Download 12–1 DNA

 Please pick up ch.12 notes: cut out and glue
into your notebooks, thank you 
 Glue new ch.12 notes into notebooks
 Lecture/Notes on section 1
 Foldable
 Vocabulary Cards
 DNA video
 No homework tonight 
12–1 DNA
Griffith and Transformation
 Griffith’s Experiment
 Transformation
Avery and DNA
The Hershey-Chase Experiment
 Bacteriophages
 Radioactive Markers
The Structure of DNA
 Chargaff’s Rules
 X-Ray Evidence
 The Double Helix
 How do genes work? What are they made of, and how do
they determine the characteristics of organisms? Are genes
single molecules, or are they longer structures made up of
many molecules? In the middle of the 1900s, questions like
these were on the minds of biologists everywhere.
 To truly understand genetics, biologists first had to
discover the chemical nature of the gene. If the structures
that carry genetic information could be identified, it might
be possible to understand how genes control the inherited
characteristics of living things.
 In 1928, British scientist Frederick Griffith was trying to
figure out how bacteria make people sick.
 Griffith had isolated two slightly different strains, or types,
of pneumonia bacteria from mice. The disease-causing
strain of bacteria grew into smooth colonies on culture
plates, whereas the harmless strain produced colonies
with rough edges. The differences in appearance made the
two strains easy to distinguish.
 When Griffith injected mice with the disease-causing
strain of bacteria, the mice developed pneumonia and
died. When mice were injected with the harmless
strain, they didn’t get sick at all. Griffith wondered if
the disease-causing bacteria might produce a poison.
 To find out, he took a culture of these cells, heated the
bacteria to kill them, and injected the heat-killed
bacteria into mice. The mice survived, suggesting that
the cause of pneumonia was not a chemical poison
released by the disease-causing bacteria.
 Griffith’s next experiment produced an amazing result. He
mixed his heat-killed, disease-causing bacteria with live,
harmless ones and injected the mixture into mice. By
themselves, neither should have made the mice sick. But to
Griffith’s amazement, the mice developed pneumonia and
many died. When he examined the lungs of the mice, he
found them filled with the disease-causing bacteria.
 Somehow the heat-killed bacteria had passed
their disease-causing ability to the harmless strain.
Griffith called this process transformation because
one strain of bacteria (the harmless strain) had
apparently been changed into another
(the disease-causing strain).
Heat-killed,
disease-causing
bacteria (smooth
colonies)
Disease-causing
bacteria (smooth
colonies)
Harmless bacteria Heat-killed, disease(rough colonies) causing bacteria
(smooth colonies)
Dies of pneumonia
Lives
Lives
Control
(no growth)
Harmless bacteria
(rough colonies)
Dies of pneumonia
Live, disease-causing
bacteria (smooth colonies)
 In 1944, a group of scientists led by Canadian biologist
Oswald Avery at the Rockefeller Institute in New York
decided to repeat Griffith’s work. They did so to
determine which molecule in the heat-killed bacteria
was most important for transformation.
 Avery and his colleagues made an extract, or juice,
from the heat-killed bacteria. They then carefully
treated the extract with enzymes that destroyed
proteins, lipids, and carbohydrates. Transformation
still occurred. Obviously these molecules were not
responsible for the transformation. If they had been,
transformation would not have occurred, because the
molecules would have been destroyed by the enzymes.
 Avery and the other scientists repeated the
experiment, this time using enzymes that would break
down DNA. When they destroyed the nucleic acid
DNA in the extract, transformation did not occur.
 Avery and other scientists discovered that DNA is the
nucleic acid that stores and transmits the genetic
information from one generation of an organism to
the next.
Bacteriophage with
phosphorus-32 in
DNA
Phage infects
bacterium
Radioactivity inside
bacterium
Bacteriophage with
sulfur-35 in protein
coat
Phage infects
bacterium
No radioactivity inside
bacterium
Bacteriophage with
phosphorus-32 in
DNA
Phage infects
bacterium
Radioactivity inside
bacterium
Bacteriophage with
sulfur-35 in protein
coat
Phage infects
bacterium
No radioactivity inside
bacterium
Bacteriophage with
phosphorus-32 in
DNA
Phage infects
bacterium
Radioactivity inside
bacterium
Bacteriophage with
sulfur-35 in protein
coat
Phage infects
bacterium
No radioactivity inside
bacterium
 1952: Two American scientists, Alfred Hershey and
Martha Chase, studied viruses- nonliving particles
smaller than a cell that can infect living organisms.
