1.
strand 1: AGTTCCGATATCGCTC
strand 2: GCAGTCGTAGCGATGTA
strand 3: ATCGCTCGATCGACTACATCGC
AGTTCCGATATCGCTC
__________ATCGCTCGATCGACTACATCGC
________________________________GCAGTCGTAGCGATGTA
2. In my oppinion I would think that the shotgun method would be more productive because there has been no previous sequencing done on this fungus and the shotgun method is a brute-force method and would probably take fewer sequencing reactions then creating a physical map first and sequencing from there. (This is just my guess though, if anyone else has an oppinion please post.)
3. The Drosophila genome is 20% exons, 16.7% introns and 63.3% intergenic sequence.
4. What probably happened was that were a known number of genes within the Drosophila genome and once the genes and their locations were identified along with the bases around them in the genome they decided there was no need to sequence 60 million exta bases of junk DNA.
Monday, February 5, 2007
Sunday, February 4, 2007
Questions from January 29, 2007
1.
a. the restriction enzyme EcoR1 would only cut the plasmid in one place creating a single 6 kb strand. When put through gel electrophoresis the strand would not travel very far because it is such a large strand.
b. the restriction enzyme BamH1 would cut the plasmid in two locations creating a 1.5 kb strand and a 4.5 kp strand. When put through gel electrophoresis the 1.5 kp strand would travel quickly through the gel while the 4.5 kp strand would move more slowly but farther than a 6 kp strand.
c. if both restriction enzymes are used the plasmid is cut in three places creating 2 1.5 kb strands and one 3 kb strand. When put through gel electrophoresis the 1.5 kb strands would move quickly through the gel while the 3 kb strand moves slightly slower, you would not be able to tell the two 1.5 kb pairs apart however.
2. Sal1 is compatible with Xho1
Xba1 is compatible with Spe1
Kas1 is not compatible with Bbe1 because the sticky ends of the two cuts are identical, not compliments, both are CGCG
a. the restriction enzyme EcoR1 would only cut the plasmid in one place creating a single 6 kb strand. When put through gel electrophoresis the strand would not travel very far because it is such a large strand.
b. the restriction enzyme BamH1 would cut the plasmid in two locations creating a 1.5 kb strand and a 4.5 kp strand. When put through gel electrophoresis the 1.5 kp strand would travel quickly through the gel while the 4.5 kp strand would move more slowly but farther than a 6 kp strand.
c. if both restriction enzymes are used the plasmid is cut in three places creating 2 1.5 kb strands and one 3 kb strand. When put through gel electrophoresis the 1.5 kb strands would move quickly through the gel while the 3 kb strand moves slightly slower, you would not be able to tell the two 1.5 kb pairs apart however.
2. Sal1 is compatible with Xho1
Xba1 is compatible with Spe1
Kas1 is not compatible with Bbe1 because the sticky ends of the two cuts are identical, not compliments, both are CGCG
Questions from January 24, 2007
1. Same as Q4 from 1/22/07
2. The actual mRNA is the shortest strand, with the coding sequence the second shortest and the pre-mRNA (junk DNA I believe) is the longest by far. (I am not positive about this one, if anyone has another oppinion please post.)
3. complimentary DNA strand:
TACATAGTCTATCTGGCCTGCCATCAACAGACC
corresponding mRNA strand:
AUGUAUCAGAUAGACCGGACGGUAGUUGUCUGG
corresponding peptide:
MetTyrGlnIleAspArgThrValValValTrp
4. If we assume there are 4 linearly arranged nucleic acids in this Martian life form like in all life forms on earth and there are 100 different amino acids in these Martian life forms then the code is based on a 4:1 relationship and is therefore a series of four non overlapping nucleic acids code for a single amino acid. Simple mathematics shows that triplets would only allow for 64 different combinations while quartets would allow for 256 combinations which is more than enough for the 100 different Martian amino acids.
5. (from highest to lowest) skin color, autism, cyctic fibrosis, sexual orientation, left-handedness, intelligence, weight, addicition, religion. (those are just my thoughts, I'd be interested to hear others)
2. The actual mRNA is the shortest strand, with the coding sequence the second shortest and the pre-mRNA (junk DNA I believe) is the longest by far. (I am not positive about this one, if anyone has another oppinion please post.)
3. complimentary DNA strand:
TACATAGTCTATCTGGCCTGCCATCAACAGACC
corresponding mRNA strand:
AUGUAUCAGAUAGACCGGACGGUAGUUGUCUGG
corresponding peptide:
MetTyrGlnIleAspArgThrValValValTrp
4. If we assume there are 4 linearly arranged nucleic acids in this Martian life form like in all life forms on earth and there are 100 different amino acids in these Martian life forms then the code is based on a 4:1 relationship and is therefore a series of four non overlapping nucleic acids code for a single amino acid. Simple mathematics shows that triplets would only allow for 64 different combinations while quartets would allow for 256 combinations which is more than enough for the 100 different Martian amino acids.
5. (from highest to lowest) skin color, autism, cyctic fibrosis, sexual orientation, left-handedness, intelligence, weight, addicition, religion. (those are just my thoughts, I'd be interested to hear others)
Questions from January 22, 2007
1. If the distance between base pairs in .34 nm and there are 6 billion base pairs in each human cell, then the total distance of all base pairs strectched end-to-end would be 2.04 billion nm or 2.04 m.
(0.34nm * 6,000,000,000 bp = 2,040,000,000nm / 1,000,000,000(nm/m) = 2.04m)
Since there are 100 trillion cells in the body, that means that if all the DNA in every was stretched end-to-end the total length would be 204 trillion meters long or 204 billion kilometers.
(2.04m * 100,000,000,000,000 cells = 204,000,000,000,000m / 1000(m/km) = 204,000,000,000km)
2. (picture)
3. Since Chargaff's rule states that %A = %T and %G = %C the B19 virus does not follow Chargaff's rule. This suggests that either the virus reproduces in a different way (highly unlikely) or that it needs to take on bases from another source to survive and duplicate.
4. The major error in the video is quickly evident. If you watch carefully when the double helix divides at the beginning of the video nucleotides begin attaching to the separated end of the DNA molecule but because DNA helices are anti-parallel nucleotides on both backbones cannot attach going the same direction. Essentially DNA can only replicate in the 5' to 3' direction but in this video it is replicating in both directions.
(0.34nm * 6,000,000,000 bp = 2,040,000,000nm / 1,000,000,000(nm/m) = 2.04m)
Since there are 100 trillion cells in the body, that means that if all the DNA in every was stretched end-to-end the total length would be 204 trillion meters long or 204 billion kilometers.
(2.04m * 100,000,000,000,000 cells = 204,000,000,000,000m / 1000(m/km) = 204,000,000,000km)
2. (picture)
3. Since Chargaff's rule states that %A = %T and %G = %C the B19 virus does not follow Chargaff's rule. This suggests that either the virus reproduces in a different way (highly unlikely) or that it needs to take on bases from another source to survive and duplicate.
4. The major error in the video is quickly evident. If you watch carefully when the double helix divides at the beginning of the video nucleotides begin attaching to the separated end of the DNA molecule but because DNA helices are anti-parallel nucleotides on both backbones cannot attach going the same direction. Essentially DNA can only replicate in the 5' to 3' direction but in this video it is replicating in both directions.
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