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Topic Five: Genetics
http://commons.wikimedia.org/wiki/File:Hall_of_Biodiversity,_American_Museum_of_Natural_History_(7356570500).jpg
Hall of Biodiversity
3.1 Genes
• Essential idea: Every living organism inherits
a blueprint for life from its parents.
http://pixabay.com/static/uploads/photo/2014/05/11/14/25/elephants-341981_640.jpg
Chromosome
Histones
DNA double helix
3.1 U.1 A gene is a heritable factor that consists of a length of DNA and
influences a specific characteristic
Chromosome is a length of DNA with associated proteins (histones).
3.1 U.1 A gene is a heritable factor that consists of a length of DNA and
influences a specific characteristic
Eukaryotic
Chromosomes
• The Photograph shows a
human chromosome,
which are made of DNA
and protein. The picture
show two chromatids
newly replicated DNA
(uncoiled chromosomes)
joined at the
centromere. Each
Eukaryotic
chromosomes is
composed of genes
which control specific
characteristics (alleles
are different forms of a
specific gene)
3.1 U.2 A gene occupies a specific position on a chromosome.
http://www.daviddarlin
g.info/images/gene.jpg
•Gene is a heritable factor that controls a specific
characteristic. There are over 20,000 genes that make up a
human. One gene codes for one polypeptide.
http://physrev.physiology.org/content/physrev/91/1/151/F11.large.jpg
A gene can exist in many different
forms, called alleles. An example
hair color: black hair, brown hair,
auburn hair, red hair, blond hair, etc.
You inherit one allele for each gene
from your mother and one from
your father. Each allele maybe a
dominant characteristic
(represented with a upper case
letter) or weak or recessive
(represented with a lower case
letter). You need only one dominant
allele to express a dominant trait.
You need two recessive alleles to
show a recessive form of a trait. See
the heredity diagram for tongue
rolling to see how dominant and
recessive alleles work. No tongue
rolling is the recessive form of the
gene.
Tongue Roller Genetics
Allele is one specific form of a gene,
differing from other alleles by one or
a few bases only and occupying the
same gene locus as other alleles of
the gene.
3.1 U.3 The various specific forms of a gene are alleles.
•Gene loci is the position of the gene on a chromosome. Alleles have the same
gene loci.
3.2 U.1 Prokaryotes have one chromosome consisting of a circular DNA
molecule.
Prokaryotes have two types of DNA: a single chromosome
and plasmids (not associated with any of the cells life
processes
• Prokaryotic DNA is circular and is not associated with any
histone proteins (“naked DNA)
• There is one copy of each gene except when the cell and
its DNA are replicating
• The single chromosome is coiled up and concentrated in
the nucleoid region.
Bacterial
DNA
Plasmids
Nucleosome Structure: consist of 8 histones. They help create super coiling of
chromatin, which creates a chromosome during cell replication
3.2 U.3 Eukaryote chromosomes are linear DNA molecules associated
with histone proteins.
http://micro.magnet.fsu.edu/cells/nucleus/images/
chromatinstructurefigure1.jpg
• Eukaryotic chromosomes are linear
and are made up of DNA and
histone proteins, important in gene
expression and replication.
• Histones are globular shaped
protein in which the DNA is
wrapped around.
• DNA wrapped around 8 histone
proteins is called a nucleosome.
• The DNA wraps twice around the
histone protein core.
• Another histone protein is attached
to the outside of the DNA strand.
This helps maintain the colloidal
structure of the nucleosome.
• DNA, because of its negative charge
is attracted to the positive charge
on the amino acids of the histone
proteins.
3.2 U.3 Eukaryote chromosomes are linear DNA molecules associated
with histone proteins.
http://upload.wikimedia.org/wikipedia/commons/
4/45/Nucleosome_organization.png
3.2 U.1 Prokaryotes have one chromosome consisting of a circular DNA
molecule.
https://classconnection.s3.amazonaws.com/730/flashcards/127
6730/jpg/cell_types1330806303645.jpg
3.2 U.2 Some prokaryotes also have plasmids but eukaryotes do not.
http://upload.wikimedia.org/wikipedia/en/6/6c/PBR322_pl
asmid_showing_restriction_sites_and_resistance_genes.jpg
Features of Plasmids:
• Naked DNA - not associated with histone
proteins
• Small circular rings of DNA
• Can be passed between prokaryotes
• Plasmids are small separate (usually
circular) DNA molecules located in some
prokaryotic cells
• Plasmids are also naked (not associated
with proteins) and are not needed for
daily life processes in the cell.
• The genes in plasmids are often
associated with antibiotic resistant and
can be transferred from one bacterial cell
to another.
• Plasmids are readily used by scientists to
artificially transfer genes from one species
to another (ie. Gene for human insulin)
3.2 U.4 In a eukaryote species there are different chromosomes that carry
different genes.
https://s-media-cache-
ak0.pinimg.com/originals/4d/d0/7
a/4dd07a61c384abefef3c521bcab
6bad8.jpg
http://dxline.info/img/new_ail/chro
mosome-4.jpg
• Chromosomes are
linear, varying in length
and in position of the
centromere that holds
the sister chromatids
together.
• In humans there are 23
types of chromosomes.
Each chromosome
carries a specific
sequence of genes
along the linear DNA
molecule. The position
where the gene is
located is called the
locus.
• Alleles are different
forms of a gene on the
chromosome.
Chromosome 1 Chromosome 4
2,000 Genes 1,000 Genes
https://public.ornl.gov/site/gallery/originals/
Eukaryotes possess multiple chromosomes. All
individuals of a species possess the same
chromosomes, with the same gene loci. For
example all humans have twenty three pairs.
Chromosomes can vary by:
• Length – the number of base pairs in the DNA molecule
• Position of the centromere
• Genes occur at a specific locus (location), i.e. it is always
found at the same position on the same chromosome
(the locus and genes possessed vary between species)
3.2 U.4 In a eukaryote species there are different chromosomes that
carry different genes.
3.1 U.6 The genome is the whole of the genetic information of an
organism.
Allows for:
• Physical mapping of chromosomes
• Used to screen for genetic diseases
• Lead to a better understanding of genetic diseases.
• May lead to the development of better drugs to fight diseases.
• Maybe be used for comparison of genomes with other species.
3.2 A.2 Comparison of genome size in T2 phage, Escherichia coli, Drosophila
melanogaster, Homo sapiens and Paris japonica. [Genome size is the total length of DNA in an
organism. The examples of genome and chromosome number have been selected to allow
points of interest to be raised
Name Genome Length
(million base pairs)
Number of Genes
T2 phage (Virus) 0.18 300
Escherichia coli
(Bacteria)
5 4,377
Drosophila
melanogaster
(Fruit Fly)
140 17,000
Paris japonica
(Woodland Plant)
150,000 Unknown
Homo sapiens
(Human)
3,000 19-23,000
3.1 A.2 Comparison of the number of genes in humans with other species. [The
number of genes in a species should not be referred to as genome size as this term is
used for the total amount of DNA. At least one plant and one bacterium should be
included in the comparison and at least one species with more genes and one with
fewer genes than a human.]
3.1 U.6 The genome is the whole of the genetic information of an
organism.
Genome Size and Number of Genes
3.2 A.2 Comparison of genome size in T2 phage, Escherichia coli, Drosophila
melanogaster, Homo sapiens and Paris japonica. [Genome size is the total length of DNA in an
organism. The examples of genome and chromosome number have been selected to allow
points of interest to be raised
Paris japonica
Largest Known Genome
3.2 A.3 Comparison of diploid chromosome numbers of Homo
sapiens, Pan troglodytes, Canis familiaris, Oryza sativa, Parascaris
equorum.
Homo sapiens (46)
3.2 A.3 Comparison of diploid chromosome numbers of Homo sapiens, Pan
troglodytes, Canis familiaris, Oryza sativa, Parascaris equorum.
Pan troglodytes (48)
http://static1.squarespace.com/static/51b89f0ce4b0000660671282/t/540785e0e4b00aec95acf4cb/1409779168825/floandfigan.jpg
3.2 A.3 Comparison of diploid chromosome numbers of Homo
sapiens, Pan troglodytes, Canis familiaris, Oryza sativa, Parascaris
equorum.
Canis familiaris 78
https://c1.staticflickr.com/5/4007/5076413065_6a42962771_b.jpg
3.2 A.3 Comparison of diploid chromosome numbers of Homo
sapiens, Pan troglodytes, Canis familiaris, Oryza sativa, Parascaris
equorum.
Species Name Number of
Chromosomes
Homo sapiens 46
Pan troglodytes 48
Canis familiaris 78
Oryza sativa 24
Parascaris
equorum
2
Rice: Oryza sativa
http://upload.wikimedia.org/wikipedia
/commons/6/65/Oryza_sativa_Kyoto_J
PN_001.JPG
Round Worm:
Parascaris equorum
http://www.vetnext.com/fotos/quizc2a4.jpg
Chromosome 16
3.1 U.6 The genome is the whole of the genetic information of an
organism.
http://physrev.physiology.org/content/physrev/91/1/151/F11.large.jpg
3.1 U.6 The genome is the whole of the genetic information of an
organism.
Advantages from the project
Advantage 1
• Screening can also be done for
adults, to see if they may be
carriers of potential genetic
conditions. Certain Jewish and
Canadian populations regularly
obtain voluntary screening for Tay-
Sachs disease, a known child-killer.
This information has been used to
help make decisions about future
marriage partners.
Advantage 2
Perhaps the greatest benefit will come
from what is called gene-based
therapy. Understanding the
molecular workings of genes and
the proteins they encode will lead
to more precise drug treatments.
The more precise the drug
treatment, the fewer and milder
will be the side effects.
3.1 U.7 The entire base sequence of human genes was sequenced in the
Human Genome Project.
