genetics
TRANSCRIPT
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Genetics
http://www.youtube.com/watch?v=CBezq1fFUEA&list=PL3EED4C1D684D3ADF
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4.1.1
State that eukaryote chromosomes are made of
DNA and proteins.
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4.1.2
Define gene, allele and genome.
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• 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
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4.1.3
Define gene mutation.
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• Sequence change• Different amino acid possible• Can be beneficial• Mutagens
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4.1.4
Explain the consequence of a base substitution mutation in relation to the processes of
transcription and translation, using the example of sickle-
cell anemia.
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• 1/655 African Americans
• Single base change• Change beta chain
shape (needles)
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4.2.1
State that meiosis is a reduction division of a diploid
nucleus to form haploid nuclei.
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4.2.2
Define homologous chromosomes.
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Same•Length•Loci for genes•Shape
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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.
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• http://www.youtube.com/watch?v=qCLmR9-YY7o&list=EC3EED4C1D684D3ADF
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Interphase
• Chromosomes not condensed• DNA replicates- S1
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Prophase I
• Nuclear membrane breakdown• DNA condensing• Spindle fibers• Centrioles move• Move towards equatorial plate• Longest phase of meiosis
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Metaphase I• Line up in center• Chromosomes most condensed• Crossing over (Metaphase and Prophase)-
chiasma• Independent assortment
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Anaphase I
• Homologous pair movement• “Arrow shape” of pairs
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Telophase I
• Sets at opposite sides• Nuclear membrane may reform (species)• Cleavage furrow• End meiosis I
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Prophase II
• Nuclear membrane breaks down• Spindle fibers reform
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Metaphase II
• Line up center• Independent assortment
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Anaphase II
• Spindle fibers contract• One chromatid per pole
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Telophase II
• Nuclear membranes form• Haploid• Crossing over variation
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4.2.4
Explain that non-disjunction can lead to changes in chromosome number,
illustrated by reference to Down syndrome (trisomy 21).
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• Can happen in Meiosis I or II
• Result in too few or too many of one chromosome
• Monosomy-more deadly, trisomy
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4.2.5
State that, in karyotyping, chromosomes are arranged in pairs according to their
size and structure.
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• Pictures during metaphase• 23 pairs
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4.2.6
State that karyotyping is performed using cells
collected by chorionic villus sampling or amniocentesis, for
pre-natal diagnosis of chromosome abnormalities.
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• 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
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4.2.7
Analyse a human karyotype to determine gender and
whether non-disjunction has occurred.
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4.3.1
Define genotype, phenotype, dominant allele, recessive allele, codominant alleles,
locus, homozygous, heterozygous, carrier and test
cross.
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4.3.2
Determine the genotypes and phenotypes of the
offspring of a monohybrid cross using a Punnett grid.
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• 2:2 ratio• Heterozygous
crosses 3:1 • Each fertilization
independent of others
• Larger the population, the closer to expected ratio
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4.3.3
State that some genes have more than two alleles
(multiple alleles).
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• 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)
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4.3.4
Describe ABO blood groups as an example of
codominance and multiple alleles.
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• I is immunoglobulin• The Allele hierarchy is IA = IB > I• When A and B present, both expressed and
both mask O
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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
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4.3.6
State that some genes are present on the X
chromosome and absent from the shorter Y
chromosome in humans.
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4.3.7
Define sex linkage.
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• 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
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4.3.8
Describe the inheritance of colour blindness and
hemophilia as examples of sex linkage.
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• 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
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• Color blindness– Red green color blindness sex linked recessive– More common in males
• Haemophelia – Sex-linked recessive– Cannot produce clotting factor
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4.3.9
State that a human female can be homozygous or
heterozygous with respect to sex-linked genes.
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4.3.10
Explain that female carriers are heterozygous for X-linked 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
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4.3.11
Predict the genotypic and phenotypic ratios of offspring
of monohybrid crosses involving any of the above
patterns of inheritance.
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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– What is the likelihood that the fourth child is a
taster?– What is the likelihood that the first three kids will
be non-tasters?
3 tasters:1 non-taster
75%
.25x.25x.25=.015=1.5%
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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.– 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?
codominance
50% black, 50% grey
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Flowers
• In snapdragons, red flower color is incompletely dominant over white flower color; the heterozygous plants have pink flowers.– If a red-flowered plant is crossed with a white-
flowered 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?
