Date post: | 02-Jan-2016 |
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Semi Conservative process of DNA Replication
The parent molecule has two complementary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C.
The first step in replication is separation of the two DNA strands.
Each parental strand now serves as a template that determines the order of nucleotides along a new, complementary strand.
Complimentary strands
5’ – AUGAAAUCCUAG – 3’ mRNA molecule3’ – TACTTTAGGATC- 5’ DNA template molecule
Repressor
DNA lacl
Regulatorygene
mRNA
5
3
RNApolymerase
ProteinActiverepressor
NoRNAmade
lacZ
Promoter
Operator
Lactose absent, repressor active, operon off
Operon and Inducers
DNA lacl
mRNA5
3
lac operon
Lactose present, repressor inactive, operon on
lacZ lacY lacA
RNApolymerase
mRNA 5
Protein
Allolactose(inducer)
Inactiverepressor
-Galactosidase Permease Transacetylase
Operon Regulation
Promoter Promoter
DNA trpR
Regulatorygene
RNApolymerase
mRNA
3
5
Protein Inactiverepressor
Tryptophan absent, repressor inactive, operon on
mRNA 5
trpE trpD trpC trpB trpA
OperatorStart codonStop codon
trp operon
Genes of operon
E
Polypeptides that make upenzymes for tryptophan synthesis
D C B A
Operon Repressor
DNA
Protein
Tryptophan(corepressor)
Tryptophan present, repressor active, operon off
mRNA
Activerepressor
No RNA made
Math Grid In Answer
• The correct answer is: around 5,500 bp Acceptable range is: 5,300 – 5, 800 bp
• HindIII RFLPS = 4,500 and 1,000 = 5,500 bp• EcoRI = 5,500 bp (This plasmid only cut once; so the DNA strand is one linear piece.)Hind + Eco = 4,500 + 800 + 200 = 5,500 bp
Short Free Response 1(4 points possible)
• Possible points awarded for:• Discussion of point mutations being one nucleotide
altered in the DNA sequence.(1pt.)This will cause a possible change in the Amino Acid coded for in the protein. (1 pt.)
• Discussion of reading frame mutations being an addition or deletion of a nucleotide(s)to the existing DNA sequence. (1 pt.) This will cause all the reading frame codons to be altered down strand from the insertion/deletion.(1 pt.)
Short Free Response 2(4 points possible)
Possible points awarded for:Discussion of using the same restriction enzyme on both the plasmid and the human DNA source.(1 pt.)Discussion of the restriction enzyme creating matching sticky ends on each DNA strand. (1pt.)Discussion of the DNA pieces being combined using ligase to solidify the connects. (1 pt.)Discussion of the recombined plasmid being inserted back into the bacteria. (1pt.)