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Major Concepts for 4th 6 weeks
• Mendel Genetics – Slides 2-25
• Pedigrees – Slides 26-36
• DNA and RNA (protein synthesis) – Slides 3773
• Genetic Disorders – Slides 74-78
• Mutations – Slides 79-101
• Genetic Engineering – Slides 102 -117
Mendel Genetics
• Objectives:
• Predict the outcome of a cross between parents
of know genotype.
• Determine the probability of a particular trait in
an offspring based upon the genotype of parents
and the particular mode of inheritance.
• Incomplete dominance, co-dominance, multiple
alleles, polygenic, complete dominance, and sexlinked
Word Wall
True-breeding
Homozygous
Phenotype
Physical Trait
Tall
Gamete
Sex Cells – Egg and Sperm
Heterozygous
Tt
Hybrid
Genotype
The actual
genetic make-up Gene
TT:Tt:tt
Allele
2 Alleles (one from each
parent that code for trait)
Form of gene (T or t)
Big Eyes are
dominant = BB or Bb
Small eyes = bb
Punnett square example
Alleles for male
Alleles for
Female
Both parents
are
heterozygous
Yy x Yy
Possible
Genotypes of
Offspring
1 YY:2 Yy: 1 yy
Phenotype –
3:1
Cross a homozygous Round
with wrinkled
R
In a
Punnett
square,
the
Alleles
always
move to
squares as
shown.
R
r
Rr
Rr
r
Rr
Rr
The actual
alleles
Physical
description of trait
Genotype =
Phenotype =
Probability =
RR or Rr= round
rr = wrinkled
Parents are RR
which is same
(homozygous)
alleles for
dominant and rr
which are same
for recessive
trait
4 Rr (heterozygous)
4 round
100% round
Cross a hybrid with a hybrid
R
In a
Punnett
square,
the
Alleles
always
move to
squares as
shown.
r
R
RR
Rr
r
Rr
rr
The actual
alleles
Physical
description of trait
RR or Rr= round
rr = wrinkled
Parents are Rr
which is
heterozygous
CLASSIC –
Mendel Hybrid
Cross
Dominant – 75%
Recessive – 25%
*Determine recessive
trait by small number
showing the trait
Genotype =
Phenotype =
Probability =
1 RR:2Rr:1rr
3 Round, 1 wrinkled
75% round, 25% wrinkled
Independent Assortment
• Alleles separate
independently
during the
formation of
gametes.
The dihybrid cross
EeTt x EeTt
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Mendel’s Peas
Dihybrid Cross
Cross:
TtYy x TtYy
TY
Ty
tY
ty
TTYY
TTYy
TtYY
TtYy
Tall, yellow
Tall, yellow
Tall, yellow
Tall, yellow
TTYy
TTyy
TtYy
Ttyy
Tall, yellow
Tall, green
Tall, yellow
Tall, green
TtYY
TtYy
ttYY
ttYy
Tall, yellow
Tall, yellow
TtYy
Ttyy
Tall, yellow
Tall, green
TY
Notice
Phenotype
Ratio
9:3:3:1
Ty
tY
Dwarf, yellow Dwarf, yellow
ttYy
ttyy
ty
Genotypes:
Phenotypes:
1 TTYY
: 2 TTYy
Dwarf, yellow Dwarf, green
: 4 TyYy : 2 TtYY : 1 TTyy
9 tall
plants with
yellow seeds
: 2 Ttyy : 1 ttYY : 2 ttYy : 1 ttyy
3 tall
plants with
green seeds
3 dwarf
plants with
yellow seeds
1 dwarf
plant with
green seeds
Incomplete Dominance
Japanese four-o-clock flowers
• Red flower plant genotype = RR
• White flower plant genotype = WW
• Pink flower plant genotype = RW
Appear blended. Incomplete, not Full
Strength.
Cross a Red flower with a
White Flower
R
In a
Punnett
square,
the
Alleles
always
move to
squares as
shown.
R
RR = Red
WW = white
RW = Pink
W
RW
RW
W
RW
RW
The actual
alleles
Physical
description of trait
Genotype =
Phenotype =
Probability =
4 RW
4 Pink
100% Pink
Parents are RR
for red and WW
for white. Both
are homozygous
or true
breeding.