 A virus that infects and kills bacteria is known as a
bacteriophage; composed of a DNA or RNA core and a
protein coat.
 Hershey and Chase reasoned that if they could
determine which part of the virus—the protein coat or
the DNA core—entered the infected cell, they would
learn whether genes were made of protein or DNA.
 To do this, they grew viruses in cultures containing
radioactive isotopes of phosphorus-32 (32P) and sulfur35 (35S). The radioactive substances could be used as
markers. If 35S was found in the bacteria, it would
mean that the viruses’ protein had been injected into
the bacteria. If 32P was found in the bacteria, then it
was the DNA that had been injected.
 All the radioactivity in the bacteria was from
phosphorus (32P), the marker found in DNA. Hershey
and Chase concluded that the genetic material of the
bacteriophage was DNA, not protein.
 Fold a piece of your notebook paper toward the spiral.
 Cut top part into 3 equal sections.
 Label the outside of the 3 sections with the 3 Scientists.
 Write the experiment they did and what their conclusion
was on the inside paper.
5
minutes
 3 Functions of DNA:
 Genes carry information from one generation to the next.
 They had to put that information to work by determining the
heritable characteristics of organisms.
 Genes had to be easily copied.
 DNA is a long molecule made up of subunits called
nucleotides.
 Each nucleotide is made up of 3 parts: a 5-carbon sugar called
deoxyribose, a phosphate group, and a nitrogen base.
 There are 4 kinds of nitrogenous bases in DNA. Two of the
nitrogenous bases, adenine (A) and guanine (G), belong to a
group of compounds known as purines. The remaining two
bases, cytosine (C) and thymine (T), are known as
pyrimidines.
 Purines have two rings, and pyrimidines have one ring.
 The backbone of a DNA chain is formed by sugar and
phosphate groups of each nucleotide. The nitrogenous
bases stick out sideways from the chain. The
nucleotides can be joined together in any order,
meaning that any sequence of bases is possible.
 Chargaff’s Rule: Erwin Chargaff, an American
biochemist, had discovered that the percentages of
guanine [G] and cytosine [C] bases are almost equal in
any sample of DNA. The same thing is true for the
other two nucleotides, adenine [A] and thymine [T].
 1950: A British scientist named Rosalind Franklin began to
study DNA. She used a technique called X-ray diffraction to
get a picture of the structure of a DNA molecule.
 By itself, Franklin’s X-ray pattern doesn’t reveal the
structure of DNA. The X-shaped pattern in the image does
shows that there are 2 strands in DNA that are twisted
around each other, a shape known as a helix.
 At the same time that Franklin was continuing her
research, Francis Crick, a British physicist, and James
Watson, an American biologist, were trying to understand
the structure of DNA by building three-dimensional
models of the molecule.
 1953: Watson was shown a copy of Franklin’s remarkable
X-ray pattern and reported the pattern’s clues to Crick.
 Within a few weeks, Watson & Crick figured out the final
structure of DNA: a two stranded double helix.
 Watson & Crick discovered that hydrogen bonds could
form between certain nitrogenous bases and provide just
enough force to hold the two strands together.
 Hydrogen bonds can form only between certain base pairs—
adenine & thymine and guanine & cytosine. Once they saw
this, it explained Chargaff’s rules. A = T and G = C.
Nucleotide
Hydrogen
bonds
Sugar-phosphate
backbone
Key
Adenine (A)
Thymine (T)
Cytosine (C)
Guanine (G)
 Please spend 2 minutes writing a summary about
what we have learned from ch.12 section 1.
 In your groups, write the definition in your own
words, and add an example and/or a picture.
 DNA Structure
 Chargaff’s Rule
 3 Functions of DNA
 Rosalind Franklin
 Griffith
 Watson & Crick
 Avery
 Purine & Pyrimidine
 Hershey-Chase
3
minutes
 Play DVD 16min. HHMI video
 Please pick up 1 warm-up card per pair 
 Glue in new notes for 12-2 also!
 Warm-up cards from section 1
 Lecture/Notes on section 2
 Vocabulary Cards
 DNA Replication Activity
12–2 Chromosomes & DNA Replication
DNA and Chromosomes
 DNA Length
 Chromosome Structure
DNA Replication
 Duplicating DNA
 How Replication Occurs
 Prokaryotic cells do not have a nucleus, instead, their
DNA is located in the cytoplasm and consists of a
single circular chromosome of genetic information.
