Download genetic ppt melanie - IB

Genetics
http://www.youtube.com/watch?v=CBezq1fFUEA&list=PL3EED4C1D684D3ADF
4.1.1
State that eukaryote
chromosomes are made of
DNA and proteins.
4.1.2
Define gene, allele and
genome.
• Gene- heritable factor that codes for a certain
trait
– Ex: eye color
• Allele- what will be expressed in that gene
– Ex: blue eyes
• Genome- Collection of all of an organism’s
genes
– usually encoded in DNA
– HGP
4.1.3
Define gene mutation.
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Sequence change
Different amino acid possible
Can be beneficial
Mutagens
4.1.4
Explain the consequence of a
base substitution mutation in
relation to the processes of
transcription and translation,
using the example of sicklecell anemia.
• 1/655 African
Americans
• Single base change
• Change beta chain
shape (needles)
4.2.1
State that meiosis is a
reduction division of a diploid
nucleus to form haploid
nuclei.
4.2.2
Define homologous
chromosomes.
Same
•Length
•Loci for genes
•Shape
4.2.3
Outline the process of meiosis,
including pairing of
homologous chromosomes
and crossing over, followed
by two divisions, which
results in four haploid cells.
• http://www.youtube.com/watch?v=qCLmR9YY7o&list=EC3EED4C1D684D3ADF
Interphase
• Chromosomes not condensed
• DNA replicates- S1
Prophase I
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Nuclear membrane breakdown
DNA condensing
Spindle fibers
Centrioles move
Move towards equatorial plate
Longest phase of meiosis
Metaphase I
• Line up in center
• Chromosomes most condensed
• Crossing over (Metaphase and Prophase)chiasma
• Independent assortment
Anaphase I
• Homologous pair movement
• “Arrow shape” of pairs
Telophase I
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Sets at opposite sides
Nuclear membrane may reform (species)
Cleavage furrow
End meiosis I
Prophase II
• Nuclear membrane breaks down
• Spindle fibers reform
Metaphase II
• Line up center
• Independent assortment
Anaphase II
• Spindle fibers contract
• One chromatid per pole
Telophase II
• Nuclear membranes form
• Haploid
• Crossing over variation
4.2.4
Explain that non-disjunction
can lead to changes in
chromosome number,
illustrated by reference to
Down syndrome (trisomy 21).
• Can happen in Meiosis I or
II
• Result in too few or too
many of one chromosome
• Monosomy-more deadly,
trisomy
4.2.5
State that, in karyotyping,
chromosomes are arranged
in pairs according to their
size and structure.
• Pictures during metaphase
• 23 pairs
4.2.6
State that karyotyping is
performed using cells
collected by chorionic villus
sampling or amniocentesis,
for pre-natal diagnosis of
chromosome abnormalities.
• Obtain fetal cells to do a karyotype to find out
if baby has disorder like Downs
• Amniocentesis- removal of amniotic fluid
containing fetal cells
– Centrifuge separates chromosomes into size
• Chorionic villus sampling- samples chorion
– Must be after 8 weeks of pregnancy
– Uses catheter
4.2.7
Analyse a human karyotype to
determine gender and
whether non-disjunction has
occurred.
4.3.1
Define genotype, phenotype,
dominant allele, recessive
allele, codominant alleles,
locus, homozygous,
heterozygous, carrier and test
cross.
4.3.2
Determine the genotypes
and phenotypes of the
offspring of a monohybrid
cross using a Punnett grid.
• 2:2 ratio
• Heterozygous
crosses 3:1
• Each
fertilization
independent of
others
• Larger the
population, the
closer to
expected ratio
4.3.3
State that some genes have
more than two alleles
(multiple alleles).
• Creates more phenotypes and genotypes
• Example: Rabbit coat colour(C) has four alleles
which have the dominance hierarchy: C > cch >
ch > c
• This produces 5 phenotypes, Dark(C_) ,
Chinchilla( cchcch), light grey (cchch ,cchc), Point
restricted (ch ch, chc) and albino (cc)
4.3.4
Describe ABO blood groups
as an example of
codominance and multiple
alleles.