Co Dominance
NOTE: Alleles can be represented
different ways. RR for Red, WW for
White,RW for Roan or RR for Red, R’R’
for white, and RR’ for Roan. Let’s look
at a Punnett Square with both
examples.
FULL Strength
RR x WW = RW or
RR X R’R’ = RR’
Roan Cow
Cross a Roan cow with white
cow. Co-Dominance
R
In a
Punnett
square,
the
Alleles
always
move to
squares as
shown.
W
RR = Red cow
WW = white cow
RW = Roan Cow
W
RW
WW
W
RW
WW
The actual
alleles
Physical
description of trait
Genotype =
Phenotype =
Probability =
Parents are RW
for Roan which
is heterozygous
WW which is
homozygous for
White
2 RW, 2 WW
2 Roan, 2 White
50% Roan, 50% White
Cross a Roan cow with white
cow. Co-Dominance
R
In a
Punnett
square,
the
Alleles
always
move to
squares as
shown.
R’
RR = Red cow
R’R’ = white cow
RR’ = Roan Cow
R’
RR’
R’R’
R’
RR’
R’R’
The actual
alleles
Physical
description of trait
Genotype =
Phenotype =
Probability =
Parents are RW
for Roan which
is heterozygous
WW which is
homozygous for
White
2 RR’, 2 R’R’
2 Roan, 2 White
50% Roan, 50% White
Multiple Alleles
• When more than two alleles (form of gene)
contribute to the phenotype.
• Human blood types are an example
• There are three possible alleles: A,B, and O
• Both A and B are dominant over O.
• O is recessive
• AB is an example of Co-Dominance
6 different genotypes, 3 different Alleles
•
•
•
•
•
•
I AI A
I Ai
I AI B
I BI B
Ibi
ii
Type A - 2 possible
genotypes
Type AB
Type B – 2 possible
genotypes
Type O
Cross a heterozygous type A with
homozygous type B
Punnett
square
the
Alleles
always
move to
squares as
shown.
A
I
i
B
I
A
B
II
B
Ii
B
I
A
B
II
B
Ii
The actual
alleles
Physical
description of trait
Genotype =
Phenotype =
Probability =
A = IAIA, IAi
B= IBIB, IBi
AB =IAIB O = ii
IAIB, IBi
2 AB, 2 B
50% AB, 50% B
Polygenic traits
• Traits controlled by two or more
genes.
• Lots of variation in trait.
• Examples:
–Human height,
eye and skin
color
Figure 11.17
Skin Color
Autosomal and Sex-Linked Traits
• Autosomal - Traits controlled by genes on
chromosomes 1 -22.
• Sex-Linked – Traits controlled by the X
chromosome or the Y chromosome.
• Most often sex-linked traits are on the X
chromosome.
• Let’s look at some of examples and work
together.
Cross a heterozygous female with
a colorblind male
n
X
Y
Female = XX
Male = XY
Normal = N, colorblind = n
N
X
N
n
X X
N
X Y
Xn
n
n
XX
n
XY
The actual
alleles
Genotype =
Phenotype =
Physical
description of trait Probability =
Work like any
other Punnett
Square.
Remember no
letter on the Y.
The trait is
connected to
the X!
XNXn,XnXn,XNY,XnY
2 Females, 1 Normal, 1 Color-blind
2 Males, 1 Normal, 1 Color-blind
50% Colorblind
Test Your Knowledge of Punnett
Square
• http://www.biology.clc.uc.edu/courses/bio10
5/geneprob.htm
Sex Cells (Gametes) from Meiosis
1N
EGG
Pedigrees
• Apply pedigree data to interpret various
modes of genetic inheritance.
 A pedigree is a chart of the genetic history of
family over several generations.
 Scientists or a genetic counselor would find
out about your family history and make this
chart to analyze.
Symbols in a Pedigree Chart
• Normal Female
• Affected female
 Female carrier Not all
Female is
represented
by a circle
pedigrees show carriers
 Normal Male
 Affected Male
Male carrier – Not possible in
Sex-linked traits (if you see
carrier male, it is autosomal)
Male is
represented as
a square
What does a pedigree chart look like?