• The Human Genome Project (HGP) was an international scientific research project with the
goal of determining the sequence of chemical base pairs which make up human DNA, and
of identifying and mapping all of the genes of the human genome from both a physical and
functional standpoint. It remains the world's largest collaborative biological project
• Key findings of the draft (2001) and complete (2004) genome sequences include:
 There are approximately 20,500 genes in human beings, the same range as in mice.
 Significantly more segmental duplications (nearly identical, repeated sections of DNA)
than had been previously suspected.
 At the time of publishing fewer than 7% of protein families appeared to be vertebrate
specific.
3.1 U.7 The entire base sequence of human genes was sequenced in the
Human Genome Project.
http://www.bioscience.heacademy.ac.uk/imagebank/search/Fullimage.aspx?IDvalues=6513
Key advances in technology:
• Biotechnology techniques such as PCR are used to prepare samples: the DNA needs
to be copied to prepare a sufficiently large pure samples to sequence
• Computers automate the sequencing process
• Fluorescent labeling techniques enable all four nucleotides to be analyzed together
• Lasers are used to fluoresce the dye markers
• Digital camera technology reads the dye markers
• Computers are used to assemble the base sequence
3.1.S1 Use of a database to determine differences in the base sequence
of a gene in two species.
One use of aligning base sequences is to determine
the differences between species: this can be used to
help determine evolutionary relationships.
http://www.ncbi.nlm
.nih.gov/genbank
GenBank
http://bitesizebio.s3.amazonaws.com/wp-content/uploads/2012/10/header-image-copy18.jpg
Your task is to analyze the differences between three or more species
(the skill asks for two species, but the online Clustal tool works better
with a minimum of three).
For each chosen species retrieve the base sequence:
• Go to GenBank website http://www.ncbi.nlm.nih.gov/genbank
• Select ‘Gene’ from the search bar
• Enter the name of a gene (e.g. AMY1A for salivary amylase 1A or COX1 for
cytochrome oxidase 1) AND the organism (use the binomial) and press ‘Search’
n.b. if you are comparing species the gene chosen needs to be the same for each
species
• Select the ‘Name/Gene ID’ to get a detailed view
• Scroll down to the ‘Genomic regions, transcripts, and products’ section and click on
‘FASTA’
• Copy the entire sequence from ‘>’ onwards
• Save the sequence – you will need to align with the other species next
3.1.S1 Use of a database to determine differences in the base sequence
of a gene in two species.
http://www.ebi.ac.uk/Tools/msa/clustalo/
Analysis:
• ‘Alignments’ allows you to visually check the results – this is easier
if the chosen gene has a short base sequence
• Under ‘Results Summary’ use the ‘Percent Identity Matrix’ to
quantify the overall similarity (0 = no similarity, 100 = identical)
• Under ‘Phylogenic Tree’ chose the ‘Real’ option for the Phylogram
to get a visual representation of how similar the species are (based
on the chosen gene).
http://bitesizebio.s3.amazonaws.com/wp-content/uploads/2012/10/header-image-copy18.jpg
To align the sequences:
• Go to the Clustal Omega website
http://www.ebi.ac.uk/Tools/msa/clustalo/
• In STEP 1 Select ‘DNA’ under ‘a set of’
• Paste the chosen sequences into the box
(each sequence must start on a new line)
• Press ‘Submit’ (and wait – depending on
the size of the sequences you may have to
wait for a couple of minutes)
3.3 Meiosis
Essential Question: Alleles segregate during meiosis allowing
new combinations to be formed by the fusion of gametes.
http://i.huffpost.com/gen/1123387/original.jpg
Grandmother Granddaughter
10.1 Meiosis
Essential idea: Meiosis leads to independent assortment of
chromosomes and unique composition of alleles in daughter cells.
http://upload.wikimedia.org/wikipedia/c
ommons/4/46/Dividing_Cell_Fluorescenc
e.jpg
Review: 1.6.U.1 Mitosis is division of the nucleus into two genetically identical daughter nuclei.
http://highered.mheducation.com/sites/0072495
855/student_view0/chapter2/animation__mitosis
_and_cytokinesis.html
Use the animated tutorials to learn about mitosis
http://www.johnkyrk.com/mitosis.html
http://www.sumanasinc.com/webcontent/animations/content
/mitosis.html
http://outreach.mcb.harvard.edu/animations/cellcycle.
swf
3.2 U.7 Haploid nuclei have one chromosome of each pair. [The two
DNA molecules formed by DNA replication prior to cell division are
considered to be sister chromatids until the splitting of the centromere
at the start of anaphase. After this, they are individual chromosomes.]
• Haploid nuclei have one copy of each chromosome or one full set of the
chromosomes (Sex cells) in that particular species eg. Human 23 chromosomes (or
n)
• These are called gametes, which are sperm and egg
• Human sperm and eggs each contain 23 chromosomes (when the fuse together the
become diploid with 46 chromosomes (or 2n, a somatic body cell developing into
any type of cell (ex. Blood cell, liver cell heart cell)
3.2 U.7 Haploid nuclei have one chromosome of each pair. [The two
DNA molecules formed by DNA replication prior to cell division are
considered to be sister chromatids until the splitting of the centromere
at the start of anaphase. After this, they are individual chromosomes.]
• Chromosome to the right
splits at the centromere.
• This occurs during Anaphase
• The chromatids move to
opposite poles of the cell to
become chromosomes in a
newly created cell.
Chromosome
With two
Chromatids
Become Two
Chromosome
3.3 U.1 One diploid nucleus divides by meiosis to produce four haploid
nuclei.
http://cis.payap.ac.th/wp-content/uploads/2011/11/142.png
• Cells divide twice
• Result: 4 daughter cells, each
with half as many chromosomes
as parent cell
• What if a complex multicellular organism (like
us) wants to reproduce?
– joining of egg + sperm
• Do we make egg & sperm by mitosis?
46 46+ 92
egg sperm zygote
What if we did, then….
Doesn’t work!
No!
3.3 U.2 The halving of the chromosome number allows a sexual life cycle
with fusion of gametes.
3.3 U.1 One diploid nucleus divides by meiosis to produce four haploid nuclei.
Edited from: https://commons.wikimedia.org/wiki/File:Diagram_of_meiosis.svg
Meiosis is a reduction division of the nucleus to form haploid gametes
One diploid (2N) body cells contain a homologous
pair of each chromosome (except for sex cells)
Four haploid (N) gametes contain
one of each chromosome
Second division of the nucleus
First
division
of the
nucleus
Chromosomes are replicated to
form sister chromatids
3.3 U.1 One diploid nucleus divides by meiosis to produce four haploid nuclei.
http://www.sumanasinc.com/webcontent/animations/c
ontent/meiosis.html
http://highered.mheducation.com/sites/dl
/free/0072437316/120074/bio19.swf
http://www.stolaf.edu/people/giannini/fla
shanimat/celldivision/meiosis.swf
http://www.biostudio.com/d_%20Meiosis.htm
The animations
are a great way
to visualize the
process –
watch and take
notes.
Meiosis is a reduction division of the nucleus to form haploid gametes
3.3 U.2 The halving of the chromosome number allows a sexual life cycle
with fusion of gametes.
http://lh6.ggpht.com/n2T0xB_8BXE/UY4bhlV52uI/AAAAAAAABxg/7ZjWF
QMBfx4/s1600-h/Humanlifecycle9.jpg
Sexual life cycle
• Meiosis: reduces #
of chromosomes
creating gametes (n)
• Fertilization:
combine gametes
(sperm + egg)
creating a fertilized
egg or zygote (2n)
3.3 U.2 The halving of the chromosome number allows a sexual life cycle
with fusion of gametes.
Stage of the sexual life
cycle
Chromosome
number (N/2N)
Haploid or
Diploid
Adults 2n = 46 D
Gametes (egg and sperm
cells)
n = 23 H
Zygote 2n = 46 D
Juvenile 2n = 46 D
Duplication of Chromosomes During S stage of Interphase
Single chromosome
(called chromatin
during Interphase)
Two identical sister
chromatids joined by
a centromere
DNA replication during
S stage of Interphase
3.3 U.3 DNA is replicated before meiosis so that all chromosomes
consist of two sister chromatids.
.
10.1 U.1 Chromosomes replicate in interphase before meiosis.
• During the S phase of
the cell cycle, so that
each chromosome has a
copy of itself and
consists of two sister
chromatids.
• During meiosis I,
chromosomes condense
and synapse to form
bivalents (homologous
chromosomes are
aligned next to each
other).
http://www.columbia.edu/cu/biology/co
urses/c200507/images/meiosis_19.gif
3.2 A.1 Cairns’ technique for measuring the length of DNA molecules by
autoradiography.
• John Cairns first supplied the E. Coli
bacteria cells with suitable radioactive
material (replaces normal hydrogen in
thymidine).
• Used to selectively label only DNA and
will not label RNA.
• Intact E. Coli bacterial chromosomes
are placed on slides. These slides are
then covered by photographic
emulsion and stored in dark.
• During this storage the particles are
emitted the radioactive material, which
exposed the film
• The photographs show the regions of
labelled DNA.
Findings
• The images showed that E. coli
possesses a single circular chromosome
which is 1,100 μm long (E. coli cells
have a length of only 2 μm)
• Cairns images also provided evidence to
support the theory of semi-conservative
replication
3.2 U.5 Homologous chromosomes carry the same sequence of genes
but not necessarily the same alleles of those genes.
https://molecularhelix.files.wordpress.com/2011/07/homolg.gif
• Homologous chromosomes are
chromosomes within each cell that
carry the same genes
• One chromosome came from an
individual’s mother and one from
the father
• They have the same shape and size
• These chromosomes pair up during
meiosis
• Even though these chromosomes
carry the same genes, they could
have different alleles (different
versions of the same gene)
Crossing Over is an exchange of genetic
information as two of the same
chromosomes move near to each other
and exchange information
3.2 U.6 Diploid nuclei have pairs of homologous chromosomes.
• Diploid nuclei have two copies of
each type of chromosome. One
chromosome comes from the
mother and one from the father.