• I is immunoglobulin
• The Allele hierarchy is IA = IB > I
• When A and B present, both expressed and
both mask O
4.3.5
Explain how the sex
chromosomes control
gender by referring to the
inheritance of X and Y
chromosomes in humans.
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23rd pair
X much longer than Y
X from mom Y from dad in boys
Theoretically 50/50 chance for gender, but
some have predisposition
• SRY gene (supposed to be found on tip of Y)
determines gender
4.3.6
State that some genes are
present on the X
chromosome and absent
from the shorter Y
chromosome in humans.
4.3.7
Define sex linkage.
• Genes on non-homologous region of X
• Females spare tire other X, males don’t have
this
• More common in males
• Boys inherit these from mom
4.3.8
Describe the inheritance of
colour blindness and
hemophilia as examples of
sex linkage.
• Genes on non-homologous part of X
• Males always get affected gene from mother
• Males cannot pass on affected gene to sons,
but can to daughters
• Color blindness
– Red green color blindness sex linked recessive
– More common in males
• Haemophelia
– Sex-linked recessive
– Cannot produce clotting factor
4.3.9
State that a human female
can be homozygous or
heterozygous with respect
to sex-linked genes.
4.3.10
Explain that female carriers
are heterozygous for Xlinked recessive alleles.
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Carriers are heterozygous women
Have both recessive and dominant alleles
Females have 2 long X chromosomes
Dominant overpowers recessive
4.3.11
Predict the genotypic and
phenotypic ratios of offspring
of monohybrid crosses
involving any of the above
patterns of inheritance.
PTC Test
• Determined by a single gene
• Ability to taste=T, inability=t
• Practice
– If heterozygous parents were crossed, predict the
genotypic and phenotypic ratios 3 tasters:1 non-taster
– What is the likelihood that the fourth child is a
taster? 75%
– What is the likelihood that the first three kids will
be non-tasters? .25x.25x.25=.015=1.5%
Roosters
• A rooster with grey feathers is mated with a
hen of the same phenotype. Among their
offspring 15 chicks are grey, 6 are black and 8
are white.
codominance
– What is the simplest explanation for the
inheritance of these colors in chickens?
– What offspring would you expect from the mating
of a grey rooster and a black hen? 50% black, 50% grey
Flowers
• In snapdragons, red flower color is
incompletely dominant over white flower
color; the heterozygous plants have pink
flowers.
All Rr, all pink
– If a red-flowered plant is crossed with a whiteflowered plant, what are the genotypes and
phenotypes of the plants of the F1 generation?
– What genotypes and phenotypes can be produced
in the F2 generation? 25% RR red, 50% Rr pink, 25% rr white
• Haemophilia or red-green color
blindness for sex linked
4.3.12
Deduce the genotypes and
phenotypes of individuals
in pedigree charts.
•Squares male, circles
female
•Red=affected,
blue=unaffected
•Horizontal line=
mated
•Vertical line denotes
offspring
• Phenylketonuria (PKU) is a metabolic disorder and a recessive genetic
condition.
• The pedigree shows the inheritance through a particular family.
• Which individuals can we be sure about their genotype?
• Since it was not possible to identify the condition of 12 and 13 suggest
their genotype and phenotype and how the diagram may need modifying?
4.4.1
Outline the use of
polymerase chain reaction
(PCR) to copy and amplify
minute quantities of DNA.
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DNA cloning
Used when DNA source is small
Temperature used to break and reform bonds
Polymerases thermostable so high temps can
be used
PCR steps,
• Initialization step: heat the reaction to a
temperature of 94–96 °C (or 98 °C if
extremely thermostable polymerases are
used) for 1–9 minutes.
• Denaturation step: heat the reaction to
94–98 °C for 20–30 seconds. It causes DNA
melting of the DNA template by disrupting
the hydrogen bonds between
complementary bases, yielding singlestranded DNA molecules.
