Transcript C - MCC Year 12 Biology

Chapter 10&11
DNA, Genes and Genomics
1.The Discovery of DNA
EL: To revise DNA structure and learn
about the discovery of DNA
Discovery of DNA
• DNA interactive interviews
• Page 377-78 of NOB
Prokaryotic DNA
• The prokaryotes usually have only one
chromosome, and it bears little
morphological resemblance to eukaryotic
chromosomes.
• Consists of single, circular DNA molecule
located in the nucleoid region of cell.
Referred to as being “naked”
• Bacterial cells may also contain multiple
plasmids - small circular fragment of DNA
separate from the main chromosome.
Eukaryotic DNA Structure
 DNA consists of two molecules that are arranged
into a ladder-like structure called a Double Helix.
 A molecule of DNA is made up of millions of tiny
subunits called Nucleotides.
 Each nucleotide consists of:
1. Phosphate group
2. Ribose sugar
3. Nitrogenous base
Nucleotides
Phosphate
Nitrogenous
Base
Ribose
Sugar
Nucleotides
 The phosphate and sugar form the backbone of the
DNA molecule, whereas the bases form the
“rungs”.
 There are four types of nitrogenous bases.
Nucleotides
A
Adenine
C
Cytosine
T
Thymine
G
Guanine
Nucleotides
 Each base will only bond with one other specific
base.
 Adenine (A)
 Thymine (T)
Form a base pair.
 Cytosine (C)
 Guanine (G)
Form a base pair.
A
DNA Structure
T
G
C
T
C
A
G
T
Because of this
complementary
base pairing, the
order of the bases
in one strand
determines the
order of the bases
in the other strand.
A
G
T
C
A
DNA replication
• DNA interactive animation
DNA Replication
1.
unfolding and unwinding of the DNA double helix at hundreds of points, known
as replication origins, along the chromosome.
2.
The enzyme helicase separates the two DNA strands, separating them like
opening a zipper, with the point of opening being termed the replication fork.
3.
Where the DNA strands are separated, a short length of RNA binds to each DNA
strand under the control of the enzyme, DNA primase. This RNA acts as a primer
(see figure 11.26a page 405).
DNA Replication
4.
A DNA polymerase enzyme can then proceed to build new DNA strands using
each of the old strands as a template (see figure 11.26b).
5.
Replication of DNA can occur only in the 5´ to 3´ direction. This is no problem
with the so-called leading strand because its new complementary strand can be
built continuously in the 5´ to 3´ direction. The other strand, known as the lagging
strand, can be built only backwards and in short discontinuous pieces (Okasaki
fragments - see figure 11.26b).
6.
When finished, the RNA primers are removed, the gaps are filled by another DNA
polymerase and the pieces are joined by the enzyme, DNA ligase.
DNA Replication
Watch DNAi clip
Mitochondrial DNA
• Mitochondria contain
mtDNA, a double
stranded circular
molecule comprising:
– 16 568 base pairs and
code for 37 genes:
– 13 genes code for
proteins that are
involved in cellular
respiration
– 2 genes code for
ribosomal RNA (rRNA)
– 22 genes code for
transfer RNAs (tRNAs).
Activity
• Activity 7.1 – Simulation of DNA replication
(handed in please)
• Ch 10 quick check qu 9-11 (pg 353), 12-14 (pg
356), 18-19 (pg 406),
• Chapter 10 review qu 2, 6, 7, 8
Reflection
• After listening to some of the interviews with
scientists, do you think you would have the
patience to see out a major scientific discovery
like that of DNA? Why or why not?
• What learning was new today?
• What learning was revision or built on what I
already know?
• What did I find most challenging and what
strategies will I put in place to help me?
• What percentage of the class did I spend on task
and how can I improve this if needed?
2. DNA to proteins
EL: To explore how protein synthesis
occurs
PROTEIN SYNTHESIS
An individuals characteristics are determined by their DNA.
The DNA determines which proteins are made.
The most important proteins are enzymes.
The sequence of bases in the DNA determines the sequence
of amino acids in the protein.
This is known as the GENETIC CODE.