• Haploid gametes (sperm and egg)
fuse during sexual reproduction
which produces zygote with a
diploid nucleus
• This cell will then divide by
mitosis to produce numerous
cells, all with a diploid nucleus
• Each nucleus has two copies of
each gene, accept the sex
chromosomes
3.2 U.8 The number of chromosomes is a characteristic feature of
members of a species.
• The chromosome number is a characteristic
feature of that species 2 (n) in a diploid cell or
n in a haploid cell.
• A chromosome number does not indicate how
complicated an organism might be
• Organisms with different numbers of
chromosomes which is why the letter n is
used because it is a variable and they might
not be able to interbred
• Chromosome number tends to remain
unchanged over millions of years of
evolution; however, sometimes through
evolution chromosomes can fuse together or
split to change the number of chromosomes
an organism contains
• During human evolution, two ancestral ape
chromosomes fused to produce human
chromosome 2
First variation created by meiosis: Crossover
• Red and Blue chromosomes are a homologous pair.
• They have replicated during the interphase and each copy is
held together by the centromere (black dot)
3.3 U.4 The early stages of meiosis involve pairing of homologous
chromosomes and crossing over followed by condensation. [The process
of chiasmata formation need not be explained.]
•The chromosomes overlap and exchange lengths of DNA.
•The chiasmata are the points at which the red and blue chromosomes
overlap.
•At a molecular level the exchange has to be exact avoiding loss of
bases. Such an error would have profound effect on the genes.
3.3 U.4 The early stages of meiosis involve pairing of homologous
chromosomes and crossing over followed by condensation. [The process
of chiasmata formation need not be explained.]
• At anaphase I the homologous pairs are separated.
• Notice that sections of DNA have been exchanged.
• These will mean that the alleles on these section have also been
exchanged
• Crossing over, Independent Assortment, combined with Fertilization:
A couple can produce over 64 trillion (8.3 million x 8.3 million)
different combinations in a zygotes during fertilization
3.3 U.4 The early stages of meiosis involve pairing of homologous
chromosomes and crossing over followed by condensation. [The process
of chiasmata formation need not be explained.]
10.1 U.4 Chiasmata formation between non-sister chromatids can result
in an exchange of alleles.
• Chiasmata are points where two
homologous non-sister chromatids exchange
genetic material during crossing over in meiosis.
• Chromosomes intertwine and break at the exact
same positions in non-sister chromatids.
• The two chromosomes are now attached at
the same corresponding position on the non-
sister chromatid.
• Once attached the non-attached portions of the
chromatids actually repel each other.
• Chiasmata refer to the actual break of the
phosphodiester bond during crossing over.
• The chiasmata are separated during anaphase 1
which can result in an exchange of
alleles between the non-sister chromatids from
the maternal and paternal chromosomes.
10.1 U.5 Homologous chromosomes separate in meiosis I.
• During meiosis I, unlike mitosis
homologous chromosomes separate
to opposite poles; however, their
sister chromatids remain attached
to each other
• Homologous chromosomes can
exchange material in a process
called crossing over
• Meiosis I is considered reduction
division because the chromosome
number is reduced by half (2n -> n
in humans)
• In Meiosis II there is no further
reduction of the chromosomes. This
stage resembles mitosis. http://i.ytimg.com/vi/RfRtVDkABPI/hqdefault.jpg
10.1 U.7 Independent assortment of genes is due to the random
orientation of pairs of homologous chromosomes in meiosis I.
Second variation created by meiosis:
Independent assortment
10.1 S.1 Drawing diagrams to show chiasmata formed by crossing over.
http://www.thestudentroom.co.uk/showthread.php?t=1561992
• The centromere is a section of
DNA on the chromosome where
the chromatids are closest, it's
generally tightly packed, the DNA
typically doesn't have a defined
sequence and is often satellite
DNA, non-coding. It's the point at
which the mitotic spindle attaches
itself.
• Chiasmata are places where the
chromosomes cross over, and
sometimes exchanges of DNA take
place. This is usually looser
packed, often functional DNA
3.3 U.5 Orientation of pairs of homologous chromosomes prior to
separation is random.
http://cnx.org/resources/9d7cd91abd65e9c7fa857cf6c6d133ec/Figure_08_03_06.jpg
Independent assortment of chromosomes alleles for a trait separate
when gametes are formed. These allele pairs are then randomly
united at fertilization.
• meiosis introduces genetic variation in several ways
• gametes of offspring do not have same combination of genes as
gametes from parents
10.1 U.2 Crossing over is the exchange of DNA material between non-
sister homologous chromatids.
10.1 U.6 Sister chromatids separate in meiosis II.
• During meiosis II sister
chromatids separate (some are
non-identical sister chromatids
due to crossing over
• This type of separation is very
similar to mitosis as the
chromatids are separated from
each other
10.1 U.7 Independent assortment of genes is due to the random
orientation of pairs of homologous chromosomes in meiosis I.
• Independent assortment is an essential
component in explaining how
chromosomes align themselves during
meiosis.
• It also explains how unlinked genes are
passed on from generation to generation.
• When homologues line up along the
equatorial plate in metaphase I, the
orientation of each pair of is random;
meaning the maternal or paternal
homologue can orient towards either pole.
• The orientation of how one set of
homologues line up has no effect on how
any of the other homologues line up.
• Two young women to the right are non
identical twin sisters, who because of
independent assortment have different
hair and skin coloring
http://hellogiggles.com/bi-racial-twins/
10.1 U.3 Crossing over produces new combinations of alleles on the
chromosomes of the haploid cells.
http://www.indiana.edu/~oso/lessons/Genetics/
figs/sheep/2_fXe_sp2.jpg
3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the
formation of four haploid cells.
http://upload.wikimedia.org/wikipedia/commons/5/54/Meiosis_diagram.jpg
Meiosis 1
• interphase
• prophase 1
• metaphase 1
• anaphase 1
• telophase 1
Meiosis 2
• prophase 2
• metaphase 2
• anaphase 2
• telophase 2
2nd division of meiosis
separates sister
chromatids
(1n  1n)
* just like mitosis *
1st division of
meiosis separates
homologous pairs
(2n  1n)
“reduction division”
3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the
formation of four haploid cells.
Meiosis: Generates haploid gametes
• Reduces the number of chromosomes by half.
• Also produces genetic variability, each gamete is different,
ensuring that two offspring from the same parents are
never identical.
• Two divisions: Meiosis I and meiosis II. Chromosomes are
duplicated in interphase prior to Meiosis I.
3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the
formation of four haploid cells.
Meiosis I: Separation of Homologous Chromosomes
1. Prophase I:
• Chromatin condenses into chromosomes.
• Nuclear membrane disappear.
• Centrosomes move to opposite poles of cell and
microtubules attach to chromatids.
• Synapsis: Homologous chromosomes pair up and
form a tetrad of 4 sister chromatids.
3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the
formation of four haploid cells.
Prophase I: Crossing Over Between Homologous Chromosomes
Crossing over combined with Fertilization: A couple can produce over 64
trillion (8.3 million x 8.3 million) different combinations in a zygotes during
fertilization.
3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the
formation of four haploid cells.
Meiosis I:
2. Metaphase I:
3. Anaphase I:
4. Telophase I and Cytokinesis:
3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the
formation of four haploid cells.
3.3 U.6 Separation of pairs of homologous chromosomes in the first
division of meiosis halves the chromosome number.
• Each individual has a pair
chromosomes both
chromosomes of a pair carry
“matching” genes control same
inherited characters
homologous = same
information
• To create sex cells the number
of chromosomes must be
reduce for Humans from 46
chromosomes  to 23.
Reduced by half.
http://41.media.tumblr.com/4c08ea694bcf30c5a9d8423
187fba2de/tumblr_mx5zd49kWN1qjofuoo1_1280.jpg
Random Assortment of Homologous Chromosomes
During Meiosis I Generates Many Possible Gametes
Meiosis II: Separation of Sister Chromatids
During interphase that follows meiosis I, no DNA replication
occurs.
Interphase may be very brief or absent.
Meiosis II is very similar to mitosis.
1. Prophase II:
• Very brief, chromosomes reform.
• No crossing over or synapsis.
• Spindle forms and starts to move chromosomes towards
center of the cell.
3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the
formation of four haploid cells.
Meiosis II: Separation of Sister Chromatids
2. Metaphase II:
• Very brief, individual chromosomes line up in the middle
of the cell.
3. Anaphase II:
• Chromatids separate and move towards opposite ends of
the cell.
3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the
formation of four haploid cells.
Meiosis II: Separation of Sister Chromatids
4. Telophase II:
• Nuclei form at opposite ends of the
cell.
• Cytokinesis occurs.
Product of meiosis:
Four (4) haploid gametes, each
genetically different from the other.
3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the
formation of four haploid cells.
Outline the differences between the behavior of
chromosomes in Mitosis and Meiosis
5 marks
Mitosis Meiosis
One division Two divisions
Diploid cells produced Haploid gametes produced
No crossing-over in prophase Crossing-over in prophase I
No chiasmata formation Chiasmata form
Homologous pairs do not associate and
line up at the equator in metaphase
Homologous pairs do line up at the
equator in metaphase I
Sister chromatids separate in anaphase
Homologous pairs separate in anaphase I
Sister chromatids separate in anaphase II
• Mendel’s Law of Independent assortment of
chromosomes
– Sexual reproduction creates genetic variation
– gametes of offspring do not have same combination of genes
as gametes from parents
• Note: Not know during Mendel’s time, random assortment in
humans produces 223 (8,388,608) different combinations in gametes
from Dadfrom Mom offspring
new gametes
made by offspring
3.3 U.7 Crossing over and random orientation promotes genetic
variation.
