Transcript Saturday Study Session 2 Theme of the day: Information Transfer

Saturday Study Session 2
Theme of the day:
Information Transfer
Session 3 – Modern Genetics
Question 1 Answer
B
Clue:3’5’
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.
Replication Bubble
5 Carbon Sugar Important Parts
(It could be DNA or RNA)
Question 2 Answer
A
Clue: recognized
Nucleotide Excision Repair
Question 3 Answer
C
Polymerase Chain Reaction (PCR)
Question 4 Answer
D
Clue: 5’ vs 3’
Complimentary strands
5’ – AUGAAAUCCUAG – 3’ mRNA molecule
3’ – TACTTTAGGATC- 5’ DNA template molecule
Question 5 Answer
A
Clue: only one ≠ poly…
As in polyribosomes
Polyribosomes
Amino Acid Codon Chart
Ribosome Structure
Question 6 Answer
B
Clue: lactose binds to the
operator
Repressor
Promoter
Regulatory
gene
Operator
lacl
DNA
lacZ
No
RNA
made
3
mRNA
5
Protein
RNA
polymerase
Active
repressor
Lactose absent, repressor active, operon off
Operon and Inducers
lac operon
DNA
lacZ
lacl
3
mRNA
5
lacA
Permease
Transacetylase
RNA
polymerase
mRNA 5
-Galactosidase
Protein
Allolactose
(inducer)
lacY
Inactive
repressor
Lactose present, repressor inactive, operon on
Operon Regulation
trp operon
Promoter
Promoter
Genes of operon
DNA
Regulatory
gene
mRNA
trpE
trpR
3
trpC
trpB
trpA
C
B
A
Operator
Start codon Stop codon
RNA
polymerase
mRNA 5
5
E
Protein
trpD
Inactive
repressor
D
Polypeptides that make up
enzymes for tryptophan synthesis
Tryptophan absent, repressor inactive, operon on
Operon Repressor
DNA
No RNA made
mRNA
Active
repressor
Protein
Tryptophan
(corepressor)
Tryptophan present, repressor active, operon off
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
Point Mutation
(A single point is changed)
Reading Frame Mutations
(Everything behind is changed)
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.)
Build and reintroduce the Plasmid