 Eukaryotes have 1,000 times more DNA than
prokaryotes! Eukaryotic DNA is found in the nucleus
and forms into many chromosomes.
 Humans have 46 chromosomes, fruit flies have 8, and
trees have 22.
 DNA Length: the single chromosome of a
prokaryotic E. coli bacteria contains 4 million base
pairs (letters), and is about 1.6mm long. DNA has to be
tightly folded to fit inside of a tiny cell.
 Eukaryotic DNA contains 30 million base pairs, and is
10x longer than a bacteria's chromosome. Eukaryotic
chromosomes contain both DNA and proteins
(histones) that help it to stick together as tightly
packed nucleosome coils.
Chromosome Structure of
Eukaryotes
Chromosome
Nucleosome
DNA
double
helix
Coils
Supercoils
Histones
 DNA stands are complementary, meaning that each
side matches the other: chargaff’s rule, so DNA can be
easily copied by following the base pairing rule.
 Before cells divide (mitosis) their DNA must be
duplicated in a process called DNA Replication.
 DNA Replication: the DNA molecule separates into 2
stands, then produces 2 new complementary strands
following the rules of base pairing. Each strand of the
double helix of DNA serves as a template/model for
the new stands.
 DNA replication is carried out by enzymes. Remember
that enzymes help speed up reactions and are very
specific (lock & key).
 The enzyme Helicase unzips a molecule of DNA first
by breaking the hydrogen bonds between the base
pairs. Each strand serves as a template for the
attachment of complementary bases.
 The enzyme DNA Polymerase adds new base pairs to
the new DNA stand, and also proofreads to make sure
it is adding the correct bases to make a perfect copy.
Original
strand
New strand
DNA
polymerase
Growth
DNA
polymerase
Growth
Replication
fork
Replication
fork
New strand
Original
strand
Nitrogenous
bases
 Each DNA molecule resulting from replication has 1
original strand and 1 new stand.
 Example: a strand that is TACGTT produces a
complementary stand with bases ATGCAA.
 Please spend 2 minutes writing a summary about what
we have learned from ch.12 section 2.
 In your groups, write the definition in your own
words, and add an example or a picture.
 Prokaryotic Chromosome
 Eukaryotic Chromosomes
 DNA Replication
 Mitosis
 Helicase
 DNA Polymerase
 In pairs:
 Pick up 1 DNA Replication Packet
 Read through the first page together, then follow
the directions on the second page.
 Take turns with different steps.
 When finished, each person glues ½ of the
replicated copy of DNA into their
notebook.
 Homework: answer the 6
questions at the end 
30
minutes
 Please pick up 1 card EACH 
 Glue in the diagram for today’s notes.
 Notes on 12-3
 Foldable
 Transcription/Translation Practice
 12-3 Summary Videos
 Snork DNA Activity
 Homework due next class.
12–3 RNA and Protein Synthesis
The Structure of RNA
 Types of RNA
Transcription
 RNA Editing
 The Genetic Code
Translation
 The double helix structure of DNA shows how easily it
can be replicated, but it doesn't explain how a gene
works to give us genic information.
 Genes are coded DNA instructions that control the
production of proteins within the cell. The first step in
decoding these genetic messages is to copy part of the
nucleotide (base pair) sequence from DNA into RNA.
 The RNA molecules are then used to make proteins.
 RNA, like DNA, consists of a long chain of nucleotides.
Remember, nucleotides are made of 3 parts: a 5-carbon
sugar, a phosphate, and a nitrogen base.
 Three differences between RNA and DNA:
 RNA contains the sugar “ribose”, where DNA contains
the sugar “deoxyribose”.
 RNA is a single stand, and DNA is a double strand.
 RNA contains the bases A,C, G, U; and DNA contains
the bases A, C, G, T.
 You can think of RNA as a disposable copy of a
segment of DNA that is only used as 1 time
instructions for a single gene.
 There are 3 types of RNA that help with protein synthesis:
 mRNA (messenger RNA = the single stand of code)
 rRNA (ribosomal RNA = what ribosomes are made of)
 tRNA (transfer RNA = brings amino acids to the ribosome in
the order of the mRNA code)
rRNA
mRNA
tRNA
RNA
can be
Messenger RNA
also called
Ribosomal
RNA
which functions to
mRNA
Carry instructions
also called
which functions to
rRNA
Combine
with proteins
from
to
to make up
DNA
Ribosome
Ribosomes
Transfer RNA
also called
which functions to
tRNA
Bring
amino acids to
ribosome
 Fold paper in ½ towards the bottom of your notebook.