Fertilization Sources of Genetic Variation
– Human Sperm + Human Egg come together to produce
8 million X 8 million = 64 trillion combinations!
3.3 U.8 Fusion of gametes from different parents promotes genetic
variation.
http://news.bbcimg.co.uk/media/images/63005000/jpg/_63005
026_p6480029-sperm_fertilizing_egg-spl.jpg
http://worms.zoology.wisc.edu/dd2/echino/fert/files/page19_2.gif
• Sperm + Egg = ?
– any 2 parents will produce a zygote with over 64
trillion (223 x 223) possible diploid combinations
http://worms.zoology.wisc.edu/dd2/e
chino/fert/files/page19_2.gif
3.3 U.8 Fusion of gametes from different parents promotes genetic
variation.
Sexual reproduction allows us to maintain both genetic
similarity & differences.
Baldwin brothers
Jonas
Brothers
http://upload.wikimedia.org/wikipedia/commons/f/ff/Jonas_Brothers_2009.jpg
Martin & Charlie Sheen, Emilio Estevez
3.3 U.8 Fusion of gametes from different parents promotes genetic
variation.
3.3 A.1 Non-disjunction can cause Down syndrome and other
chromosome abnormalities.
Accidents During Meiosis Can Cause Chromosomal Abnormalities
– Nondisjunction: Chromosomes fail to separate.
– Gametes (and zygotes) will have an extra chromosome, others will be
missing a chromosome.
• Trisomy: Individuals with one extra chromosome, three instead of
pair. Have 47 chromosomes in cells.
• Monosomy: Missing a chromosome, one instead of pair. Have 45
chromosomes in cells.
http://upload.wikimedia.org/wikipedia/commons/3/31/Mitotic_nondisjunction.png
3.2 U.9 A karyogram shows the chromosomes of an organism in
homologous pairs of decreasing length.
https://commons.wikimedia.org/wiki/File:NHGRI_human_male_karyotype.png
3.2 U.9 A karyogram shows the chromosomes of an organism in homologous pairs of
decreasing length.
http://learn.genetics.utah.edu/content/chromosomes/karyotype/
Karyotype is a property of the cell described by the number and type of chromosomes
present in the nucleus (of a eukaryote cell).
Karyogram is a diagram or photograph of the chromosomes present in a nucleus (of a
eukaryote cell) arranged in homologous pairs of decreasing length.
a Karyogram is a diagram
that shows, or can be used to
determine, the karyotype.
Normal Male
3.2 A.4 Use of karyograms to deduce sex and diagnose Down syndrome
in humans.
http://commons.wikimedia.org/wiki/File:NHGRI_human_male_karyotype.png
Normal Female
3.2 A.4 Use of karyograms to deduce sex and diagnose Down syndrome
in humans.
X and y Chromosomes
y
X Females have a sex chromosome
arrangement of XX, both the same.
Males have a sex chromosome arrangement
Of Xy (about 200 genes
Male with Down Syndrome: Trisomy 21; mental deficiencies ,
as well as physical abnormalities
3.2 A.4 Use of karyograms to deduce sex and diagnose Down syndrome
in humans.
http://upload.wikimedia.org/wikipedia/commons/6/6f/Trisomie_21_Genom-Schema.gif
3.3 A.3 Description of methods used to obtain cells for karyotype
analysis e.g. chorionic villus sampling and amniocentesis and the
associated risks.
http://www.hopkinsmedicine.org/healthlibrary/GetImage.aspx?ImageId=161385
• Amniocentesis medical
procedure used in prenatal
diagnosis of chromosomal
abnormalities and also used
for sex determination.
• A small amount of amniotic
fluid, which contains fetal
tissues, is sampled from
the amniotic
sac surrounding a
developing fetus, and the
fetal DNA is examined for
genetic abnormalities.
3.3.A3 Description of methods used to obtain cells for karyotype analysis e.g. chorionic villus sampling and
amniocentesis and the associated risks.
Can be carried out in the 16th week of the pregnancy with around
a 1% chance of a miscarriage
http://www.medindia.net/animation/amniocentesis.asp
3.3.A3 Description of methods used to obtain cells for karyotype analysis e.g. chorionic villus sampling and
amniocentesis and the associated risks.
Can be carried out in the 11th week of the
pregnancy with around a 2% chance of a
miscarriage
3.3 A.2 Studies showing age of parents influences chances of non-
disjunction.
http://knowgenetics.org/wp-content/uploads/2012/12/DownSyndromeRisk.png
• Weakening of
cohesive ties holding
together
chromosomes in sex
cells may contribute
to maternal age-
related errors.
• The loss of cohesion
may contribute to
incorrect microtubule
attachment during
meiotic divisions
3.4 Inheritance
Essential Question: The inheritance of genes follows patterns.
http://upload.wikimedia.org/wikipedia/commons/1/11/Peas_in_pods_-_Studio.jpg
Gregor Mendel
• Austrian monk who published results of garden pea
plants inheritance in 1865
• Used artificial pollination in a series of experiments by
using a small brush to place the pollen on the
reproductive parts of the flowers
3.4 U.1 Mendel discovered the principles of inheritance with experiments
in which large numbers of pea plants were crossed.
Key terminology
1. Genotype – symbolic representation of pair of alleles possessed by
an organism, typically represented by two letters
• Ex: Bb, GG, tt
2. Phenotype – characteristics or traits of an organism
• Ex: five fingers on each hand, color blindness, type O blood
3. Dominant allele – an allele that has the same effect on the
phenotype whether it is paired with the same allele or a different
one; always expressed in phenotype
• Ex: Aa give dominant trait A because the a allele is masked; the
a allele is not transcribed and translated during protein
synthesis
3.4 U.1 Mendel discovered the principles of inheritance with
experiments in which large numbers of pea plants were crossed.
4. Recessive allele – an allele that has an effect on the
phenotype only when present in the homozygous state
• Ex: aa gives rise to the recessive trait because no
dominant allele is there to mask it
5. Codominant allele – pairs of alleles that both affect the
phenotype when present in a heterozygote
• Ex: parent with curly hair and parent with straight
hair can have children with different degrees of
curliness as both alleles influence hair condition
when both are present in the genotype
6. Locus – particular position on homologous chromosomes
of a gene
3.4 U.1 Mendel discovered the principles of inheritance with experiments in
which large numbers of pea plants were crossed.
7. Homozygous – having two identical alleles of a gene
Ex: AA is a genotype which is homozygous dominant
whereas aa is the genotype which is homozygous
recessive
8. Heterozygous – having two different alleles of a gene
Ex: Aa is a heterozygous genotype
9. Carrier – an individual who has a recessive allele of a
gene that does not have an effect on their
phenotype
3.4 U.1 Mendel discovered the principles of inheritance with
experiments in which large numbers of pea plants were crossed.
10. Test cross – testing a suspected heterozygote plant
or animal by crossing it with a known homozygous
recessive (aa). Since a recessive allele can be
masked, it is often impossible to tell if an organism
is AA or Aa until they produce offspring which
have the recessive trait.
3.4 U.1 Mendel discovered the principles of inheritance with
experiments in which large numbers of pea plants were crossed.
3.4 U.1 Mendel discovered the principles of inheritance with
experiments in which large numbers of pea plants were crossed.
3.4 U.1 Mendel discovered the principles of inheritance with
experiments in which large numbers of pea plants were crossed.
3.4 U.2 Gametes are haploid so contain only one allele of each gene.
• Gametes are sex cell.
• Sex cells contain one chromosome
of each type, as an example Humans
have 23 types.
• Parents pass information in
the form of genes in gametes (sex cell)
• These cell will fuse together with the
cell of the opposite sex to create a
zygote.
Review
Meiosis = reduction
division
• Cells divide twice
• Result: 4 daughter
cells, each with half
as many
chromosomes as
parent cell
3.4 U.3 The two alleles of each gene separate into different haploid
daughter nuclei during meiosis.
Review
Life cycle: reproductive history of organism, from
conception  production of own offspring
• Fertilization and meiosis alternate in sexual life cycles
• Meiosis: cell division that reduces # of chromosomes (2n 
n), creates gametes
• Fertilization: combine gametes (sperm + egg)
– Fertilized egg = zygote (2n)
• Zygote divides by mitosis to make multicellular diploid
organism
3.4 U.4 Fusion of gametes results in diploid zygotes with two alleles of
each gene that may be the same allele or different alleles.
.
Review
Human Life Cycle
3.4 U.4 Fusion of gametes results in diploid zygotes with two alleles of
each gene that may be the same allele or different alleles.
.
Review
3.4 U.4 Fusion of gametes results in diploid zygotes with two alleles of
each gene that may be the same allele or different alleles.
.
Review
Mendel’s Law’s
Three parts
• Law of SegregationFor each characteristic, an organism inherits two
alleles, one from each parent.
• Law of Dominance Alternative versions of genes dominant or
recessive that account for variations in inherited characteristics.
Organized into three possible combinations. Homozygous dominant,
Homozygous recessive or Heterozygous dominant.
• Law of Independent Assortment The two alleles for each
characteristic segregate during gamete production
3.4 S.1 Construction of Punnett grids for predicting the outcomes of
monohybrid genetic crosses.
Principle of Segregation: Each Parent or Gamete
Contributes One Allele to Offspring
3.4 S.1 Construction of Punnett grids for predicting the outcomes of
monohybrid genetic crosses.
Punnet Square:
Used to determine the outcome of a cross between two individuals.
