CHAPTER 10
MOLECULAR GENETICSNUCLEIC ACIDS to
PROTEINS: •STRUCTURE
•REPLICATION
•PROTEIN SYNTHESIS
DNA is the unit of heredity.it is transmitted fromparents to offspring in reproductive cells calledgametes. (meiosis)
CHROMOSOMES are…DNA coiled around HISTONEproteins then supercoiled into a compact shape. This is important for cell division.
FRANKLIN, WATSON & CRICK discovered the structure of DNA
1953
X-ray diffraction crystallography is a technique that was usedto help scientists learn the structure of DNA. (Rosiland Franklin)
(NOT ON TEST… just thought you’d like to know)
DNA STRUCTURE
Double stranded– 2 polymer chains of
nucleotide monomers– Covalent bonds
joining amino acid monomers “sugar phosphate backbone”
– Hydrogen bonds joining the two strands through nitrogen-containing-bases “A”, “T”, “G” or “C
FUNCTION
Unit of Heredity– Easily copied
(Sphase)– 2 template strands– Gene copied into
mRNA is instructions to assemble protein.
base-pairsare the “rungs” of the ladder
sugar-phosphatebackbone or “uprights” of theladder
THE DOUBLE HELIX
DNA NUCLEOTIDES
DeoxyriboNucleic Acids
are made of 3 parts:
1. Phosphate Group
2. Sugar = deoxyribose
3. Nitrogen-containing base:
• A (adenine)• T (thymine)• G (guanine)• C (cytosine)
RNA NUCLEOTIDES
RiboNucleic Acid
are made of 3 parts:
1. Phosphate Group
2. Sugar = RIBOSE
3. Nitrogen-containing base:
• A (adenine)• U (uracil) instead of
T• G (guanine)• C (cytosine)
The nucleotide is the chemical building block of NUCLEIC ACIDS
sugar
phosphategroup
NitrogenContainingbase
There is no THYMINE base in RNA
URACIL takes it’s place
URACIL is a molecular mimic of THYMINE and does it’s same
job
Base-Pair Rules or Chargaff’s LAWS:
1. purine-pyrimidine2. Adenine-Thymine Thymine-Adenine3. Guanine-Cytosine Cytosine-GuanineRNAUracil-AdenineAdenine-Uracil
purines(A & G)pyrimidines
(T & C)
Each nucleotide’s partner is called it’sCOMPLEMENT
•A
•T
•G
•C
•A
•T
•C
•C
•A
•G
•C
•T
•T
Tell me the complements to this nucleotide sequence….
Each nucleotide’s partner is called it’sCOMPLEMENT
•A
•T
•G
•C
•A
•T
•C
•C
•A
•G
•C
•T
•T
A A G C T G G A T G C A T (DNA)
A A G C U C C A U C G A U (RNA)
if the DNA is 35% A, then…What are the values for T, C, G?
Chargaff’s Rule… The amounts of A & T match, so do the Amounts of C & G (where the base-pair rules came from).
• If A = 35% and A = T… then T=35%• Since 35+35=70 and 100-70=30, G+C=30• If G=C… then G = 15% and C = 15%• ANSWER:A= 35%, T= 35%, G = 15%, C= 15%
• QUESTION: What are the values of A,T,G, and C if the DNA is 20% G?
• ANSWER: A=30%, T=30%, G=20%, C=20%
• QUESTION: What are the values of ATC&G if the DNA IS 10% T?
• ANSWER: A=10%, T=10%, G=40%, C=40%
DNA REPLICATION- copying the DNA.ENZYMES USED:•HELICASE-open DNA•DNA POLYMERASE-build polymer•DNA LIGASE-connect fragments
NUCLEOTIDES USED:•DNA NUCLEOTIDES base-pair rule
SEMICONSERVATIVEREPLICATION:•2 strands: each is 1/2 old & 1/2 new•Leading strand- continuous•Lagging strand- okazaki fragments•Polymerase moves along the template strand in a 3’ to 5’ direction.•New strand is built 5’ to 3’
The 5’ end has the phosphate
Notice, the two strands runanti-parallel (opposite).