All Rr, all pink
25% RR red, 50% Rr pink, 25% rr white
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• Haemophilia or red-green color blindness for sex linked
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4.3.12
Deduce the genotypes and phenotypes of individuals
in pedigree charts.
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•Squares male, circles female•Red=affected, blue=unaffected•Horizontal line= mated•Vertical line denotes offspring
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• 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?
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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
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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 single-stranded 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 DNA-DNA 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.
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• 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 single-stranded 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.
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4.4.2
State that, in gel electrophoresis, fragments of DNA move in an electric
field and are separated according to their size.
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• 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
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4.4.3
State that gel electrophoresis of DNA is
used in DNA profiling.
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• 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.
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Main uses
• Forensic crime investigations• Parentage Issues• Animal breeding pedigrees• Disease detection
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4.4.4
Describe the application of DNA profiling to determine
paternity and also in forensic investigations.
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4.4.5
Analyse DNA profiles to draw conclusions about
paternity or forensic investigations.
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• 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.
Parentage
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Forensics
• Compare bands of specimen to suspects
• Not enough to convict someone but can narrow the search
• Match suspect one
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4.4.6
Outline three outcomes of the sequencing of the
complete human genome.
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• 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
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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.
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• 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
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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.
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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.
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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
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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
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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
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Stage 5: Product manufacture
• 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.
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4.4.9
State two examples of the current uses of genetically modified crops or animals.
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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.
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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
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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
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4.4.10
Discuss the potential benefits and possible harmful effects of one
example of genetic modification.
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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
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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
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4.4.11
Define clone.
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• Clone: a group of genetically identical organisms or a group of cells derived from a single parent
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4.4.12
Outline a technique for cloning using differentiated
animal cells.
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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 cells-
implanted in surrogate• Normal gestation period• Sheep 1 and 4 genetically identical
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4.4.13
Discuss the ethical issues of therapeutic cloning in
humans.
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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
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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
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Genetics HL
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10.1.1
Describe the behavior of the chromosomes in the phases of meiosis.
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• 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
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• Anaphase I– Spindle fibers shorten– Chromosomes pulled in opposite directions
• Telophase I– Chromosomes uncoil– Nuclear membrane may reform– May enter an interphase period
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• Meiosis II the same but nucleus reforms and there are half the number of chromatids
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10.1.2
Outline the formation of chiasmata in the process of
crossing over.
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• 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
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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.
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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
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Random assortment• Homologue from either parent can go in
either direction• Over 8 million possible orientations (2 )• Combined with crossing over, the possibilities
are extremely large
23
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10.1.4
State Mendel’s law of independent assortment.
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• The transmission of genes to each child is independent of the others
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10.1.5
Explain the relationship between Mendel’s law of independent assortment
and meiosis.
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• Orientation of chromosomes at prophase I random
• Direction the chromosome is facing is also random and independent
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10.2.1
Calculate and predict the genotypic and phenotypic
ratio of offspring of dihybrid crosses involving
unlinked autosomal genes.
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• 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)
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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
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10.2.2
Distinguish between autosomes and sex
chromosomes.
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• Sex chromosomes determine gender (X and Y)– Not always same size (X much larger
than Y)• Autosomes- other 22 pairs
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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.
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10.2.4
Define linkage group.
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• A set of genes on a chromosome that tend to be inherited together
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10.2.5
Explain an example of a cross between two linked
genes.
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• 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
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Flower color and pollen length
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10.2.6
Identify which of the offspring are recombinants in a dihybrid cross involving
linked genes
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• Combinations not found in parents
• Lower frequency
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10.3.1
Define polygenic inheritance.
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• A single characteristic determined by two or more genes
• Follow a bell curve generally• Phenotype combinations increase-
continuous
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10.3.2
Explain that polygenic inheritance can contribute to
continuous variation using two examples, one of which must be human skin color.
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Human skin color
• Determined by amount of melanin present• At least four genes involved• One allele codes for melanin the other for
nonproduction
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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
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Wheat grain color
• Varies from white to dark red• Controlled by three genes• Pigment producing vs no pigment alleles