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Annealing step: The reaction temperature is lowered to 50–65 °C for 20–40 seconds
allowing annealing of the primers to the single-stranded DNA template. Stable DNADNA hydrogen bonds are only formed when the primer sequence very closely
matches the template sequence. The polymerase binds to the primer-template hybrid
and begins DNA formation.
Extension/elongation step: commonly a temperature of 72 °C is used with Taq
polymerase. At this step the DNA polymerase synthesizes a new DNA strand
complementary to the DNA template strand by adding dNTPs that are complementary
to the template in 5' to 3' direction, condensing the 5'-phosphate group of the dNTPs
(deoxyribonucleotide triphosphates ) with the 3'-hydroxyl group at the end of the
DNA strand. The extension time depends both on the DNA polymerase used and on
the length of the DNA fragment to be amplified. As a rule-of-thumb, at its optimum
temperature, the DNA polymerase will polymerize a thousand bases per minute.
Under optimum conditions, i.e., if there are no limitations due to limiting substrates
or reagents, at each extension step, the amount of DNA target is doubled, leading to
exponential (geometric) amplification of the specific DNA fragment.
• Final elongation: This single step is
occasionally performed at a
temperature of 70–74 °C for 5–15
minutes after the last PCR cycle to
ensure that any remaining singlestranded DNA is fully extended.
• Final hold: This step at 4–15 °C for an
indefinite time may be employed for
short-term storage of the reaction.
4.4.2
State that, in gel
electrophoresis, fragments
of DNA move in an electric
field and are separated
according to their size.
• Sample of fragmented DNA is placed in one of
the wells on the gel.
• An electrical current is passed across the gel.
• Fragment separation based on charge and size
• Large fragments move slowly
4.4.3
State that gel
electrophoresis of DNA is
used in DNA profiling.
• Satellite (Tandem repeating) DNA are highly
repetitive sequences of DNA from the non
coding region of DNA.
• Different individuals have a unique length to
their satellite regions.
• These can be used to differentiate between
one individual and another.
Main uses
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Forensic crime investigations
Parentage Issues
Animal breeding pedigrees
Disease detection
4.4.4
Describe the application of
DNA profiling to determine
paternity and also in
forensic investigations.
4.4.5
Analyse DNA profiles to
draw conclusions about
paternity or forensic
investigations.
Parentage
• The DNA fragments in the child comes
from the mother and father.
• A band present in the child must come
either from the mother or from the
father
• The bands on the child's fragments are
either found on the mother or the
male1.
Forensics
• Compare bands of specimen to
suspects
• Not enough to convict
someone but can narrow the
search
• Match suspect one
4.4.6
Outline three outcomes of
the sequencing of the
complete human genome.
• It is now easier to study how genes influence
human development
• It helps identify genetic diseases
• It allows the production of new drugs based
on DNA base sequences of genes or the
structure of proteins coded for by these genes
• It will give us more information on the origins,
evolution and migration of humans
4.4.7
State that, when genes are
transferred between species,
the amino acid sequence of
polypeptides translated from
them is unchanged because
the genetic code is universal.
• All known organisms use the same genetic
code
• Transferring a gene from one species to
another would result in the production of the
same protein
• Some prokaryotes differ slightly, but not
dramatically
4.4.8
Outline a basic technique used
for gene transfer involving
plasmids, a host cell
(bacterium, yeast or other
cell), restriction enzymes
(endonucleases) and DNA
ligase.
Stage 1: obtaining the gene for
transfer
• Restriction enzymes are used to cut out the
useful gene that is to be transferred.
• Note the 'sticky ends' of unattached hydrogen
bonds.
Stage 2: Preparing a vector for the
transferred gene
• Plasmids are small circular DNA molecules found in bacteria
• These can be cut with the same restriction enzyme before
• This leaves the same complementary 'sticky ends' in the
plasmid
• The plasmid can be cut at particular sites. These are called
restriction sites and some are named in the diagram
Stage 3: Recombinant DNA
• (a) plasmid that will be the vector
• (b) plasmid cut at restriction site Pstl
• (c) Source DNA cut with same restriction enzyme
as plasmid to (d)
• (e) Recombinant DNA
• (f) unaffected plasmid
Stage 4: Isolation of transformed cells
• Recombinant DNA is introduced into the host cells
• Many cells remain untransformed
• Some cells are transformed to contain the recombinant
DNA.