THE GENETIC CODE
TRIPLET CODE
3 bases in the DNA code for one amino acid in the protein. Each
triplet is known as a CODON.
UNIVERSAL
Found in all organisms.
DEGENERATE
More than one codon for each amino acid.
NON-OVERLAPPING
START AND STOP CODONS
PROTEIN SYNTHESIS
TRANSCRIPTION
AMINO ACID ACTIVATION
TRANSLATION
Protein Synthesis Summary
TRANSCRIPTION
U
U
A
C
U
U
C
A
G
C
A
G
U
G
A
A
U
A
TA C GG CT AA CA TA C AATC
G
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
U
A
C
U
U
C
A
G
C
A
G
U
G
A
A
U
A
TA C GG CT AA CA TA C AATC
G
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
U
A
C
U
U
C
A
G
C
A
G
G
A
A
U
AU
TA C GG CT AA CA TA C AATC
G
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
U
A
C
U
U
C
A
G
C
A
G
A
A
U
AU G
TA C GG CT AA CA TA C AATC
G
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
C
U
U
A
U
A
G
C
A
G
A
A
U
A U GC
TA C GG CT AA CA TA C AATC
G
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
C
U
U
A
U
A
G
A
G
A
A
U
A U GC C
TA C GG CT AA CA TA C AATC
G
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
C
U
U
A
U
A
G
A
A
A
U
A U GC C G
TA C GG CT AA CA TA C AATC
G
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
C
U
U
A
U
A
G
A
A
U
A U GC C GA
TA C GG CT AA CA TA C AATC
G
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
C
U
A
U
A
G
A
A
U
A U GC C GA U
TA C GG CT AA CA TA C AATC
G
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
C
A
U
A
G
A
A
U
A U GC C GA U U
TA C GG CT AA CA TA C AATC
G
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
C
A
U
A
G
A
A
U
A U GC C GA U U G
TA C GG CT AA CA TA C AATC
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
C
A
G
A
A
A
U
A U GC C GA U U GU
TA C GG CT AA CA TA C AATC
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
U
C
A
G
A
A
U
A U GC C GA U U GU A
TA C GG CT AA CA TA C AATC
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
C
A
G
A
A
U
A U GC C GA U U GU A U
TA C GG CT AA CA TA C AATC
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
C
A
A
A
U
A U GC C GA U U GU A U G
TA C GG CT AA CA TA C AATC
U
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
G
C
A
A
A
U
A U GC C GA U U GU A U GU
TA C GG CT AA CA TA C AATC
G
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
C
A
A
A
A U GC C GA U U GU A U GUU
TA C GG CT AA CA TA C AATC
G
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
C
A
A
A U GC C GA U U GU A U GUU A
TA C GG CT AA CA TA C AATC
G
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
C
A
A
A U GC C GA U U GU A U GUU A G
TA C GG CT AA CA TA C AATC
HELICASE unwinds and unzips the relevant part of the DNA helix.
RNA POLYMERASE attaches to the DNA.
One DNA strand acts as the template (SENSE STRAND), the other is redundant (ANTISENSE
STRAND).
As RNA POLYMERASE moves along the sense strand, ribonucleotides are assembled in a precise
order due to complementary base pairing (A <-> T/U ; G <-> C).
Fully formed mRNA peels off the DNA and leaves the nucleus via a nuclear pore.
The DNA rewinds.
Fully formed mRNA peels off the DNA and leaves the nucleus via a nuclear pore.
The DNA rewinds.
AMINO ACID ACTIVATION
ACTIVATION OCCURS WHEN
THE tRNA COMBINES WITH A
SPECIFIC AMINO ACID
U
A
C
THE ANTICODON DETERMINES
WHICH SPECIFIC AMINO ACID IS
ATTACHED
TRANSLATION
U A C
G G C
A U G C C G A U U G U A U G U UA G
A ribosome binds to the mRNA near the START CODON.
G G C
U A C
A U G C C G A U U G U A U G U UA G
tRNA with the complementary ANTICODON (UAC) binds to the
start codon (AUG) held in place by the large subunit of the
ribosome. It brings with it the amino acid methione.