In the example we have two parents that are heterozygous
dominant for a trait
Offspring:
Genotype: 1/4 PP, 1/2 Pp, and 1/4 pp
Phenotype: 3/4 Purple and 1/4 white
3.4 S.1 Construction of Punnett grids for predicting the outcomes of
monohybrid genetic crosses.
3.4 S.2 Comparison of predicted and actual outcomes of genetic crosses
using real data.
3.4 S.2 Comparison of predicted and actual outcomes of genetic crosses
using real data.
F0
F1
Genotype: R ? r r
Phenotypes: All red
Test Cross Used to determine the genotype of an unknown individual.
The unknown is crossed with a known homozygous recessive.
Homozygous recessiveunknown
Phenotype:
Key to alleles:
R = Red flower
r = white
Some white, some red
Unknown parent = RR Unknown parent = Rr
Possible outcomes:
gametes r r
R Rr Rr
R Rr Rr
gametes r r
R Rr Rr
r rr rr
3.4 S.2 Comparison of predicted and actual outcomes of genetic crosses using real data.
3.4 U.5 Dominant alleles mask the effects of recessive alleles but co-
dominant alleles have joint effects.
.
http://gestblog.scientopia.org/wp-content/uploads/sites/35/2012/07/10-04.gif
Problems with predictions
I. Codominance
II. Multiple
alleles
III. Sex linked
genes
3.4 U.5 Dominant alleles mask the effects of recessive alleles but co-dominant alleles
have joint effects.
3.4 A.1 Inheritance of ABO blood groups. [The expected notation for ABO blood group
alleles: O = i, A=IA, B = IB.]
There are 4: A, B,
AB (codominant example)and
O. A & B refer to 2 genetically
inherited A and B antigens on
the surface of red blood cells.
IA – codes for A
IB – codes for B
i - codes for no antigen = type
O blood
http://www.anatomybox.com/tag/erythrocytes/
3.4.U5 Dominant alleles mask the effects of recessive alleles but co-dominant alleles have joint effects.
Dominant alleles have the same effect on
the phenotype whether it is present in the
homozygous or heterozygous state
IAi
Type “O” allele present and blood
type is not O therefore the type “O”
allele is recessive to type “A”
Type “A” allele present and blood
type is A therefore the type “A”
allele is dominant to type “O”
IAIB Type “A” and “B” alleles are
present and blood type is AB
therefore type “A”and “B” alleles
are codominant
Codominant alleles are pairs of different
alleles that both affect the phenotype when
present in a heterozygote
Recessive alleles only have an effect on the
phenotype when present in the homozygous
state
http://www.anatomybox.com/tag/erythrocytes/
3.4.U5 Dominant alleles mask the effects of recessive alleles but co-dominant alleles have joint effects.
Dominant alleles have the same effect on
the phenotype whether it is present in the
homozygous or heterozygous state
IAi
Type “O” allele present and blood
type is not O therefore the type “O”
allele is recessive to type “A”
Type “A” allele present and blood
type is A therefore the type “A”
allele is dominant to type “O”
IAIB Type “A” and “B” alleles are
present and blood type is AB
therefore type “A”and “B” alleles
are codominant
Codominant alleles are pairs of different
alleles that both affect the phenotype when
present in a heterozygote
Recessive alleles only have an effect on the
phenotype when present in the homozygous
state
Multiple Alleles: There are three alleles
That control blood type, type A, type B and Type O
This produce 4 different Blood Groups
Blood type O: Universal donor. Blood type AB: Universal acceptor
3.4 U.5 Dominant alleles mask the effects of recessive alleles but co-dominant alleles
have joint effects.
3.4 A.1 Inheritance of ABO blood groups. [The expected notation for ABO blood group
alleles: O = i, A=IA, B = IB.]
3.4 A.1 Inheritance of ABO blood groups. [The expected notation for
ABO blood group alleles: O = i, A=IA, B = IB.]
http://image.slidesharecdn.com/incompletecodominancemultiplealleles-120127095123-phpapp02/95/incomplete-codominance-multiplealleles-7-728.jpg
3.4 U.6 Many genetic diseases in humans are due to recessive alleles of
autosomal genes, although some genetic diseases are due to dominant
or co-dominant alleles
3.4 U.6 Many genetic diseases in humans are due to recessive alleles of
autosomal genes, although some genetic diseases are due to dominant
or co-dominant alleles
Inheritance characterized by full expression of both alleles in
the heterozygote.
Seen in:
• Roan Cattle
• Tay Sacs disease
• Blood Types
You have a brown Bull and a
white Cow. You cross them and
get a mix of the two colors.
BB = Brown Bull/Cow
WW = white Bull/Cow
BW = Mixture of the two colors
3.4 U.6 Many genetic diseases in humans are due to recessive alleles of
autosomal genes, although some genetic diseases are due to dominant
or co-dominant alleles
https://jeanurquharthighlandsandislandsmsp.files.wordpress.com/2014/06/cavans-bourbon-orkney-by-robert_scarth-on-flickr.jpg
Sickle Cell Anemia: (example of a
codomainant gene mutation and
its consequences through protein
synthesis)
The Genetics of Sickle Cell Anemia
•HBA HBA Suceptible to malaria without
anemia
• HBA HBs Increase resistance to malaria
with mild anemia
• HBs HBs Sickle cell shaped cell suceptible
to malaria with severe anemia
3.4 U.6 Many genetic diseases in humans are due to recessive alleles of
autosomal genes, although some genetic diseases are due to dominant
or co-dominant alleles
This may cause:
•Change in the base sequence of the gene
•Change in the mRNA made in transcription
•Change in the amino acids sequence / primary structure of the protein
•Change in the secondary/ tertiary or quaternary structure of the
protein
•Change the shape of the protein
•Change the function of the protein.
http://evolution.berkeley.edu/evolibrary/images/interviews/pax_mouse_fly.gif
3.1 U.5 New alleles are formed by mutation.
Review
Examples of the mutations to the gene
Deletion
Duplication
Inversion
Translocation
3.1 U.5 New alleles are formed by mutation. [Deletions, insertions and
frameshift mutations do not need to be included.]
http://www.goldiesroom.org/Multimedia/Bio_Images/19%20Applie
d%20Genetics/05%20Chromosome%20Mutations.jpg
The Disease:
•Sickle cell anemia is an inherited disorder
that affects hemoglobin, a protein that
enables red blood cells to carry oxygen
to all parts of the body.
•The disorder caused by a base
substiution mutation, produces
abnormal hemoglobin, which along
with the abnormalities of the cells
membrane cause damage from be
trapped and freed in the capillaries.
This shortens the blood cells life to as
little as 4 days.
Disease Symptoms:
•chronic anemia,
•acute chest syndrome,
•stroke,
•spleen and renal dysfunction,
•pain crises and susceptibility to bacterial
infections, particularly in children.
•Sickle cell disease is also associated with
significant mortality.
3.1 A.1 The causes of sickle cell anemia, including a base substitution mutation, a change to the
base sequence of mRNA transcribed from it and a change to the sequence of a polypeptide in
hemoglobin
3.1 U.4 Allleles differ from each other by one or only a few bases.
Review
3.1 A.1 The causes of sickle cell anemia
•The HBA gene carries the
instructions for the manufacture
of a protein that is a component
of hemoglobin.
•Hemoglobin is the protein
containing a sequence GAG
which codes for the amino acid
glutamic acid.
•In a base substitution mutation of
Hemoglobin proteins in Sickle
Cell Anemia. GAG has been
changed into GTG which codes
for the amino acid valine. It is
given the symbol HBs for the
gene.
http://upload.wikimedia.org/wikipedia/commons/a/ac/Sickle_cell_01.jpg
Alleles, genotypes and Phenotypes of Sickle Cell:
HbA Normal allele producing normal protein using the amino acid
glutamic acid
Hbs Sickle allele which is the abnormal mutation of the normal allele. Is
produced using the replacement amino acid valine.
3.1 A.1 The causes of sickle cell anemia
http://www.zo.utexas.edu/faculty/sjasper/images/17.23.gif
3.1.A1 The causes of sickle cell anemia, including a base substitution mutation, a change to the base
sequence of mRNA transcribed from it and a change to the sequence of a polypeptide in hemoglobin.
•In these regions sickle cell trait ( HbAHbs ) are resistant to the
malaria parasite. This is because the Hbs allele makes it difficult
for the parasite to live inside the red cells. Sickle cell trait
(carriers) therefore survive malaria infection.
•HbAHbA normal hemoglobin people are susceptible to malaria
infection and do not survive well.
•HbsHbs have sickle disease and do not survive well.
HbAHbs no or reduce sickle cell symptoms and resistant to
malaria
•The sickle allele survives well in malaria regions accounting for its
high frequency in these regions.
•With migration the allele has spread around the world.
Compare the frequency distribution of sickle allele with the
distribution of malaria.
3.4 U.6 Many genetic diseases in humans are due to recessive alleles of
autosomal genes, although some genetic diseases are due to dominant
or co-dominant alleles
3.4 U.6 Many genetic diseases in humans are due to recessive alleles of
autosomal genes, although some genetic diseases are due to dominant
or co-dominant alleles
World Sickle allele Distribution:
Compare the frequency distribution of sickle allele with the
distribution of malaria.
•Notice that the high frequency of sickle coincides with
malaria regions
3.4 U.6 Many genetic diseases in humans are due to recessive alleles of
autosomal genes, although some genetic diseases are due to dominant
or co-dominant alleles
3.4 U.6 Many genetic diseases in humans are due to recessive alleles of
autosomal genes, although some genetic diseases are due to dominant
or co-dominant alleles
3.4 U.6 Many genetic diseases in humans are due to recessive alleles of
autosomal genes, although some genetic diseases are due to dominant
or co-dominant alleles
3.4 U.8 Many genetic diseases have been identified in humans but most
are very rare.
