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Concept 24
The RNA message is sometimes edited.
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HI! Phil Sharp was one of the first to use restriction enzymes to "map" DNA molecules. You have a piece of bacterial DNA that is 15,000 base pairs long (15 kb). Youââ¬â¢re going to map this 15-kb DNA using three restriction enzymes: BamHI, EcoRI and HindIII. You set up three reactions to cut the 15-kb DNA sample, each with one of the three enzymes. You then run the digested results on an agarose gel. Click on the 7 kb EcoRI DNA fragment. No, this is the 8-kb EcoRI piece. No, this is the 7-kb BamHI piece. No, this is the 3-kb BamHI piece. No, this is the 5-kb BamHI piece. No, this is the 13-kb HindIII piece. No, this is the 2-kb HindIII piece. This is the 7 kb EcoRI fragment. EcoRI cuts the 15 kb DNA once to give a 7 kb and an 8 kb fragment. Based on the information you have, which of the following is NOT a possible map? This is a possible map. Map# 1 can't be right because it shows two HindIII sites and the sizes of the bands in the gel do not match the sizes of the HindIII pieces on the map. Since all of the other maps are consistent with the data, how are you going to decide which, if any of these three, is correct? No decision needs to be made; each map is correct. (No, the maps canââ¬â¢t all be correct.) You need to set up additional digests using combinations of two restriction enzymes: BamHI/EcoRI, BamHI/HindIII and EcoRI/HindIII. (That is correct.) You need to cut the DNA using completely new enzymes. (You can, but this doesnââ¬â¢t help you resolve the map with BamHI, EcoRI and HindIII.) By combining the data from double enzyme digests with the single digests, you can map restriction sites relative to each other. The bands in the EcoRI/HindIII lane add up to 15 kb ââ¬â the length of the starting DNA. Why donââ¬â¢t the bands in the BamHI/EcoRI and BamHI/HindIII lanes add up to 15 kb? There is no reason that the bands should add up to 15 kb. (No, if 15-kb DNA "goes in," 15 kb of DNA should "come out." The pieces must add up to 15 kb.) The bands do add up to 15 kb. Some of the bands are the same size. (That is correct.) Cutting with two enzymes chews up the DNA and there are pieces missing. (No, restriction enzymes cut at specific sites. They don't degrade DNA.) In each lane the fragments do add up to 15 kb. Some of the bands are actually two DNA fragments of the same size migrating to the same place. These are known as doublets. For example, in the BamHI/EcoRI lane, there are three bands: 5 kb, 3 kb and 2 kb. The 5 kb band must be a doublet ââ¬â two 5 kb bands migrating to the same place ââ¬â in this example, this is how the bands add up to 15 kb (5 + 5 + 3 + 2). Which of the three bands in the BamHI/HindIII lane is a doublet? 7 kb (No, if the 7 kb is a doublet that would be 7 + 7 + 3 + 2 = 19 kb ââ¬â too much DNA.) 3 kb (That is correct.) 2 kb (No, if the 2 kb is a doublet that would be 7 + 3 + 2 + 2 = 14 kb ââ¬â not enough DNA.) The 3 kb band must be a doublet of two 3 kb DNA pieces migrating to the same place: 7 + 3 + 3 + 2 = 15 kb. With the additional data from the double digests, which of the following is the correct map? No, the EcoRI site is missing. No, there is one BamHI site instead of two and two EcoRI sites instead of one. No, the map fits the data for the double digests but not for the single digest. This map is consistent with all the data from the single- and double-enzyme digests. When the 15 kb DNA is cut with the two enzymes BamHI and EcoRI, we see three bands: 5 kb, 3 kb and 2 kb. The 5 kb band is a doublet. When the 15 kb DNA is cut with the two enzymes BamHI and HindIII, we see three bands: 7 kb, 3 kb and 2 kb. The 3 kb band is a doublet. When the 15 kb DNA is cut with the two enzymes EcoRI and HindIII, we see three bands: 8 kb, 5 kb and 2 kb. Here is an electron micrograph of a DNA/RNA hybrid. The DNA is single-stranded and is outlined in red. The gene on this piece of DNA codes for an RNA outlined in green. How many R-loops are there in this DNA/RNA hybrid? None (No, there are R-loops in this DNA/RNA hybrid.) 1 (No, there is more than one R-loop in this DNA/RNA hyrid.) The number of R-loops varies for any one DNA/RNA hybrid. (No, there is a set number of R-loops for any one DNA/RNA hybrid.) 7 (That is correct.) 10 (No, there are less than ten R-loops in this DNA/RNA hybrid.) There are seven R-loops in this DNA/RNA hybrid. They are labelled A-G. If there are seven R-loops in this DNA/RNA hybrid, how many introns are there in this gene? 1 (There is more than one intron in this gene.) 6 (There are more than six introns in this gene.) 7 (That is correct.) 8 (There are less than eight introns in this gene.) DNA regions that do not hybridize to the RNA loop out. Since in this example, there are seven R-loops, there must be seven introns in this gene. How many exons are there in this gene? 1 (There is more than one exon in this gene.) 6 (There are more than six exons in this gene.) 7 (This is the number of R-loops and the number of introns.) 8 (That is correct.) Exons are regions where the DNA and the RNA hybridized. In this example there are seven R-loops, so there are eight points of contact between the DNA and the RNA. There are eight exons. We can show the exon/intron arrangement in a linear map. CONGRATULATIONS!!! YOU'RE SO SMART!
CLASSICAL GENETICS
DNA and proteins are key molecules of the cell nucleus.
One gene makes one protein.
A gene is made of DNA.
Bacteria and viruses have DNA too.
The DNA molecule is shaped like a twisted ladder.
A half DNA ladder is a template for copying the whole.
RNA is an intermediary between DNA and protein.
DNA words are three letters long.
A gene is a discrete sequence of DNA nucleotides.
The RNA message is sometimes edited.
Some viruses store genetic information in RNA.
RNA was the first genetic molecule.
Mutations are changes in genetic information.
Some types of mutations are automatically repaired.
GENETIC ORGANIZATION AND CONTROL