G G C
U A C
A U G C C G A U U G U A U G U UA G
The ribosome now slides along the mRNA to “read” the next
codon.
U A C G G C
A U G C C G A U U G U A U G U UA G
A second tRNA now bind to this codon, bringing a second amino
acid with it.
U A C G G C
A U G C C G A U U G U A U G U UA G
A peptide bond is formed between the two amino acids.
G G C
A U G C C G A U U G U A U G U UA G
The tRNA which carried the first amino acid is released but leaves
its amino acid behind as a DIPEPTIDE.
G G C
A U G C C G A U U G U A U G U UA G
The ribosome now slides along the mRNA to “read” the next
codon.
G G C
A U G C C G A U U G U A U G U UA G
One by one each codon is read as the ribosome moves along the
mRNA.
G G C
A U G C C G A U U G U A U G U UA G
Each time the growing polypeptide is linked to the amino acid on
the incoming tRNA.
A U G C C G A U U G U A U G U UA G
A U G C C G A U U G U A U G U UA G
A U G C C G A U U G U A U G U UA G
A U G C C G A U U G U A U G U UA G
A U G C C G A U U G U A U G U UA G
A U G C C G A U U G U A U G U UA G
A U G C C G A U U G U A U G U UA G
STOP !
A U G C C G A U U G U A U G U UA G
The polypeptide is complete when the ribosome reaches the
STOP codon.
A U G C C G A U U G U A U G U UA G
The polypeptide is released.
A U G C C G A U U G U A U G U UA G
The polypeptide is released.
The polypeptide may combine with other polypeptides and will
become variously coiled/folded to produce a protein.
The tRNAs are recycled.
A U G C C G A U U G U A U G U UA G
The mRNA may be used again in this form, or it may be broken
down into nucleotides which can be reassembled to produce a
different polypeptide.
The ribosome is free to move along another mRNA.
Ribosomes work in groups so that many slide along a mRNA
molecule simultaneously.
These groups are called POLYRIBOSOMES.
Each ribosome takes about 1 minute to travel along a mRNA
molecule.
Transcription and translation
• DNAi animations
QUIZ
What sort of chemical is helicase?
Why is DNA double stranded if one strand is redundant?
Where in the cell are the ribosomes?
What is the start codon?
Give the three alternative stop codons.
Give the primary structure (sequence of amino acids) of the polypeptide made
in this animation.
What is the difference between a polypeptide and a protein?
What is the advantage of ribosomes operating as polyribosomes?
What are the similarities and differences between DNA replication and protein
synthesis?
ANSWERS
Helicase is an enzyme and therefore also a protein.
DNA is double stranded to permit replication.
Ribosomes are located in the cytoplasm.
The start codon is AUG.
The three stop codons are UGA, UAG and UAA.
Give the primary structure of the polypeptide is methionine, proline, isoleucine,
valine, cysteine.
Polypeptides have less than 100 amino acids, protein have more. A protein may
consist of several polypeptides.
Polysomes increase efficiency, they enable one mRNA molecule to produce
many polypetides simultaneously.
HINT - Think about the enzymes and nucleotides used, the end product and the
location of the process.
Really good website
• http://www.le.ac.uk/ge/genie/vgec/he/index.
html
Activity
• In pairs, complete activity 11.1
• Quick check qu 4-8 (pg 391)
• Ch 11 ch review qu 2, 3, 4, 9 (pg 414-416)
Reflection
• Did the hands-on activity help cement your
understanding of transcription and translation? If not,
how else do you think you could ensure your
understanding?
• What learning was new today?
• What learning was revision or built on what I already
know?
• What did I find most challenging and what strategies
will I put in place to help me?
• What percentage of the class did I spend on task and
how can I improve this if needed?
3.Genomics
EL: What a genome is and how gene
expression is regulated
What is a genome?
• All of the genetic material (the base pairs) found in one
complete set of an organism’s chromosomes.
• The study of genomes is called genomics.
Does genome size matter?
