BamHI is a restriction enzyme that recognizes the specific DNA sequence "GGATCC" and cuts between the G and the A. The number of DNA fragments produced by BamHI cutting a DNA molecule depends on the number of BamHI recognition sites present in that molecule. Each recognition site will result in one additional fragment; thus, if there are n cut sites, the DNA will be divided into n+1 fragments.
If a plasmid is cut at more than one site by restriction enzymes, it would result in multiple DNA fragments. These fragments can be ligated back together in different combinations, resulting in plasmids with different sizes or configurations. This can lead to the creation of recombinant plasmids with altered properties compared to the original plasmid.
EcoR1 cuts double-stranded DNA at specific recognition sites generating two fragments, so to generate 4 fragments, EcoR1 would need to cut the DNA twice.
On a gel electrophoresis setup, small DNA pieces are expected to migrate further down the gel compared to larger DNA fragments. This is because smaller fragments can move more easily through the gel matrix, while larger fragments encounter more resistance. As a result, the smallest DNA pieces will be located closer to the bottom of the gel, while larger fragments will remain nearer to the wells where the samples were loaded.
It depends on the specific cleavage sites for trypsin and V8 protease within the polypeptide sequence, as well as the length and composition of the polypeptide. Generally, these proteases cleave at specific amino acid residues, resulting in smaller peptide fragments. The number of fragments would need to be determined by analyzing the sequence of the polypeptide and the cleavage specificity of the proteases.
It depends on the length of your four fragments. If the 4 fragments are same in length, you can not distinguis the band on the gel as all the (4x500) molecules run at the same length. If they are different in size let say 100,200,300,400 you can see four distinct bands. The intensity of the bands may be stronger at 400 and drops down to the lower fragments because of the higher molecular mass.
Three.To see why, cut a piece of string in two places! Of course, strictly you would not be able to see only three fragments. You would amplify the DNA before carrying out electrophoresis. That way, you would get perhaps 200 million copies of each fragment, and they would show up. Also, you would only be able to distinguish the fragments if they were different lengths. Electrophoresis separates pieces of DNA by length.
1. Which enzyme(s) would cut the human DNA shown in Part A on both sides of the vgp gene, but not inside the gene? Answer: BamHI, HaeIII, and HindIII 2. Which enzymes(s) would cut the plasmid without disrupting the function of the amp^R gene? Answer: BamHI, EcoRI, and HaeIII 3. Which enzyme(s) would produce sticky ends when cutting both the human DNA and the plasmid? Answer: BamHI, EcoRI, and HindIII 4. Which one restriction enzyme satisfies all three of the requirements listed above? Answer: BamHI only
The sites would not be able to trade, putting everything at a stand-still.
If a plasmid is cut at more than one site by restriction enzymes, it would result in multiple DNA fragments. These fragments can be ligated back together in different combinations, resulting in plasmids with different sizes or configurations. This can lead to the creation of recombinant plasmids with altered properties compared to the original plasmid.
EcoR1 cuts double-stranded DNA at specific recognition sites generating two fragments, so to generate 4 fragments, EcoR1 would need to cut the DNA twice.
all the idol fragments would be by in the building by the bridge
If the plasmid has 3 recognition sequences for a given restriction endonuclease, then 4 linear DNA fragments are obtained because, if the DNA is linear then the number of fragments obtained is (N+1) whereas if the DNA is circular then the number of fragments obtained will be N for N recognition sequences for the given restriction endonuclease in a plasmid.
On a gel electrophoresis setup, small DNA pieces are expected to migrate further down the gel compared to larger DNA fragments. This is because smaller fragments can move more easily through the gel matrix, while larger fragments encounter more resistance. As a result, the smallest DNA pieces will be located closer to the bottom of the gel, while larger fragments will remain nearer to the wells where the samples were loaded.
It depends on the specific cleavage sites for trypsin and V8 protease within the polypeptide sequence, as well as the length and composition of the polypeptide. Generally, these proteases cleave at specific amino acid residues, resulting in smaller peptide fragments. The number of fragments would need to be determined by analyzing the sequence of the polypeptide and the cleavage specificity of the proteases.
It depends on the length of your four fragments. If the 4 fragments are same in length, you can not distinguis the band on the gel as all the (4x500) molecules run at the same length. If they are different in size let say 100,200,300,400 you can see four distinct bands. The intensity of the bands may be stronger at 400 and drops down to the lower fragments because of the higher molecular mass.
Rock fragments can take different paths in the rock cycle due to various factors such as erosion, deposition, heat, and pressure. These processes can result in the transformation of rock fragments into different types of rocks, ultimately leading to their divergent pathways in the cycle. Additionally, the influence of external factors like water, wind, and temperature variations can also contribute to the unique journey of rock fragments in the rock cycle.
It would have to be weathered & the fragments cemented together.