X NY
X NX n
1st
X NX N
generation
X nY
2nd generation
3rd generation
•Does this pedigree show a sex-linked trait?
•Yes, males are affected more than females, and females are carriers.
•How many children were born in generation 2 to couple with affected male?
•3, 2 boys and a girl.
•What is the genotype of the female in generation 3?
•XNXN
•What are genotypes for generation 1?
If carriers are not shown, genotype could be homozygous or
heterozygous even though trait is not shown.
X NY
XNXN or XNXn
XNXN or
1st generation
X NX n
X nY
2nd generation
3rd generation
This is the same pedigree without female carriers
being shown. The large affect it has on males, tells
us it is sex-linked and since it is not showing up in
females, it is recessive. NOT all pedigrees will show
carriers, so be careful with analyzing!
Interpreting a Pedigree Chart
1. Determine if the pedigree chart shows an
autosomal or X-linked disease.
– If most of the males in the pedigree are
affected the disorder is most likely X-linked
– If it is a 50/50 ratio between men and
women the disorder is most likely
autosomal
• When interpreting a pedigree chart of a family with a
disease like muscular dystrophy, it is important to
consider two steps. The first is to determine if the
disorder is autosomal or X-linked.
• If the disorder is X-linked most of the males will have
the disorder because the Y-chromosome cannot mask
the affects of an affected X-chromosome. A female
can have the disorder, but it would be a very low
percentage. For a female to be affected, she would
have had to receive an affected gene from both the
mother and the father. This means that the father
would have the disorder and the mother was a carrier.
• In an autosomal disorder, the disorder is not found on
the X or Y chromosome. It is found on the other 22
chromosomes in the human body. This means that
men and women have an equal chance of having the
disorder.
 Is it Autosomal or X-linked?
Autosomal because it affects males and females equally
Interpreting a Pedigree Chart
2. Determine whether the disorder is dominant
or recessive.
– If the disorder is dominant, one of the
parents must have the disorder.
– If the disorder is recessive, neither parent
has to have the disorder because they can
be heterozygous.
It is important to find out if a disorder is dominant or recessive. For
example, Huntington’s disease is a dominant disorder. If you have
only one dominant gene you will have Huntington’s disease, which
is a lethal disorder. The disorder does not show up until a person is
in their middle ages such as 45. It will quickly decrease their motor
skills and the brain will begin to deteriorate.
• If a disorder is dominant, one parent must have the disorder (either
homozygous dominant (TT) or heterozygous recessive (Tt). Both
parents do not have to have the disorder. One parent might not
have the disorder or be a carrier. If a disease is dominant, it does
not skip a generation unless one parent is heterozygous dominant
(Tt) and the other parent is homozygous recessive (tt). In this case
the child has a chance of not receiving the dominant gene.
• If the disorder is recessive, a parent does not have to have the
disorder, but could still pass it to their offspring. This would happen
when a parent is heterozygous recessive (Tt) and passes on the
recessive (t) gene. This means this disorder can skip generations.
An example of a recessive disorder would be sickle cell anemia.
Dominant or Recessive?
It is dominant because a parent in every generation has the disorder.
Remember if a parent in every generation has the disorder, the disorder
has not skipped a generation. If the disorder has not skipped a generation,
the disorder is dominant.
Practice Analyzing Pedigrees
• http://www.zerobio.com/drag_gr11/pedigree
/pedigree_overview.htm
Dominant or Recessive?
It is recessive, because a parent in every generation does not have the disorder. If a disorder
Skips a generation, then the disorder is recessive. If a carrier is shown, it is recessive also.
Scientists call this the:
DNA
RNA
Protein
DNA Nucleotide
Deoxyribose Nucleic Acid
Phosphate
Group
O
O=P-O
O
5
CH2
O
N
C1
C4
Sugar
(deoxyribose)
C3
C2
Nitrogenous base
(A, G, C, or T)
Watson and Crick constructed a Model
of DNA showing the double helix.