COMMON NAME
SPECIES NAME
Approx GENOME SIZE
(millions of base pairs)
Fruit fly
Drosophila melanogaster
180
Snake
Boa constrictor
2100
Human
Homo sapiens
3100
Onion
Allium cepa
18000
Lungfish
Protopterus aethiopicus
140000
Amoeba
Amoeba dubia
670000
Why would a single celled animal like the amoeba need a genome that is
about 200 times larger than the human genome?
Ans: They carry a lot of junk DNA!
What is a gene?
• Segment of DNA that codes
for formation of a protein
– Structural genes express
structural and/or functional
proteins.
– Regulatory genes are short
nucleotide sequences that
express proteins that control
the activity of structural genes
by feedback mechanisms.
Number of genes
COMMON NAME
SPECIES NAME
No. GENES
Human
Homo sapiens
25000
Mustard plant
Arabidopsis thaliana
27000
Fruit fly
Drosophila melanogaster
14000
Baker’s yeast
Saccharomyces cerevisiae
6000
Gut bacterium
Escherichia coli
4000
Should we be offended that a mustard plant has
as many genes as a human?
An Overview of Gene Structure
Coding Region
DNA sequence that will be transcribed
from the template strand.
5’
3’
3’
Regulatory
region
S
T
A
R
T
Promoter
region
S
T
O
P
Terminator
region
5’
Gene Expression
• The expression of genetic information is one
of the fundamental activities of all cells.
• Instructions stored in DNA are transcribed and
translated into various RNA molecules.
Introns and Exons
• The coding region in eukaryotes contain:
– introns - non-coding regions of DNA
– exons - coding regions of DNA
• Prokaryotes do not have introns – why?
– They don’t carry “junk DNA” due to short
replication cycles
RNA Processing in Eukaryotic Cells
DNA Template Strand
INTRON
EXON
INTRON
EXON
EXON
Spliceosome
Pre-mRNA transcript of DNA template strand
EXON
EXON
Spliceosome
EXON
Introns are spliced out by spliceosomes leaving only the sequences that will be
expressed. This is an example of RNA processing. The introns usually are degraded.
The result is a mature mRNA strand that will leave the nucleus to be translated.
Genome to proteome
• The human genome has about 25,000 genes but
our proteome (the total number of different
proteins) is much larger (~100,000)
– How can this occur?
• Many genes can produce more than one protein
because the mRNA transcript contains different
combinations of exons. This process is called
alternative splicing.
Alternative splicing
INTRONS
Pre-mRNA transcript
EXON 1
EXON 2
EXON 3
EXON 4
Possible mRNAs using different combinations of exons
EXON 1
EXON 1
EXON 2
EXON 2
EXON 2
EXON 3
EXON 4
EXON 3
EXON 4
EXON 4
PROTEIN 1
PROTEIN 2
PROTEIN 3
Result: when each mRNA is translated, a different protein is produced.
Activity:
• In pairs, complete activity 10.2 “Sequencing a
genome”
• Quick check qu 15-17 (pg 362), 18-23 (pg 364),
24-27 (pg 370), 16&17 (pg 403)
• Chapter review qu 3, 5, 9 (pg 381-382)
Reflection
• What learning was new today?
• What learning was revision or built on what I
already know?
• What did I find most challenging and what
strategies will I put in place to help me?
• What percentage of the class did I spend on
task and how can I improve this if needed?
4.Gene Regulation and Genetic
Mutations
EL: How genes are switched on and off
The effects genetic mutations can have
on organisms
Gene Regulation
• Each cell contains an entire organism’s genome.
• All cells of an organism have the same genome,
but can have different phenotypes.
• For example, cells in your eye have the gene for
producing fingernail protein (keratin) but this
gene is not expressed.
• How do genes get switched on or switched off?
Why regulate gene expression?
• Cells conserve energy and materials by blocking
unneeded gene expression.
• If a substrate is absent in the environment why produce
the enzyme for that substrate!
• Repressor molecules keep the cell from wasting energy
by not transcribing mRNA or making enzyme molecules
that have no use.
• The cell can control its metabolism – resources are
used only when there is a metabolic need and can be
redirected to other metabolic pathways.
