Draw a dna molecule that has five randomly spaced restriction sites for a spedific palindrome. how many fragments would be produced if each sile were cut by that specific restriction enzyme? 5. 6. label each fragment. 7. rank them in order of size from largest to smallest in this diagram a and b are different palindrome sequences on a dna strand. only the restriction enzyme that recognizes site b is present. 8. explain why only two fragments would be produced.

Figures attatched

Explanation:

1. Restriction enzymes cut the DNA in specific places of a specific sequence. Each restriction enzyme has different sequences of recognition.

2. Palindromic sequences are sequences that can be read the same in both senses (3' and 5'), for that reason restriction enzymes can cut both DNA strands

Now, for the first image: if a molecule has 5 restriction sites that are known for a single restriction enzyme, the enzyme will cut the DNA 5 times. If you see the image, the number of fragments is the number of restriction sites +1. In this case, it is 5+1=6 DNA fragments.

In the case of two different palindromic sequences, two different restriction enzymes recognize different sequence. If just one enzyme is present, the only cutting site will be the one that has the palindromic sequence recognizable. So, the number of fragments will be 1+1=2 DNA fragments (figure 2)

prophase: we can observe the chromosomes to coil up and condense. they become visible. nuclear membrane and envelope disintigrate.s

metaphase: chromosomes align at the center, the metaphase plate. spindle fibers attach to centromeres of the sister chromatids. they meet at kinetochores.

anaphase: sister chromatids are pulled towards each opposing centromere, set on different poles.

telophase: chromosomes are now clustered up only on either sides of the cell. nuclear membranes reform, and a new cell wall forms along the cell plate. this will lead to cytokinesis, the process that divides the cell membrane and forms the cell wall.

b c a d mark me