• James Watson and Francis Crick worked out the
three-dimensional structure of DNA, based on
work by Rosalind Franklin
Figure 10.3A, B
DNA Double Helix
“Rungs of ladder”
Nitrogenous
Base (A,T,G or C)
“Legs of ladder”
Phosphate &
Sugar Backbone
Chargaff’s Rule
• Adenine must pair with Thymine
• Guanine must pair with Cytosine
• Their amounts in a given DNA molecule will be about
the same.
T
A
G
C
DNA
Nucleotides
joined
together
DNA Double Helix
5
O
3
Notice base pairing
A+T
G+C
3
P
5
O
C
G
1
O
P
5
3
2
4
4
2
3
P
1
T
5
A
P
3
O
O
P
5
O
3
5
P
The Code of Life…
• The “code” of the chromosome is the SPECIFIC ORDER
that bases occur. Proteins are built from the code.
A T C G T A T G C G G…
DNA Replication
• DNA must be copied so new cells will
have complete instructions for making
the RIGHT proteins.
• The DNA molecule produces 2
IDENTICAL new complementary strands
following the rules of base pairing:
A-T, G-C
•Each strand of the
original DNA serves as a
template for the new
strand
Each DNA molecule
contains one original and
one new complementary
strand
DNA Replication
• Complementary base pairs form new strands.
• …DNA control cell functions by serving as a
template for PROTEIN structure.
• RNA uses base pairing, but the T is replaced
with U for Uracil. A + U, G + C
• 3 Nucleotides = a triplet or CODON
(which code for a specific AMINO ACID
• AMINO ACIDS are the building blocks of
proteins.
• Proteins regulate cell activity and express
traits controlled by genes.
DNA – Blueprint for Life
DNA
RNA – Ribosome –
Amino Acid
Protein
Trait
Expresses Trait
Protein Synthesis – Building
Proteins
DNA contains the instructions for the proteins
that are needed for life. If the DNA does not
replicate correctly, the wrong protein could be
made.
DNA and RNA Comparison
Double
Strand
Single
Strand
A+T
G+C
A+U
G+C
Both have
Phosphate
Deoxyribose
Ribose
DNA always STAYS in Nucleus
RNA is in nucleus during transcription, moves in
cytoplasm, and on ribosome during translation.
Table 14.2
Types of RNA
Type of RNA
Functions in
Messenger RNA
(mRNA)
Nucleus,
migrates
to ribosomes
in cytoplasm
Transfer RNA
(tRNA)
Cytoplasm
Provides linkage
between mRNA
and amino acids;
transfers amino
acids to ribosomes
Ribosomal RNA
(rRNA)
Cytoplasm
Structural
component
of ribosomes
Function
Carries DNA
sequence
information to
ribosomes
DNA makes RNA during Transcription
• DNA can “unzip” itself
and RNA nucleotides
match up to the DNA
strand.
• Both DNA & RNA are
formed from
NUCLEOTIDES and are
called NUCLEIC acids.
• The information constituting an organism’s
genotype is carried in its sequence of bases
– The DNA is transcribed into RNA, which is
translated into the polypeptide
DNA
TRANSCRIPTION
RNA
TRANSLATION
Protein
Figure 10.6A
Transcription produces genetic
messages in the form of mRNA
RNA
polymerase
RNA nucleotide
Direction of
transcription
Template
strand of DNA
Figure 10.9A
Newly made RNA
RNA polymerase
• In transcription, DNA
helix unzips
– RNA nucleotides line
up along one strand
of DNA, following the
base-pairing rules
– single-stranded
messenger RNA peels
away and DNA
strands rejoin
DNA of gene
Promoter
DNA
Initiation
Elongation
Terminator
DNA
Area shown
in Figure 10.9A
Termination
Growing
RNA
Completed RNA
Figure 10.9B
RNA
polymerase
Eukaryotic RNA is processed
before leaving the nucleus
• Noncoding
segments,
introns, are
spliced out
• A cap and a tail
are added to
the ends
Exon Intron
Exon
Intron
Exon
DNA
Cap
RNA
transcript
with cap
and tail
Transcription
Addition of cap and tail
Introns removed
Tail
Exons spliced together
mRNA
Coding sequence
NUCLEUS
CYTOPLASM
Figure 10.10
RNA builds Proteins from Amino
Acids during Translation
• The cell uses
information from
“messenger” RNA to
produce proteins
mRNA leaves the nucleus
to go to ribosome
Amino
Acids
tRNA
Anti-codon
Proteins – Express
Traits
codon
rRNA and tRNA translate
The message to make proteins
Translation of nucleic acids into
amino acids
• The “words” of the DNA “language” are triplets
of bases called codons
• The codons in a gene specify the amino acid
sequence of a polypeptide
• RNA Transcription copies the DNA onto mRNA.