Gene regulation in prokaryotes
• Bacteria have groups of genes that are controlled together and are
turned on/off as required.
• E.g. the lac operon is a set of genes in bacteria used for lactose
metabolism.
• Bacteria produce the enzymes to break down lactose to glucose and
galactose only when lactose is present.
• http://www.sumanasinc.com/webcontent/animations/content/lacop
eron.html
• http://pages.csam.montclair.edu/~smalley/LacOperon.mov
Gene regulation in Eukaryotes
• Still being investigated
• Proteins involved (like prokaryotes) – but
more complicated
– Enhancers: act as binding sites for activator proteins and
increase number of DNA polymerase molecules
transcribing genes
– Chemical modification: eg presence or absence of histone
proteins
Changing the Genetic Information
Mutations
• A mutation refers to any
permanent change in the DNA
nucleotide base sequence of an
organism.
• Mutations occur spontaneously
and randomly throughout the
lifetime of all organisms
This red delicious apple illustrates a somatic
mutation. A mutation to the ovarian wall
gives rise to a sector of yellow colored fruit.
The mutation does not affect the seeds
(germline) which give rise to the standard
red delicious type.
• The effects of mutations vary
depending on their location both
within the chromosome (or
gene) and the body of the
organism.
Acquiring Mutations
• Mutations in the DNA of an
organism can be caused by:
- Mistakes in DNA replication.
• This is a natural process
• 1 mistake in 1,000,000,000 bases
• Proof reading enzymes correct most
mistakes
- Environmental factors that
increase the rate of mutations are
called mutagens.
• Radiation
• Various chemicals
• High temperatures
Can you inherit a mutation?
• Yes! If a mutation occurs in the cells that
produce gametes (germ-line cells) the change
will be passed onto the offspring.
• If a mutation occurs in any other cell of the
body (somatic cells) it will not be inherited,
but it may affect the individual during their
lifetime.
The biological consequences of
mutations
• Mutations may be beneficial, neutral or harmful!
• Mutations are a source of genetic variation – new
alleles in a population – that may be selected for
by environmental factors and confer an
advantage on the organism.
• Mutations are a source of biological novelty for
evolution.
Types of Mutations
• Point Mutations (i.e. spelling mistakes)
- Changes in a single DNA nucleotide
- These can occur within a gene’s coding region or within regulatory
regions of genes
• Block Mutations
-
Changes in a segment of a chromosome
These changes usually involve the rearrangement of a number of genes
• Chromosome Number Mutations
- Changes in the number of chromosomes
Point Mutations
• Single nucleotide substitutions may result in:
1. Changed amino acid sequence
2. No amino acid change because genetic code is
degenerate
3. Results in “stop” instruction and formation of a new
allele.
Mutation: Substitute T instead of C
Original
DNA
Mutant
DNA
Point Mutations
As a reference for the following screens, the diagram below
illustrates the transcription and translation of DNA without a
point mutation.
Original Unaltered Code
Original
DNA
Transcription
mRNA
Translation
Amino
acids
Amino acid sequence forms a normal polypeptide chain
1. Changed amino acid sequence
•
A single base is substituted by another.
• Usually results in coding for a new amino acid in the
polypeptide chain.
Mutation: Substitute T instead of C
Original DNA
Mutant DNA
mRNA
Amino acids
Polypeptide chain with wrong amino acid
2. No amino acid change
A single base is
Normal DNA
substituted by another.
mRNA
Called silent or neutral Amino acids
mutations and produce
little or no change in the
phenotype.
A change in the third
base of a codon still
codes for the same
amino acid.
Amino acid sequence from
the non-mutated DNA forms
a normal polypeptide chain
Mutation: Substitute C instead of T
Mutant DNA
mRNA
Amino acids
Despite the change in the last
base of a triplet, the amino acid
sequence is unchanged
3. Results in “stop” instruction and
formation of a new allele
•
A single base is substituted by another.
•
This results in a new triplet that does not code for an amino acid.
•
The resulting triplet may be an instruction to terminate the synthesis of the
polypeptide chain.