• These transformed cells must be separated from
untransformed
Stage 5: Product manufacture
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The transformed bacterial cells are isolated.
They are introduced into a Fermenter to be cloned.
The bacterial population grows by asexual reproduction.
The Recombinant DNA is copied along with the rest of the
bacterial genome.
• In a fermenter the conditions for growth and reproduction
are controlled.
• Once the bacteria express the transformed gene the
product is produced.
• The next (long ) step is to isolate and purify the product.
This is called downstream processing.
4.4.9
State two examples of the
current uses of genetically
modified crops or animals.
Factor IX
• produced by genetically modified sheep
• expressed in milk from which it must be isolated before use by
haemophiliacs
• A ewe is treated with fertility drugs to create super-ovulation.
• Eggs are inseminated.
• Each fertilised egg has the transgene injected.
• A surrogate ewe has the egg implanted for gestation.
• Lambs are born which are transgenic, GMO for this factor IX gene.
• Each Lamb when mature can produce milk.
• The factor IX protein is in the milk and so must be isolated and
purified before use in human.
Herbicides
• Weeds growing near a crop use up soil nutrients that would
otherwise be used by the crop plant
• This competition of resources reduces the productivity of
the crop plant and therefore the efficiency of farming
• Herbicides can be used prior to crop planting to kill weeds.
• The herbicide cannot be used after crops have been sown
as they will also kill the crop.
• However, Cotton, Corn and Soybeans have been genetically
modified to contain an enzyme that breaks down
glyphosate
• This makes these crops resistant to the herbicide
• Herbicide can then be use after the crop has grown to
prevent the reoccurrence of weed competition
Retinol Rice
• Retinol deficiencies can cause stinted growth or blindness
• Third world countries
• Rice does not contain retinol or beta-carotene (used to
make retinol), but contains molecule normally used to
make beta-carotene
• The gene and enzymes to manufacture are missing from
rice
• Source of the gene is either Erwinia bacterium or the
common daffodil
• The transgenic rice is usually yellow in color because of the
accumulation of beta-carotene
• This transgenic rice is then crossed with local strains of rice
4.4.10
Discuss the potential
benefits and possible
harmful effects of one
example of genetic
modification.
Adding Bt toxin to corn crops
• Used to keep insects from eating crop
• Benefits
– Higher crop yield
– Less land needed for more crops
– Reduce use of pesticides
• Costs
– Unsure of consequences to human health
– Kill other insects or animals on crop or surrounding
plants (contain toxin in pollen)
– Other plants obtain the advantage and crowd out
those who do not have it
General
• Benefits
– Reduce spoilage
– Increase crop or animal byproduct yields
– Increase sources of essential dietary needs
• Costs
– May be unsafe for human
– Considered “unnatural”
– “Contaminate” other organisms
4.4.11
Define clone.
• Clone: a group of genetically
identical organisms or a group of
cells derived from a single parent
4.4.12
Outline a technique for
cloning using differentiated
animal cells.
Dolly
• Udder cell removed from parent
sheep
• Remove egg cell nucleus of second
sheep
• Cells are fused
• Egg cell can still divide by mitosis
• Cell grown IV until contains 16 cellsimplanted in surrogate
• Normal gestation period
• Sheep 1 and 4 genetically identical
4.4.13
Discuss the ethical issues of
therapeutic cloning in
humans.
Benefits
• Can replace tissues/ organs saving lives and
stopping pain
• Cells can be taken from embryos no longer
developing so they would have died anyway
• Can be taken when embryos cannot feel due
to lack of nerve cells
Costs
• Embryos should be given the chance to
develop
• Often too many embryos are produced and
are simply killed
• Embryonic stem cells may become cancerous
Genetics HL
10.1.1
Describe the behavior of the
chromosomes in the
phases of meiosis.