• Translation takes place in the cytoplasm on the
ribosomes.
• tRNA picks up the correct amino acid and builds
a protein on the rRNA from the mRNA.
Types of RNA
• mRNA contains codons which code for amino acids.
3 Letter
Code for
amino acids
What
amino
acid will
the code
CAU
make?
His
Virtually all organisms share the same genetic
code “unity of life” Second Base
C
U
UUU
UUC
UUA
UUG
C
CUU
CUC
CUA
CUG
A
AUU
AUC
ile
AUA
AUG met (start)
ACU
ACC
ACA
ACG
G
GUU
GUC
GUA
GUG
GCU
GCC
GCA
GCG
phe
leu
leu
val
UCU
UCC
UCA
UCG
CCU
CCC
CCA
CCG
A
ser
UAU
UAC
UAA
UAG
pro
CAU
CAC
CAA
CAG
thr
AAU
AAC
AAA
AAG
ala
GAU
GAC
GAA
GAG
G
tyr
stop
stop
his
gln
asn
lys
asp
glu
UGU
UGC
UGA
UGG
CGU
CGC
CGA
CGG
AGU
AGC
AGA
AGG
GGU
GGC
GGA
GGG
64 possible combinations – 20 specific amino acids
cys
stop
trp
arg
ser
arg
gly
U
C
A
G
U
C
A
G
U
C
A
G
U
C
A
G
Third Base
First Base
U
What signals the ribosome to start translating the mRNA
Into a new amino acid sequence and signals it to stop?
An initiation codon marks the
start of an mRNA message
AUG = methionine
Start of genetic message
End
Figure 10.13A
• An exercise in translating the genetic code
Transcribed strand
DNA
Transcription
RNA
Start
codon
Polypeptide
Translation
Stop
codon
Figure 10.8B
Gene 1
Gene 3
DNA molecule
Gene 2
Proteins
are built
from
chains of
amino
acids
DNA strand
TRANSCRIPTION
RNA
Codon
TRANSLATION
Polypeptide
Amino acid
Ribosomes build polypeptides
(chain of amino acids)
Next amino acid
to be added to
polypeptide
Growing
polypeptide
tRNA
molecules
P site
A site
Growing
polypeptide
Large
subunit
tRNA
P
A
mRNA
mRNA
binding
site
Codons
mRNA
Small
subunit
Figure 10.12A-C
• mRNA, a specific tRNA, and the ribosome
subunits assemble during initiation
Large
ribosomal
subunit
Initiator tRNA
P site
A site
Start
codon
mRNA
1
Figure 10.13B
Small ribosomal
subunit
2
Amino acid
Polypeptide
A
site
P site
Anticodon
mRNA
1
Codon recognition
mRNA
movement
Stop
codon
New
peptide
bond
3
Translocation
2
Peptide bond
formation
Figure 10.14
Overview of
Protein
Synthesis
• Let’s look at it
ONE more
time!
TRANSCRIPTION
DNA
mRNA
RNA
polymerase
Stage 1 mRNA is
transcribed from a
DNA template.
Amino acid
TRANSLATION
Enzyme
Stage 2 Each amino
acid attaches to its
proper tRNA with the
help of a specific
enzyme and ATP.
tRNA
Initiator
tRNA
mRNA
Figure 10.15
Anticodon
Large
ribosomal
subunit
Start
Codon
Small
ribosomal
subunit
Stage 3 Initiation of
polypeptide synthesis
The mRNA, the first
tRNA, and the
ribosomal subunits
come together.