SEMICONSERVATIVE REPLICATION…
each strand of the DNA double helix acts as the “template” or instructions to build the matching strand
Mutations dooccur but arefixed by enzymes.
1/100,000 nucleotides are incorrectly placed
Enzymes fix mostmistakes reducingthe number to
1/ billion base pairs
Mutations (changes in the DNAsequence) can mean changes in phenotypes. They are the fuel forevolution.
What is the connection between your genes (genotype) and how
you look (phenotype)???
PROTEIN
PROTEIN SYNTHESIS
Central dogma = DNA --> RNA --> protein
DNA (instructions) REMAIN IN THE NUCLEUS
PROTEINS ARE ASSEMBLED OUTSIDE THE NUCLEUS on RIBSOMES
DNA… the instructions• double stranded• twisted shape
RNA… the workersmRNA- single strandedrRNA- globular form
w/ proteinstRNA- single stranded,
hairpin loop
PROTEIN… the productAmino acids strung together
molecularmodel oftRNA I made while working at my oldJob.
TRANSCRIPTION
making mRNA, tRNA, and rRNA
The promoter has the sequence TAC which causes the RNA polymeraseto attach to the beginning of the template chain and mRNA to be made.
The terminator sequence causes the RNA polymerase to release boththe DNA and the newly formed mRNA. There are 3 different terminator signals: ACT, ATT, and ATC.
TRANSCRIPTION
TRANSLATION
•3 types of RNA work together to produce proteins
• proteins are chains of amino acids bonded together by peptide bonds
TRANSLATION • Ribosome attaches to start codon on mRNA (AUG).
• 1st tRNA w/ anticodon (UAC) binds to start codon w/in ribosome.
• 2nd tRNA w/ matching anticodon to 2nd codon binds.
• Peptide bond is formed between amino acid #1 and amino acid #2.
• First tRNA leaves. • Ribosome moves to next codon
on mRNA and tRNA #3 moves in…. Peptide bond.REPEAT
• When stop codon is reached (UAA, UGA, UAG) release factor breaks binds instead of tRNA.
• All parts separate.
How can 4 nucleotides code for 20 different
proteins??????????????read the nucleotides as sets of three
3 (4)
4 x 4 x 4 = 64 combinations
anticodon
codon
Nucleotides have meaning in sets of THREE.
The CODE can be cracked using this table. Use it to determine whichof the 20 amino acids each mRNA codon- codes for.
Don’t Betricked
WHICH AMINO ACID?1) DNA Triplet is TTT 2) Codon is AUG3) Anticodon is CCC4) ACT
WHICH AMINO ACID?1) DNA Triplet is TTT 2) Codon is AUG3) Anticodon is CCC4) ACT
OK… I will!ANSWERS:1) Lys2) Met3) Gly4) STOP
…G A T T A C A T A……C T A A T G T A T…
DNA TRIPLET
mRNA CODON
tRNA ANTICOD
ON
Amino Acid
GAT
TAC
ATA
…G A T T A C A T A…
DNA TRIPLET
mRNA CODON
tRNA ANTICOD
ON
Amino Acid
GAT
TAC
ATA
RNA polymeraseU
U
UA
A
A
AG C RNAnucleotides
…G A T T A C A T A…
DNA TRIPLET
mRNA CODON
tRNA ANTICOD
ON
Amino Acid
GAT CUA
TAC AUG
ATA UAU
RNA polymeraseU
U
UA
A
A
AG C RNAnucleotides
…G A T T A C A T A…
DNA TRIPLET
mRNA CODON
tRNA ANTICOD
ON
Amino Acid
GAT CUA GAU
TAC AUG UAC
ATA UAU AUA
RNA polymeraseU
U
UA
A
A
AG C RNAnucleotides
1. WRITE THE CORRECT LETTER BELOW EACH SYMBOL.
2. USE “BASE PAIR” RULES TO WRITE THE mRNA
3. TRANSFER THE CODONS BELOW.
4. USE “BASE PAIR RULES” TO WRITE THE ANTICODONS.
5. USE CODONS & THE CHART TO FIND AMINO ACIDS.
CHECK YOUR WORK.
HOW DID YOU DO?
ANY QUESTIONS?
YOUR Classwork/ homework… plus the Other sheets.