15 7 14 6 11 3 168
RsrII CG/GWCCG 2 5 0 0 0 1 1 0 0 0 1
SacI § GAGCT/C 16 2 1 0 2 1 2 0 1 1 0
SacI-HF® GAGCT/C 16 2 1 0 2 1 2 0 1 1 0
SacII CCGC/GG 33 4 0 0 2 0 1 1 1 0 0
SalI § G/TCGAC 3 2 1 1 1 1 1 4 1 1 0
SalI-HF® G/TCGAC 3 2 1 1 1 1 1 4 1 1 0
SapI GCTCTTC(1/4) 7 10 0 1 2 1 3 1 1 1 4
Sau3AI /GATC 87 116 6 22 35 23 31 24 29 15 6
Sau96I G/GNCC 164 74 4 15 14 20 21 26 24 6 79
SbfI § CCTGCA/GG 3 5 1 0 1 1 1 0 1 1 0
SbfI-HF® CCTGCA/GG 3 5 1 0 1 1 1 0 1 1 0
ScaI-HF® AGT/ACT 5 5 0 1 2 1 1 2 1 4
ScrFI CC/NGG 233 185 11 16 25 27 28 46 29 12 11
SexAI A/CCWGGT 9 5 0 0 3 0 0 0 0 0 0
SfaNI GCATC(5/9) 85 169 7 22 18 20 17 23 19 8 96
SfcI C/TRYAG 47 38 7 4 9 4 10 6 8 4 48
SfiI GGCCNNNN/NGGCC 3 0 0 0 0 0 1 0 0 0 1
SfoI GGC/GCC 20 1 1 4 1 1 1 1 4 1 2
SgrAI CR/CCGGYG 6 6 0 1 0 0 0 1 1 0 0
SmaI CCC/GGG 12 3 1 0 1 0 1 0 0 1 0
SmlI C/TYRAG 29 17 4 6 8 5 10 8 9 4 75
SnaBI TAC/GTA 0 1 1 0 1 0 1 0 0 0 13
SpeI § A/CTAGT 3 0 0 0 0 0 1 1 1 0 2
SpeI-HF® A/CTAGT 3 0 0 0 0 0 1 1 1 0 2
SphI § GCATG/C 8 6 1 1 2 0 2 2 1 1 0
SphI-HF® GCATG/C 8 6 1 1 2 0 2 2 1 1 0
SrfI GCCC/GGGC
SspI § AAT/ATT 5 20 6 1 6 2 1 3 5 1 6
SspI-HF® AAT/ATT 5 20 6 1 6 2 1 3 5 1 6
StuI AGG/CCT 11 6 0 0 1 0 0 1 1 0 1
StyD4I /CCNGG 233 185 11 16 25 27 28 46 29 12 11
StyI-HF® C/CWWGG 44 10 0 1 4 1 4 2 4 0 36
SwaI ATTT/AAAT 1 0 1 0 0 0 1 1 1 0 1
TaqI-v2 T/CGA 50 121 12 7 32 16 15 22 17 4 111
TfiI G/AWTC 32 87 18 6 14 4 5 5 10 2 103
TseI G/CWGC 179 199 10 21 27 25 20 27 29 12 116
Tsp45I /GTSAC 73 81 9 9 9 7 5 12 13 4 108
TspMI C/CCGGG 12 3 1 0 1 0 1 0 0 1 0
TspRI NNCASTGNN/ 83 119 9 11 22 14 16 16 15 10 94
Tth111I GACN/NNGTC 12 2 0 1 1 1 1 1 2 0 1
XbaI T/CTAGA 5 1 1 0 1 0 1 1 1 1 3
XcmI CCANNNNN/NNNNTGG 14 12 0 0 1 3 0 3 4 0 8
XhoI C/TCGAG 6 1 0 0 1 0 1 1 0 0 0
XmaI C/CCGGG 12 3 1 0 1 0 1 0 0 1 0
XmnI GAANN/NNTTC 5 24 2 2 3 1 3 7 2 1 12
ZraI GAC/GTC 3 10 0 1 0 0 5 1 0 1 1

§ An HF version of this enzyme is available.

* For M13mp18, only double-stranded regions will be cut.
** Refers to the wild-type DNA substrate Hind III has 6 restriction sites on the wild-type lambda phage DNA, while NEB’s lambda phage mutant (Lambda DNA, NEB #N3011) has 7 Hind III sites.