New
peptide
bond
forming
Growing
polypeptide
Codons
Stage 4 Elongation
A succession of tRNAs
add their amino acids to
the polypeptide chain as
the mRNA is moved
through the ribosome,
one codon at a time.
mRNA
Polypeptide
Stop Codon
Figure 10.15 (continued)
Stage 5 Termination
The ribosome recognizes
a stop codon. The polypeptide is terminated and
released.
DNA – Blueprint for Life
DNA
RNA – Ribosome –
Amino Acid
Protein
Trait
Expresses Trait
1. Why is transcription necessary?
Transcription makes messenger RNA (mRNA) to
carry the code for proteins out of the nucleus to
the ribosomes in the cytoplasm.
2. Describe transcription.
RNA polymerase binds to DNA, separates the
strands, then uses one strand as a template to
assemble mRNA.
3. Why is translation necessary?
Translation assures that the right amino acids are
joined together by peptides to form the correct
protein.
4. Describe translation.
The cell uses information from mRNA to
produce proteins. The tRNA brings the right
amino acid to ribosome, rRNA to produce a
specific amino acid chain that will later become
an active protein.
5. What are the main differences between DNA
and RNA.
DNA has deoxyribose, RNA has ribose; DNA has
2 strands, RNA has one strand; DNA has
thymine, RNA has uracil.
6. Using the chart on page 303, identify the amino
acids coded for by these codons: UGG CAG UGC
tryptophan-glutamine-cysteine
Genetic Disorders
Autosomal Recessive
Normal = N
Genetic
Disorders
nn = cystic fibrosis
Both parents Must be Carriers
Nn X Nn
Sickle Cell
Anemia
Autosomal recessive
Both parents must be carriers
To pass to children.
Nn X Nn
Or one is carrier and other
has condition. Nn x nn
Would not show in parents if
Carriers
Tay-Sachs
Autosomal
Recessive
Huntingdon’s Disease
Autosomal Dominant
What Are Mutations?
• Changes in the nucleotide sequence
of DNA
• May occur in somatic cells (body
cells,aren’t passed to offspring)
• May occur in gametes (eggs &
sperm) and be passed to offspring
• May be chromosomal or gene
mutations.
DNA – If there is a mutation in the DNA strand,
then the RNA strand will be changed
DNA
If the mRNA brings the wrong
instructions, may result in
Gene
wrong protein
– Ribosome –
Amino Acid
Protein
Many mutations
do not change
the amino acid,
so NO mutation
will occur.
Trait
Expresses Trait
Mutation – wrong protein
Protein Translation
• Modified genetic code is “translated” into
proteins
• Codon code is specific, but redundant!
– 20 amino acids
– 64 triplet (codon) combinations
Which is why some mutations don’t matter!
Gene Mutations
• Change in the
nucleotide sequence of
a gene
• May only involve a single
nucleotide
• May be due to copying
errors, chemicals,
viruses, etc.
Point Mutation
• Change of a single
nucleotide
• Includes the deletion,
insertion, or substitution
of ONE nucleotide in a
gene
• Sickle Cell disease is the
result of one nucleotide
substitution
• Occurs in the
hemoglobin gene
Frameshift Mutation
• Inserting or
deleting one or
more nucleotides
• Changes the
“reading frame”
like changing a
sentence
• Proteins built
incorrectly
Normal hemoglobin DNA
mRNA
Mutant hemoglobin DNA
mRNA
Normal hemoglobin
Sickle-cell hemoglobin
Glu
Val
Example of
Sickle Cell
mutation
• Illustration of mutations
NORMAL GENE
mRNA
Protein
Met
Lys
Phe
Gly
Ala
Lys
Phe
Ser
Ala
BASE SUBSTITUTION
Met
Missing
BASE DELETION
Met
Lys
Leu
Ala
His
Figure 10.16B
•Chromosomal changes can be large or small
Deletion
Homologous
chromosomes
Duplication
Inversion
Reciprocal
translocation
Nonhomologous
chromosomes
Figure 8.23A, B
Chromosome Mutations
• May Involve:
– Changing the
structure of a
chromosome
– Can cause
abnormal
development of
offspring. of part
Deletion
• Due to breakage
• A piece of a
chromosome is lost
Inversion
• Chromosome segment
breaks off
• Segment flips around
backwards
• Segment reattaches
Duplication
• Occurs when a gene
sequence is
repeated
Translocation
• Involves two
chromosomes that
aren’t homologous
• Part of one
chromosome is
transferred to
another chromosomes
Nondisjunction
• Failure of chromosomes to
separate during meiosis
• Causes gamete to have too many
or too few chromosomes
• Disorders:
– Down Syndrome – three 21st chromosomes
– Turner Syndrome – single X chromosome
– Klinefelter’s Syndrome – XXY chromosomes
Normal Male Karotype
2n = 46
96
Normal Female Karotype
2n = 46
97
Male, Trisomy 21 (Down’s)
Can you spot the problem?
