Saturday Study Session 2 Theme of the day: Information Transfer Session 3 – Modern Genetics.

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Presentation transcript:

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 DNA lacl Regulatory gene mRNA 5 3 RNA polymerase Protein Active repressor No RNA made lacZ Promoter Operator Lactose absent, repressor active, operon off

Operon and Inducers DNAlacl mRNA 5 3 lac operon Lactose present, repressor inactive, operon on lacZ lacYlacA RNA polymerase mRNA 5 Protein Allolactose (inducer) Inactive repressor  -Galactosidase Permease Transacetylase

Operon Regulation Promoter DNA trpR Regulatory gene RNA polymerase mRNA 3 5 Protein Inactive repressor Tryptophan absent, repressor inactive, operon on mRNA 5 trpE trpD trpC trpBtrpA Operator Start codon Stop codon trp operon Genes of operon E Polypeptides that make up enzymes for tryptophan synthesis D C B A

Operon Repressor DNA Protein Tryptophan (corepressor) Tryptophan present, repressor active, operon off mRNA Active repressor 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, = 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.)

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