• Medical research has identified
over 4,000 genetic diseases,
however many individuals do
not suffer from one.
• Most genetic diseases are
caused by rare recessive
alleles. Making the chance of
inheritance very small.
• Genetic sequencing of the
human genome current
estimates are that there
maybe as little as 75-200
genes out of over 20,000
genes in the genome that
contain these traits.
3.2 U.10 Sex is determined by sex chromosomes and autosomes are chromosomes that do not
determine sex.
3.2 U.10 Sex is determined by sex chromosomes and autosomes are
chromosomes that do not determine sex.
• The X and Y chromosome determine the
sex of an individual
• If an individual has two homologous X
chromosomes they will be a female and if
they have an two non homologous X and
a Y chromosome they will be a male
• All other chromosomes are called
autosomes and do not affect the sex of
an individual
• The X chromosome has many genes
located on it essential to human
development, while the Y chromosome
has a small number of genes (some of
these are shared with the X
chromosome). The rest of the genes on
the Y chromosome are only necessary for
male development.
• Color Blindness and hemophilia are
two examples of sex linked traits
http://images.zeit.de/wissen/gesundheit/2014
-01/y-chromosom/y-chromosom-540x304.jpg
Sex Linkage Examples:
Hemophilia
• Hemophilia is an example of a
sex linkage condition.
• The hemophilia allele is
recessive to the normal allele.
• The gene is located on the non-
homologous region of the X
chromosome.
• The disease is associated with
an inability to produce a clotting
factor in blood.
• Internal bleeding takes longer to
stop.
3.4 A.2 Red-green color blindness and hemophilia as examples of sex-
linked inheritance.
http://blog.nz-online.de/lieb/wp-content/uploads/sites/8/2010/07/blut.jpg
Female carriers of sex linked alleles
• Female heterozygote's for sex linked alleles e.g. Hemophilia
XHXh or Color Blindness XBXb are carriers of the allele.
• They are unaffected by the condition.
• They do pass on the allele which may result in a
homozygous female or a male with the sex linked recessive
allele.
3.4 A.2 Red-green color blindness and hemophilia as examples of sex-
linked inheritance.
Sex Chromosomes
3.4 U.7 Some genetic diseases are sex-linked. The pattern of inheritance
is different with sex-linked genes due to their location on sex
chromosomes. [Alleles carried on X chromosomes should be shown as
superscript letters on an upper case X, such as Xh.]
• The X chromosome in humans
spans more than 153 million base
pairs (the building material
of DNA). It represents about 1400
out of 20,000 - 25,000 genes.
• The Y chromosome containing
200genes, out of the estimated
20,000 to 25,000 total genes in the
human genome. Genetic disorders
that are due to mutations in genes
on the X chromosome are
described as X linked.
• Because males have so many
fewer genes, they are much more
likely to express sex linked
disorders then females
3.4 U.7 Some genetic diseases are sex-linked. The pattern of inheritance
is different with sex-linked genes due to their location on sex
chromosomes. [Alleles carried on X chromosomes should be shown as
superscript letters on an upper case X, such as Xh.]
Male sex chromosomes
• There are non-homologous
region males in which there
is only one allele per gene
and that is inherited from
the female on the X-
chromosome
• In the homologous region
the male inherited two
copies of an allele per gene.
Female sex chromosomes
• All regions of the X
chromosome are
homologous.
• There are two alleles per
gene as with all other genes
on all other chromosomes
This difference in x and y chromosomes plays a large role in
determining rates of genetic inherited defects
3.4 U.7 Some genetic diseases are sex-linked. The pattern of inheritance
is different with sex-linked genes due to their location on sex
chromosomes. [Alleles carried on X chromosomes should be shown as
superscript letters on an upper case X, such as Xh.]
Sex Linkage Alleles on the non-homologous region of the
X chromosome are more common in females than in
males
• A gene with two alleles where one is dominant and one is recessive.
• Female has three possible genotypes and one is the homozygous
recessive.
• In a population the chance of being homozygous recessive is 33.3 %.
• Males have two possible genotypes.
• There is a 50% chance of the homozygous recessive condition in the
population.
• In sex linked conditions the recessive condition is more common in
males than females.
3.4 U.7 Some genetic diseases are sex-linked. The pattern of inheritance
is different with sex-linked genes due to their location on sex
chromosomes. [Alleles carried on X chromosomes should be shown as
superscript letters on an upper case X, such as Xh.]
3.4 U.7 Some genetic diseases are sex-linked. The pattern of inheritance
is different with sex-linked genes due to their location on sex
chromosomes. [Alleles carried on X chromosomes should be shown as
superscript letters on an upper case X, such as Xh.]
• Red Green Color Blindness is an
example of a sex linked
condition.
• Red Green Color blindness is a
recessive condition.
• The color blind allele is recessive
to the normal allele.
• Female homozygous recessives
XbXb are color blind.
• Males with the genotype XbY are
color blind.
• Notice that in a population the
probability of having a Red Green
color blind genotype in males is
higher.
3.4 A.2 Red-green color blindness and hemophilia as examples of sex-
linked inheritance.
http://en.wikipedia.org/wiki/Color_blindness#/media/File:Ishihara_9.png
Above is a color test
plate.[The numeral "74"
should be clearly visible to
viewers with normal color
vision.
3.4 A.2 Red-green color blindness and hemophilia as examples of sex-
linked inheritance.
http://upload.wikimedia.org/wikipedia/commons/a/a3/XlinkRecessive.jpg
Normal color vision
Red/green color blindness
• The
homozygous
genotype(*) in
females has a
high mortality.
• The genotype
XnY in males
has a high
mortality.
3.4 A.2 Red-green color blindness and hemophilia as examples of sex-
linked inheritance.
Hemophilia
Hemophilia
Cystic fibrosis (CF) Non Sex link recessive genetic trait found
on Chromosome 7. Cystic fibrosis is a genetic disorder that affects
mostly the lungs, Long-term issues include difficulty breathing and
coughing up mucus as a result of frequent lung infections.
Example: Cross
The couple below are heterozygous for CF
3.4 A.3 Inheritance of cystic fibrosis and Huntington’s disease.
http://www.bbc.co.uk/staticarchive/088e5fc50b3c51cfb49ebc4b6eaf203b18b93bbc.gif
Huntington’s Disease Non Sex link dominant genetic trait which
causes a decline in thinking and reasoning skills, including memory,
concentration, judgment, and ability to plan and organize
The couple two couples below are examples
• Couple 1: 1 heterozygous (has trait) with 1 homozygous (without the
trait)
• Couple 2: Both parents are heterozygous with Huntington's
3.4 A.3 Inheritance of cystic fibrosis and Huntington’s disease.
http://vanhornhuntingtonsdisease.weebly.com/uploads/1/3/7/4/13740905/4993818.jpg?1347964948
Couple 1 Couple 2
Pedigree Chart
• Another way to visualize a
monohybrid crosses or
determining a genotype is by
using a pedigree chart
• Knowing the phenotype of
individuals in a family will
sometimes allow genotypes to
be determined.
• In genetic counseling this
enables probabilities to be
determined for the inheritance
of characteristics in children.
3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance
of genetic diseases.
Pedigree Chart
• White circle : Normal
female
• White Square: Normal
male
• Black Circle: affected
female
• Black square: affected male
• (1) and (2)..Normal Parents
• (3) affected female
• (4),(5) and (6) normal
3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance
of genetic diseases.
3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance
of genetic diseases.
3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance
of genetic diseases.
1. Phenylketonuria (Pku)
• Using the allele key provided
state the genotype of parents
1 and 2?
• Give the genotype and
phenotype of individual 5 ?
• Is it possible that the
condition is sex linked ?
• What is the genotype and
phenotype of individuals 7
and 8?
• Which two individuals have
the incorrect pedigree
3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance
of genetic diseases.
2. Muscular Dystrophy
• What type of genetic disease
is muscular dystrophy?
• Give the genotype and
phenotype of 1?
• Give the genotype and
phenotype of 2?
• Give the genotype and
phenotype of 8 ?
• Give the genotype and
phenotype of 5 and 6 ?
3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance
of genetic diseases.
Review: 1.6.U6 Mutagens, oncogenes and metastasis are involved in the development of primary
and secondary tumors.
mutation in a oncogene
If a mutation occurs in an oncogenes it can become cancerous. In normal cells
oncogenes control of the cell cycle and cell division. Cancer is uncontrolled cell
growth.
http://en.wikipedia.org/wiki/Oncogene#mediaviewer/File:Oncogenes_illustration.jpg
uncontrolled cell division
tumor formation
malfunction in the control
of the cell cycle
3.4 U.9 Radiation and mutagenic chemicals increase the mutation rate
and can cause genetic diseases and cancer.
• A mutagen is a physical
(radiation), viruses, or
ultraviolet light. They can
also be chemical agent like
Nitrosamines, found in
tobacco. These mutagens
change the genetic
material, usually DNA, of
an organism and increases
the frequency
of mutations above the
natural background level.
Many mutations
cause cancer, mutagens are
therefore also likely to
be carcinogens.
• Radiation-induced cancers do not appear until at least 10 years after exposure (for
tumors) or 2 years after exposure (for leukemia).
• The risk of cancer after exposure can extend beyond this latent period for the rest
of a person’s life for tumors or about 30 years for leukemia.
• Risk is calculations are based on:
– The type of radiation.
• Each type of radiation is different and affects tissues differently.
– The energy that it leaves in the body.
• More energy means a higher probability of an effect.
– Where in the body the energy remains.
• Radiation exposure to a non-sensitive area of the body (i.e., wrist) really
has no actual effect. Radiation exposure to a sensitive area of the body (i.e.,
blood-forming organs) can have an effect if the amount of energy left is
high enough.