2n = 47
98
Female Down’s Syndrome
2n = 47
99
Klinefelter’s Syndrome
2n = 47
100
Genetic Engineering
• Evaluate the scientific and
ethical issues associated
with gene technologies.
• Genetic Engineers refers to
the alteration of an
organism’s genes for
practical purposes.
• Recombinant DNA
• Transgenic Organisms
• Cloning
• Stem Cell Research
• Gel Electrophoresis/DNA
fingerprinting
Recombinant Bacteria
1.
Remove bacterial DNA (plasmid).
2.
Cut the Bacterial DNA with
“restriction enzymes”.
3.
Cut the DNA from another
organism with “restriction
enzymes”.
4.
Combine the cut pieces of DNA
together with another enzyme and
insert them into bacteria.
5.
Reproduce the recombinant
bacteria.
6.
The foreign genes will be expressed
in the bacteria.
Benefits of Recombinant Bacteria
1. Bacteria can make human insulin or human
growth hormone.
1. Bacteria can be engineered to “eat” oil spills.
Recombinant DNA
• The ability to combine
the DNA of one
organism with the DNA
of another organism.
• Recombinant DNA
technology was first
used in the 1970’s with
bacteria.
Genetically modified organisms are called
transgenic organisms.
TRANSGENIC ANIMALS
1.
Mice – used to study human
immune system
2.
Chickens – more resistant to
infections
3.
Cows – increase milk supply and
leaner meat
4. Goats, sheep and pigs – produce
human proteins in their milk
Transgenic Goat
Human DNA in a
Goat Cell
Carries a foreign
gene that has been
inserted into its
genome.
.
This goat contains a human gene
that codes for a blood clotting
agent. The blood clotting agent
can be harvested in the goat’s
milk.
How to Create a
Transgenic Animal
Desired DNA
is
added to an
egg cell.
The DNA of plants and animals
can also be altered.
PLANTS
1. disease-resistant and
insect-resistant crops
2. Hardier fruit
3. 70-75% of food in
supermarket is genetically
modified.
How to Create a Genetically
Modified Plant
1.Create recombinant
bacteria with desired
gene.
2. Allow the bacteria to
“infect" the plant cells.
3. Desired gene is inserted
into plant
chromosomes.
DNA Cloning
• Transfer of DNA
fragment from one
organism to a selfreplicating genetic
element such as
bacterial plasmid
Reproductive Cloning
• Generate an animal that
has the same nuclear
DNA as another existing
animal.
Therapeutic Cloning
• Also called “embryo cloning”,
is the production of human
embryos for use in research.
• Stem Cell Collection:
• Are unspecialized cells
capable of renewing
themselves through cell
division.
• Under certain experimental
conditions, they can be
induced to become tissue or
organ specific cells with
special functions.
What do you think about eating genetically
modified foods?
Polymerase Chain Reaction
PCR
•
PCR allows scientists to
make many copies of a
piece of DNA.
1. Heat the DNA so it
“unzips”.
2. Add the complementary
nitrogenous bases.
3. Allow DNA to cool so the
complementary strands
can “zip” together.
Steps Involved in Gel Electrophoresis
1. “Cut” DNA sample with
restriction enzymes.
2. Run the DNA fragments through
a gel.
3. Bands will form in the gel.
4. Everyone’s DNA bands are unique
and can be used to identify a
person.
5. DNA bands are like “genetic
fingerprints”.