Mutation: Substitute A instead of C
Original DNA
Mutant DNA
mRNA
Amino acids
Mutated DNA creates a STOP
codon which prematurely ends
synthesis of the polypeptide chain
Reading Frame Shift by Insertion or Deletion
A single base is inserted, upsetting the reading sequence for all those after it.
•
A reading frame shift results in new amino acids in the polypeptide chain from the
point of insertion onwards.
•
The resulting protein will be grossly different from the one originally encoded (it is
most likely to be non-functional).
Mutation: Insertion of C
Original DNA
Mutant DNA
mRNA
Amino acids
Large scale frame shift results in a
new amino acid sequence. The
resulting protein is unlikely to have
any function.
Trinucleotide Repeat Expansions
• Many normal human genes contain multiple copies of a
three base sequence called a trinucleotide.
• These repeating sequences can expand in number. This
mutation gives rise to several inherited conditions.
• The mutant allele that causes “fragile X syndrome” has
200 to 2000 repeats of the trinucleotide CGG, in
contrast to 6 to 50 repeats in a normal person, in the
untranslated region of the FMR1 gene.
Fragile X Syndrome
• Occurs 1 out of every 4000 males and 1 out
of every 6000 females.
• Mutation of the FMR1 gene on the X
chromosome leads to loss of the fragile Xmetal retardation protein, FMRP. This FMRP
protein is involved in the translation of a
number of essential neuronal mRNA’s.
• Characteristics of this disease include:
-
Metal retardation
Shyness and limited eye contact
Elongated face
Large or protruding ears
Large testicles (macroorchidism)
Low muscle tone
Activity
• Quick check qu 28&29 (pg 373), 30&31 (pg
376), 20-24 (pg 412)
• Ch 10 ch review qu 11, 12, 13 pg 382-383
Reflection
• What learning was new today?
• What learning was revision or built on what I
already know?
• What did I find most challenging and what
strategies will I put in place to help me?
• What percentage of the class did I spend on
task and how can I improve this if needed?
5. Genetic Mutations
EL: Chromosomal mutations and
defects
Block Mutations
• The rearrangement of blocks of
genes within a chromosome. Can
occur during crossing over in
meiosis
• The rearrangement of blocks of
genes between non-homologous
chromosomes (translocation). A
piece of one chromosome is
broken off and joined to another
chromosome.
• Block mutations result in a new
gene order along a chromosome.
They can be highly disruptive!
translocations
Normal
Deletion: Pieces of chromosome are lost.
Duplication: Pieces of chromosome are repeated so there are duplicate segments
Inversion: Pieces of chromosome are flipped so the genes appear in reverse order.
Translocation: Pieces of chromosome are moved from one chromosome onto another
.
Changes to the
Chromosome Number
Examples of Polyploid
Plants
Name
Number
• Aneuploidy – changes
to the number of
specific chromosomes
Common
wheat
6N = 42
Tobacco
4N = 48
• Polyploidy- changes to
the number of whole
sets of chromosomes
Potato
4N = 48
Banana
3N = 27
Boysenberry
7N = 49
Strawberry
8N = 56
Aneuploidy
• Change to the number of specific
chromosomes.
• The extra or missing chromosome
may be an autosomal or a sex
chromosome.
• Such changes are due to nondisjunctions. These events are due
to errors in chromosome
segregation in meiosis.
• Pairs of homologous chromosomes
may fail to separate in meiosis I or
the centromere may fail to
separate the sister chromatids in
meiosis II.
Down’s Syndrome
An extra chromosome 21
Non-Disjunction: Meiosis I
Non-disjunction
•
•
Meiosis I
A non-disjunction in meiosis I
occurs when homologous
chromosomes fail to
separate properly during
anaphase I.
Meiosis II
One gamete receives two of
the same sort of
chromosome and the other
gamete receives no copy.
n+1
Gametes
n+1
n–1
n–1
Non-Disjunction: Meiosis II
•
•
•
Meiosis I
A non-disjunction in
meiosis II occurs when
sister chromatids fail to
separate properly during
anaphase II.
One gamete receives
two of the same sort of
chromosome and the
other gamete receives
no copy.