3.4 U.9 Radiation and mutagenic chemicals increase the mutation rate
and can cause genetic diseases and cancer.
• Indirect damage
– Water molecule is ionized, breaks apart,
and forms OH free radical.
– OH free radical contains an unpaired
electron in the outer shell and is highly
reactive: Reacts with DNA.
– 75 percent of radiation-caused DNA
damage is due to OH free radical.
• Direct damage
– DNA molecule is struck by radiation,
ionized, resulting in damage.
3.4 U.9 Radiation and mutagenic chemicals increase the mutation rate
and can cause genetic diseases and cancer.
Chromosome Damage
Formation of a ring and fragments followed
by replication of chromosomes.
3.4 U.9 Radiation and mutagenic chemicals increase the mutation rate
and can cause genetic diseases and cancer.
Chromosome Damage
Interchange between two chromosomes
forms a chromosome with two centromeres
and fragment, followed by replication.
3.4 U.9 Radiation and mutagenic chemicals increase the mutation rate
and can cause genetic diseases and cancer.
Commonly Encountered Radiation Doses
Effective Dose Radiation Source
<= 0.01 rem annual dose living at nuclear power plant panoramic, or full-mouth dental
x rays; skull or chest x ray
<=0.1 rem single spine x ray; abdominal or pelvic x ray; hip x ray; mammogram
<=0.5 rem kidney series of x rays; most barium-related x rays; head CT; any spine x-ray
series; annual natural background radiation dose; most nuclear
medicine brain, liver, kidney, bone, or lung scans
<=1.0 rem barium enema (x rays of the large intestine); chest, abdomen, or pelvic CT
<=5.0 rem cardiac catheterization (heart x rays); coronary angiogram (heart x rays);
other heart x-ray studies; most nuclear medicine heart scans
CT = computerized tomography; a specialized x-ray exam.
3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima
and accident at Chernobyl.
Radiation Doses and Expected Effects (cont.)
General radiation doses to the entire body and expected effects:
• 100-200 rem received in a short time will cause nausea and fatigue.
• 100-200 rem received over a long period will increase a person’s chances of getting cancer.
• 200-300 rem received in a short time will cause nausea and vomiting within 24-48 hours.
Medical attention should be sought.
• 300-500 rem received in a short time will cause nausea, vomiting, and diarrhea within hours.
Loss of hair and appetite occurs within a week. Medical attention must be sought for
survival; half of the people exposed to radiation at this high level will die if they receive no
medical attention.
• 500-1,200 rem in a short time will likely lead to death within a few days.
• Greater than 10,000 rem in a short time will lead to death within a few hours.
3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima
and accident at Chernobyl.
3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima
and accident at Chernobyl.
http://inapcache.boston.com/universal/site_graphics/blogs/bigpicture/hiroshima_08_05/h17_04.jpg
3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima
and accident at Chernobyl.
http://www.nucleardarkness.org/include/nucleardarkness/images/cityonfire/hiroshima_after_02_full.jpg
Up to one mile from the blast site:
36,000 rem
3.4.A4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl.
nuclear bombing of Hiroshima
https://upload.wikimedia.org/wikipedia/commons/e/e9/The_patient%27s_skin_is_burned_in_a_pattern_corresponding_to_the_dark_porti
ons_of_a_kimono_-_NARA_-_519686.jpg
https://upload.wikimedia.org/wikipedia/commons/f/f6/Hiroshima_girl.jpg
• Elevated rate of Leukemia (with the
greatest impact in children and young
adults)
• Elevated rates of other cancers
• No evidence of stillbirth or mutations in
the children of those exposed to radiation
http://i.telegraph.co.uk/multimedia/archive/02446/hiroshima-bomb_2446747b.jpg
3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima
and accident at Chernobyl.
http://www.zap-actu.fr/wp-content/uploads/2013/11/pripyat-une-des-villes-fantomes-pres-de-tchernobyl-01.jpg
Average thyroid radiation dose observed in Chernobyl
evacuees : 49,000 rem
3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima
and accident at Chernobyl.
3.4.A4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl.
accident at Chernobyl nuclear power station
https://upload.wikimedia.org/wikipedia/commons/1/16/VOA_Markosian_-_Chernobyl02.jpg
• A large area of pine forest downwind of
the reactor turned brown and died.
• Horses and cattle near the plant died
from radiation damage to their thyroid
glands.
• Bioaccumulation of radioactive caesium
in fish (Scandinavia and Germany) and
lamb (Wales) - contaminated meat was
banned from sale for years afterward.
http://i.guim.co.uk/img/static/sys-images/Guardian/Pix/pictures/2014/6/27/1403890449199/933cb303-bf75-4e9a-8b0b-806bbfa6a37b-
2060x1373.jpeg?w=620&q=85&auto=format&sharp=10&s=abe2802021d01fe090859454e9020a44
• Drinking water (and milk) contaminated with radioactive iodine - at least 6,000
thyroid cancer attributed to radioactive iodine.
• No clear evidence to support an increase in the rate of leukemia other cancers – in
part due to the widely dispersed variable radiation and measures taken in
European populations.
10.2 Inheritance
Essential idea: Genes may be linked or unlinked and are
inherited accordingly.
http://upload.wikimedia.org/wikipedia/c
ommons/4/46/Dividing_Cell_Fluorescenc
e.jpg
http://ies.rayuela.mostoles.educa
.madrid.org/deptos/dbiogeo/recu
rsos/Apuntes/ApuntesBioBach2/i
magenes/genetica/droso.png
10.2 U.1 Gene loci are said to be linked if on the same chromosome.
• Genes have specific locations (loci) on chromosomes
• Chromosomes segregate and assort independently
http://web.csulb.edu/~kmacd/361-6-Ch2.htm
10.2 U.2 Unlinked genes segregate independently as a result of meiosis.
• Mendel’s law of independent
assortment states allele pairs
separate independently from
other allele
pairs during gamete formation
(meiosis).
• Therefore, traits on different
chromosomes are transmitted
to the offspring independently
of traits on other
chromosomes.
• An exception to this rule is
linked genes
https://biologywarakwarak.files.wordpress.com/2011/12/crossingover-rec.jpg
10.2 U.3 Variation can be discrete or continuous.
• Discrete Variation is controlled by alleles of
a single gene or a small number of genes.
The environment has little effect. In this case
you either have the characteristic or you
don't. Cystic fibrosis is a good example for
this; either you have cystic fibrosis or you
don’t.
• Chi-squared calculations work well when
using examples with discrete variation
• Continuous Variation is a complete range of
phenotypes that can exist from one extreme
to the other. Height is an example of
continuous variation. Continuous variation is
the combined effect of many genes (known
as polygenic inheritance) and is often
significantly affected by environmental
influences. Skin color is another example of
continuous variation. http://articles.latimes.com/2013/nov/11/entertainment/la-et-
cm-princess-bride-disney-musical-20131111
10.2 A.3 Polygenic traits such as human height may also be influenced
by environmental factors.
• Polygenic inheritance occurs when two or more genes control the expression of
a phenotype.
• As the amount of genes that control one trait increase, the number of phenotypes increases.
• Each additional gene has an additive affect, increasing the phenotypes. This is called
continuous variation.
• Example: human height, which varies from person to person within the same race, and varies
between identical twin . Below the effect of environmental facts on the height of these
identical twin due to diet.
http://http://www.dailymail.co.uk/femail/article-2591087/Double-Poignant-portraits-50-
year-old-identical-twins-reveal-differences-lifestyle-affected-way-age.html
North side, upper branches South side upper branches
North side, lower branches South side, lower branches
© Horticultural Research International
Apple Size
• What environmental conditions in the different parts of the tree might have
caused the differences in size of the apples ?
• The upper branches will receive more sunlight than the lower branches, but on
the North side the sunlight would have been less intense
• The lower branches will receive less sunlight because of the shade cast by the
upper branches. This effect will be more pronounced on the North side
Skin color
• controlled by 3 or 4 genes and environmental factors leading to a wide range of
phenotypes
• Exposure to the sun can alter skin color and obscure genotypic differences.
Mendel's Second Law of Independent Assortment
• This law states that allele pairs separate independently during the
formation of gametes. Therefore, traits are transmitted to offspring
independently of one another
• Polygenic traits are those traits that are controlled by more than one
gene. Such traits may even be controlled by genes located on entirely
different chromosomes. Human height, eye and hair color are
examples of polygenic traits. Skin color is another polygenic trait for
humans and a variety of other animals. Multiple alleles are several
variant forms of one gene (ex. Genes for blood type A, O, B)
10.2 A.2 Completion and analysis of Punnett squares for dihybrid traits
Dihybrid Crosses Monohybrid crosses is one trait. Consider two traits, each
carried on separate chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea color and pea surface.
What is the predicted phenotype ratio for a cross between
two pea plants which are heterozygous at both loci?
F1
gametesPunnet Grid:
F0
Genotype:
Phenotype:
Heterozygous at both loci Heterozygous at both loci
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
What is the predicted phenotype ratio for a cross between
two pea plants which are heterozygous at both loci?
F1
gametesPunnet Grid:
F0
Genotype: SsYy SsYy
Phenotype:
Heterozygous at both loci Heterozygous at both loci
Smooth, yellow Smooth, yellow
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
What is the predicted phenotype ratio for a cross between
two pea plants which are heterozygous at both loci?
F1
gametes SY Sy sY sy
SY SSYY SSYy SsYY SsYy
Sy SSYy SSyy SsYy Ssyy
sY SsYY SsYy ssYY ssYy
sy SsYy Ssyy ssYy ssyy
Punnet Grid:
F0
Genotype: SsYy SsYy
Phenotype:
Heterozygous at both loci Heterozygous at both loci
Smooth, yellow Smooth, yellow
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
What is the predicted phenotype ratio for a cross between
two pea plants which are heterozygous at both loci?