 Cut 2 vertical sections to make 3 areas to write:
 Outside: list & draw the 3 types of RNA.
 Inside: Describe what their job and location.
Adenine (DNA and RNA)
Cystosine (DNA and RNA)
Guanine(DNA and RNA)
Thymine (DNA only)
Uracil (RNA only)
RNA
polymerase
DNA
RNA
 A process in the nucleus, where mRNA is made by
copying part of the nucleotide sequence (a gene) in
DNA. Transcription requires an enzyme called RNA
Polymerase (acts just like DNA Polymerase by adding
RNA nucleotides: A, U, G, C).
 How does RNA Polymerase know where to start? There
is a region on DNA known as a Promoter, which
signals where to bind and start transcription.
 Like a writer’s first draft of a paper, RNA molecules
require editing before they are ready to leave the
nucleus and make protein.
 There are 2 main parts to mRNA:
 Introns: “in between” pieces that are not needed.
 Exons: “exactly” the genes to be read to make protein.
 The introns get cut out and we are left with only exons
in our mRNA strand. It can now leave the nucleus and
go to a ribosome to be read.
 Proteins are made by joining amino acids together into
chains called polypeptides. There are 20 different types of
amino acids.
 By reading the nucleotide sequence of mRNA, we can
“translate” the letters into amino acids, and join them
together one after another to make a protein.
 We read 3 mRNA letters at a time to make 1 amino acid:
each set of 3 letters is called a Codon.
 Ex: UCGCACGGU would be read as: UCG-CAC-GGU, and
would make the amino acids: serine-histidine-glycine
 Because there are 4 nucleotide bases in RNA (A, U, G, C),
there are 64 possible codons combinations.
 Some amino acids can be made from more than 1 codon.
 All proteins start with the amino acid Methionine - AUG.
 All proteins end with a “stop” codon (multiple codons).
 The process of making proteins from mRNA, occurs at
the ribosome.
 In the first step we edited the mRNA into exons only, then
it left the nucleus and went to a ribosome. At the
ribosome, we know that the mRNA gets read and 1 amino
acid is attached after reading a codon (3 letters). So how
does the amino acid get there?
 The tRNA molecule has an anticodon (3 complementary
letters to the mRNA codon) on one side, and a specific
amino acid on the other side.
Nucleus
Messenger RNA
Messenger RNA is transcribed in the nucleus.
Phenylalanine
tRNA
Lysine
mRNA
Transfer RNA
Methionine
The mRNA then enters the cytoplasm and
attaches to a ribosome. Translation begins at
AUG, the start codon. Each tRNA has an
anticodon whose bases are complementary to a
codon on the mRNA strand.
Ribosome
mRNA
Start codon
The Polypeptide “Assembly Line”
The ribosome joins the two amino acids,
Then their tRNA floats away, allowing the
ribosome to bind to another tRNA. The
ribosome moves along the mRNA, binding
new tRNA molecules and amino acids.
Lysine
Growing polypeptide chain
Ribosome
tRNA
tRNA
mRNA
Completing the Polypeptide
mRNA
Ribosome
Translation direction
The process continues until the ribosome
reaches a stop codon. The result is a
polypeptide chain = protein.
 Transcribe your DNA into mRNA, then translate into
codons and anticodons to determine the amino acids.
 Glue into your notebook when finished.
 Please spend 2 minutes writing a summary in your
notes about what we learned from section 3.
 Please study the processes of:
Transcription and Translation.
 Be familiar with:
 DNA  mRNA  Protein
 Vocabulary: codon, anticodon, polypeptide,
intron, exon, 3 types of RNA.
 You are given a Snork’s mRNA sequence, your job is to
determine the Snork’s traits that are coded for in the
genes you read by Translating them into Amino Acids.
 After determining the traits, draw your Snork!
Be creative, use colors, show Ms. Lutz when you’re finished.
 Choose
1 for homework:
 Snicker Snork
 Snuffle Snork
 Snapple Snork
 Snoopy Snork
 In your groups: write the definition in your own
words, and add an example and picture.
 3 types of RNA
 Transcription
 RNA Polymerase & Promoter
 Introns & Exons
 Translation
 Amino Acids (# number)
4 minutes
Pin to board when
finished!
 Please close and put away your notebook.