Some gametes are
unaffected.
Meiosis II
Nondisjunction
n+1
Gametes
n–1
n
n
Turner’s Syndrome 44 + X
• Occurs in 1 out of every
25,000 births
• Only one X chromosome
is present and is fully
functional.
• Common symptoms
include:
–
–
–
–
–
Short stature
Swelling of hands an feet
Broad chest
Low hairline
Webbed neck
Klinefelter’s Syndrome
44 + XXY
Mildly impaired IQ
(intelligence)
Chest
hair is
sparse
Poor
beard
growth
Frequently some
breast development
(low levels of
testosterone)
Osteoporosis
Female
type pubic
hair pattern
Penis and testes
underdeveloped, low
levels of testosterone.
Always infertile.
Limbs tend
to be longer
than
average
Sex chromosomes: XXY
•
•
•
•
•
•
•
•
Occurs in about 1 in 500 to 1000 births.
Characteristics vary widely.
Males are normally sterile
Some degree of language impairment
Youthful build
Rounded body type
Some degree of gynecomastia
Hypogonadism
Polyploidy
• Organisms that have more
than two sets of
chromosomes
• Very common in plants
because they can
reproduce asexually, but
rarer in animals.
• Polyploidy can result in
“instantaneous
speciation”.
• Polyploid plants are usually more
robust and sturdy than diploid
plants.
• In our long history of plant
cultivation we have selected out
such plants because they
produced a higher yield and were
less subject to disease.
• As a result many of our crops
today have been bred to a high
level of ploidy.
• Wheat was the first crop to be
domesticated originating in SW
Asia about 10,000 years ago.
• Today, bread wheat is a hexaploid.
An Expert in Polyploidy
• The twenty different species
worldwide differ widely in
their chromosome number as
they exhibit a range of
polyploidy.
• The haploid number is 7
As a general rule, strawberry species
with more sets of chromosomes tend to
be more robust and produce larger
plants with larger berries.
• Strawberry species an be:
–
–
–
–
–
Diploid
Tetraploid
Hexaploid
Octoploid
decaploid
Fitness of
Mutations
•
The fitness of a mutation describes its value to
the survival and reproductive success of the
organism. A mutation may turn out to be:
•
Lethal: Many mutations are lethal and
embryos are non-viable.
•
Harmful: Non-lethal mutations, e.g. Down
syndrome and sickle cell disease, may be
expressed as effects that lower fitness.
•
Silent (neutral): Most point mutations are
probably harmless, with no noticeable effect
on the phenotype.
•
Beneficial (useful): Occasionally mutations
may be useful, particularly in a new
environment, e.g. insecticide resistance in
insects, antibiotic resistance in bacteria.
Mutations: Key Points
•
All new alleles originate by
mutation.
•
New alleles introduce genetic
variation: the raw material on
which natural selection can
act.
•
Most mutations occur in
somatic cells and are not
inherited.
•
Only mutations in gametes can
be inherited.
Evolutionary Significance
of Mutations
• Polyploidy can result in the
formation of “instant
species” by creating a
barrier to chromosome
pairing at meiosis
(common in plants).
• Fusion of chromosomes (a
form of translocation) may
reduce chromosome
number. This can result in
reproductive isolation and
a new species.
Possible fusion of
two chromosomes
to create the No. 2
chromosome in
humans.
Note the similar
banding
patterns of
chromosomes
from related
primate species.
2
12
12
12
13
11
11
Human Chimpanzee Gorilla
Orangutan
Example: Fusion of chromosomes may have
taken place during the course of human
evolution.
The chromosome number in the great apes is
2N = 48, whereas in humans 2N = 46.
Activity:
• Complete Activity 11.2 “Changing your genes”
• Quick check qu 14&15 (pg 400)
• Biochallenge pg 413
• Ch 11 ch review qu 5, 7, 8, 11, 13, 14, 15
Reflection
• What learning was new today?
• What learning was revision or built on what I
already know?
• What did I find most challenging and what
strategies will I put in place to help me?
• What percentage of the class did I spend on
task and how can I improve this if needed