• Prophase I
– Chromosomes coil tightly
– Homologous chromosomes pair up
– Crossing over
– Nuclear membrane dissolves
• Metaphase I
– Spindle fibers attach to centromeres
– Homologous line up at the middle
– Chromosomes most condensed
• Anaphase I
– Spindle fibers shorten
– Chromosomes pulled in opposite directions
• Telophase I
– Chromosomes uncoil
– Nuclear membrane may reform
– May enter an interphase period
• Meiosis II the same but
nucleus reforms and there
are half the number of
chromatids
10.1.2
Outline the formation of
chiasmata in the process of
crossing over.
• Interphase
– DNA replicates
• Prophase I
– Molecules (cohesins) hold the pairs
together
– DNA is exchanged
• Metaphase I
– Homologous pairs repel at
centromere but are held together at
chiasma
– Cohesins break down
• Anaphase I
– Pulling apart breaks chiasma
10.1.3
Explain how meiosis results in
an effectively infinite genetic
variety in gametes through
crossing over in prophase I
and random orientation in
metaphase I.
Crossing over
• Exchange maternal and paternal chromosome
info
• Chromatids with new combinations
• Different from both parents- recombinants
• Can occur at any locus and more than one
could form
Random assortment
• Homologue from either parent can go in
either direction
23
• Over 8 million possible orientations (2 )
• Combined with crossing over, the possibilities
are extremely large
10.1.4
State Mendel’s law of
independent assortment.
• The transmission of genes to
each child is independent of the
others
10.1.5
Explain the relationship
between Mendel’s law of
independent assortment
and meiosis.
• Orientation of chromosomes at
prophase I random
• Direction the chromosome is facing is
also random and independent
10.2.1
Calculate and predict the
genotypic and phenotypic
ratio of offspring of
dihybrid crosses involving
unlinked autosomal genes.
• Crossing the F1 generation of
homozygous dominant and
recessive P1 for true cross
(RRYYxrryy or RRyyxrrYY)
• Two characteristics controlled by
two different genes
• 16 square Punnet square
• 9:3:3:1 ratio phenotypes in
double heterozygous (RrYy)
Test Cross for Heterozygotes
• Cross expected heterozygote with double
homozygous recessive
• Homozygous dominant can only produce
dominant offspring
• 1:1:1:1 phenotype ratio should be found if
heterozygous
10.2.2
Distinguish between
autosomes and sex
chromosomes.
• Sex chromosomes determine gender
(X and Y)
– Not always same size (X much larger
than Y)
• Autosomes- other 22 pairs
10.2.3
Explain how crossing over
between non-sister
chromatids of a homologous
pair in prophase I can result
in an exchange of alleles.
10.2.4
Define linkage group.
• A set of genes on a chromosome
that tend to be inherited together
10.2.5
Explain an example of a
cross between two linked
genes.
• Will follow the phenotypic ratio for a
monohybrid because they are inherited
together
• Recombinants will only occur if crossing over
in prophase I
• 3:1 rather than 9:3:3:1
Flower color and pollen length
10.2.6
Identify which of the
offspring are recombinants
in a dihybrid cross involving
linked genes
• Combinations not found in
parents
• Lower frequency
10.3.1
Define polygenic
inheritance.
• A single characteristic determined by
two or more genes
• Follow a bell curve generally
• Phenotype combinations increasecontinuous
10.3.2
Explain that polygenic
inheritance can contribute to
continuous variation using
two examples, one of which
must be human skin color.
Human skin color
• Determined by amount of melanin present
• At least four genes involved
• One allele codes for melanin the other for
nonproduction
Finch beak depth
• A= add depth (1
unit)
• a= no depth added
• B= add depth (1
unit)
• b= no depth added
• C= add depth (1
unit)
• c= no depth added
Wheat grain color
• Varies from white to dark red
• Controlled by three genes
• Pigment producing vs no pigment alleles