 Have a pencil out!
 Hand in quiz to your block’s bin when done.
 Pick up and glue in notes for 12-4 & 12-5 
 Please pick up 1 card EACH! 
 Warm-up: pick up 1 card each please 
 Pick up graded quizzes
 Notes on 12-4 and 12-5
 2 short video clips
 Unit 5 Review Packet
12–4 Mutations
Gene Mutations
Chromosome Mutations
 Every now and then cells make mistakes in copying their
own DNA - inserting an incorrect base or sometimes even
skipping a base as the new strand is put together. These
mistakes are called mutations.
 Mutations are changes in the DNA sequence that affect
genetic information. Gene mutations result from changes
in a single nucleotide base. Chromosomal mutations
involve changes in whole chromosomes.
Substitution
Insertion
Deletion
 2 Types of gene mutations:
 Point Mutations result in the substitution of a base
 Frameshift Mutations result in the insertion or deletion
of a base, causing different amino acids to be made.
Deletion
Duplication
Inversion
Translocation
 4 types of chromosome mutations:
 Deletion, duplication, inversion, translocation.
 This is NOT in your notes.
 Please spend the next 2 minutes writing a
summary in your notes about what we learned
from section 4.
12–5 Gene Regulation
Prokaryotic & Eukaryotic Gene Regulation
Gene Enhancement
Regulation and Development
Regulatory
sites
Promoter
(RNA polymerase
binding site)
Start transcription
DNA strand
Stop transcription
As we’ve seen, there is a promoter to one side of the gene. But what
are the “regulatory sites” next to the promoter? These are places
where other proteins can regulate transcription. The actions of these
proteins help to determine whether a gene is turned on or off.
 Only some of the genes in a cell are expressed at any
given time. An expressed gene is a gene that is
transcribed into RNA (turned on/being used). How
does the cell determine which genes will be expressed
and which will remain off?
 Molecular biologists have found that certain DNA
sequences serve as promoters (binding sites for RNA
polymerase). Others serve as start and stop signals for
transcription. Cells are filled with DNA-binding
proteins that attach to specific DNA sequences and
help to regulate gene expression (turn on/off).
 Lac operon genes are found in certain bacteria and help
the bacteria to use lactose as food.
 If lactose is present the lac operon will turn the gene on
as RNA Polymerase binds to the promoter.
 After the lactose food is used up, the gene is turned off
by a Repressor Protein binding to the operon.
 Repressor proteins stop gene transcription by twisting the
DNA strand so that RNA Polymerase cannot read it.
 The lac operon turns on when lactose is present because
the DNA strand will straighten out, causing the repressor
protein to fall off. RNA polymerase attaches to the
promoter and can easily read the straight DNA strand.
 Unlike prokaryotes, eukaryotic genes do not contain
operons to regulate gene expression, instead, they have
a short nucleotide sequence of –TATA- for a start site.
 The TATA Box helps position where RNA polymerase
needs to bind to begin transcription.
 Short DNA Promoter sequences are found just before
the TATA box to help position it correctly.
 There are special proteins that bind to enhancer
sequences of a gene that help to regulate gene
expression in several ways:
 Many proteins can bind to different enhancer sequences
at the same time = more efficient.
 Some proteins enhance transcription by opening up
tightly coiled chromatin.
 Help attract RNA polymerase to begin transcription.
 Video
 Gene regulation in eukaryotes is more complex than in
prokaryotes because:
 Cell specialization requires genetic specialization.
 All eukaryotic cells carry the entire genetic code, each
type of cell performs a different function in a different
part of the body.
 Hox genes control the differentiation of cells and
tissues of developing embryos.
 Animals share many common patterns of development
because we all share the same bases (ATGC) but have
different numbers and arrangements of chromosomes.
 Example: Fruit flies can grow legs on their head if the
hox gene is tampered with during development.
 In humans the hox gene acts similar to fruit fly’s.
 Please write definition in your own words, add a
picture and an example. ALL group members work
on card together, we will go over responses after
posting to card board.
 Mutations
 Gene Regulation
 Lac Operon
 TATA Box
 Gene Enhancement
 Hox Genes
 Please spend the next 2 minutes writing a
summary about what we learned from section 5.
 Hox Genes Video
 Unit 5 Review Packet will be due the day of
the test:
 B-day the test is Friday 2/13
 A-day the test is Tuesday 2/17
 Next class we will spend the entire block
reviewing ch.12 material.