F1
gametes SY Sy sY sy
SY SSYY SSYy SsYY SsYy
Sy SSYy SSyy SsYy Ssyy
sY SsYY SsYy ssYY ssYy
sy SsYy Ssyy ssYy ssyy
Punnet Grid:
F0
Genotype: SsYy SsYy
Phenotype:
Heterozygous at both loci Heterozygous at both loci
Smooth, yellow Smooth, yellow
Phenotypes: 9 Smooth, yellow : 3 Smooth, green : 3 Rough, yellow : 1 Rough, green
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
F1
Punnet Grid:
F0
Genotype:
Phenotype:
Heterozygous at both loci Heterozygous for S, homozygous dominant for Y
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
F1
Punnet Grid:
F0
Genotype: SsYy SsYY
Phenotype:
Heterozygous at both loci Heterozygous for S, homozygous dominant for Y
Smooth, yellow Smooth, yellow
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
F1
gametes SY sY
SY
Sy
sY
sy
Punnet Grid:
F0
Genotype: SsYy SsYY
Phenotype:
Heterozygous at both loci Heterozygous for S, homozygous dominant for Y
Smooth, yellow Smooth, yellow
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
F1
gametes SY sY
SY SSYY SsYY
Sy SSYy SsYy
sY SsYY ssYY
sy SsYy ssYy
Punnet Grid:
F0
Genotype: SsYy SsYY
Phenotype:
Heterozygous at both loci Heterozygous for S, homozygous dominant for Y
Smooth, yellow Smooth, yellow
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
F1
gametes SY sY
SY SSYY SsYY
Sy SSYy SsYy
sY SsYY ssYY
sy SsYy ssYy
Punnet Grid:
F0
Genotype: SsYy SsYY
Phenotype:
Heterozygous at both loci Heterozygous for S, homozygous dominant for Y
Smooth, yellow Smooth, yellow
Phenotypes: 3 Smooth, yellow : 1 Rough, yellow Present the ratio in the simplest
mathematical form.
6 Smooth, yellow : 2 Rough, yellow
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Common expected ratios of dihybrid crosses.
SY Sy sY sy
SY SSYY SSYy SsYY SsYy
Sy SSYy SSyy SsYy Ssyy
sY SsYY SsYy ssYY ssYy
sy SsYy Ssyy ssYy ssyy
Heterozygous at
both loci
Heterozygous at
both loci
SsYySsYy
9 : 3 : 3 : 1
SY sY
SY SSYY SsYY
Sy SSYy SsYy
sY SsYY ssYY
sy SsYy ssYy
Heterozygous at
both loci
Heterozygous at one locus,
homozygous dominant at the other
SsYySsYy
3 : 2
Sy sy
SY SSYy SsYy
Sy SSyy Ssyy
sY SsYy ssYy
sy Ssyy ssyy
Heterozygous at
both loci
SsyySsYy
4 : 3 : 1
Heterozygous/
Homozygous recessive
ssYYSSyy = All SsYy
ssYySsyy = 1 : 1 : 1 : 1
ssyySSYY = all SyYy
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
Punnet Grid:
F0
Genotype:
Phenotype: Rough, yellow
F1
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
Punnet Grid:
F0
Genotype: ssYy
Phenotype: Rough, yellow
F1
gametes sY sy
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
F1
gametes sY sy sY sYPunnet Grid:
F0
Genotype: ssYy or ssYY
Phenotype: Rough, yellow
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
F1
gametes sY sy sY sY
All sy
Punnet Grid:
F0
Genotype: ssYy or ssYYssyy
Phenotype: Rough, yellow
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
F1
gametes sY sy sY sY
All sy ssYy ssyy ssYy ssYy
Punnet Grid:
F0
Genotype: ssYy or ssYYssyy
Phenotype: Rough, yellow
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
F1
gametes sY sy sY sY
All sy ssYy ssyy ssYy ssYy
Punnet Grid:
F0
Genotype: ssYy or ssYYssyy
Phenotype: Rough, yellow
Phenotypes:
Some green peas will be present in
the offspring if the unknown parent
genotype is ssYy.
No green peas will be present in the
offspring if the unknown parent
genotype is ssYY.
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
F1
Punnet Grid:
F0
Genotype:
Phenotype: Smooth, green
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
F1
gametes
All sy
Punnet Grid:
F0
Genotype: ssyy
Phenotype: Smooth, green
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
F1
gametes Sy Sy
All sy Ssyy Ssyy
Punnet Grid:
F0
Genotype: SSyyssyy
Phenotype: Smooth, green
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
F1
gametes Sy Sy Sy sy
All sy Ssyy Ssyy Ssyy ssyy
Punnet Grid:
F0
Genotype: SSyy or Ssyyssyy
Phenotype: Smooth, green
Phenotypes:
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
Dihybrid Crosses Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
F1
gametes Sy Sy Sy sy
All sy Ssyy Ssyy Ssyy ssyy
Punnet Grid:
F0
Genotype: SSyy or Ssyyssyy
Phenotype: Smooth, green
Phenotypes:
No rough peas will be present in the
offspring if the unknown parent
genotype is SSyy.
The presence of rough green peas in the
offspring means that the unknown genotype
must be Ssyy.
The expected ratio in this cross is 3 smooth green : 1 rough green. This is not the same as the outcome.
Remember that each reproduction event is chance and the sample size is very small. With a much larger
sample size, the outcome would be closer to the expected ratio, simply due to probability.
10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted
genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of
offspring of dihybrid crosses involving unlinked autosomal genes
Phenotype Key:
Phenotypic ratio:
– 9 Smooth Yellow:
– 3 Smooth green:
– 3 Rough Yellow:
– 1 Rough Green
10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of
offspring of dihybrid crosses involving unlinked autosomal genes
10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of
offspring of dihybrid crosses involving unlinked autosomal genes
10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of
offspring of dihybrid crosses involving unlinked autosomal genes
10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of
offspring of dihybrid crosses involving unlinked autosomal genes
10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of
offspring of dihybrid crosses involving unlinked autosomal genes
10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of
offspring of dihybrid crosses involving unlinked autosomal genes
10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of
offspring of dihybrid crosses involving unlinked autosomal genes
Example: Human skin color
• This trait is controlled by
multiple alleles, as many as 6
genes each with its own alleles.
• As the number of genes
increases the amount of
phenotypic variation increases.
• The alleles control the
production of melanin which is
a pigment that colors skin.
• In this example the calculation
is performed with 3 genes each
with 3 alleles. The cross is
between two individuals
heterozygous at both alleles
Allele Key
A= add melanin
a= no melanin added
B= adds melanin
b= no melanin added
10.2 U.4 The phenotypes of polygenic characteristics tend to show
continuous variation.
http://www2.estrellamountain.edu/faculty/farabee/BIOBK/BioBookgeninteract.html
10.2 U.4 The phenotypes of polygenic characteristics tend to show
continuous variation.
10.2 U.4 The phenotypes of polygenic characteristics tend to show
continuous variation.
10.2 A.1 Morgan’s discovery of non-Mendelian ratios in Drosophila.
Thomas Hunt Morgan
Drosophila melanogaster –
fruit fly
Fast breeding,
chromosomes (XX/XY)
Discovered sex-linked gene,
located on X or Y chromosome.
These genes characteristic are
not expressed in the same
ratios as traits located on other
chromosomes.
10.2 A.1 Morgan’s discovery of non-Mendelian ratios in Drosophila.
• Link genes are located on the same chromosome and do not sort independently
• Genes inherited together. Differing from sex linked genes which are genes found on
the sex chromosomes.
• These genes do not show the expected Mendelian ratios as predicted by the Laws
Independent Assortment.
Linked Genes
10.2 S.2 Identification of recombinants in crosses involving two linked
genes.
10.2 S.2 Identification of recombinants in crosses involving two linked
genes.
10.2 S.2 Identification of recombinants in crosses involving two linked
genes.
10.2 S.2 Identification of recombinants in crosses involving two linked
genes.
10.2 S.2 Identification of recombinants in crosses involving two linked
genes.
10.2 S.2 Identification of recombinants in crosses involving two linked
genes.
10.2 S.2 Identification of recombinants in crosses involving two linked
genes.
10.2 S.3 Use of a chi-squared test on data from dihybrid crosses.
• For an example let’s use Mendel’s results from his pea plant crosses
• When he did a dihybrid cross between two heterozygotes RrYy x RrYy, the expected
phenotypic ratio due to independent assortment would be 9:3:3:1. Look at the
chart below to see his actual results.
10.2 S.3 Use of a chi-squared test on data from dihybrid crosses.
Chi-Square Analysis Practice
• Two true-breeding Drosophila are crossed: a normal-winged, red-
eyed female and a miniature-winged, vermillion-eyed male. The F1
offspring all have normal wings and red eyes. When the F1 offspring
are crossed with miniature-winged, vermillion-eyed flies, the
following offspring resulted:
– 233 normal wing, red eye
– 247 miniature wing, vermillion eye
– 7 normal wing, vermillion eye
– 13 miniature wing, red eye
• What type of conclusions can you draw from this experiment? Explain
your answer.
10.2 S.3 Use of a chi-squared test on data from dihybrid crosses.
10.2 U.5 Chi-squared tests are used to determine whether the
difference between an observed and expected frequency distribution is
statistically significant.
• A chi-square test is a statistical test that
can be used to determine whether
observed frequencies are significantly
different from expected frequencies
• These statistical tests enable us to
compare observed and expected
frequencies empirically and to decide if
the results we see are statistically
significant. Statistical significance in this
case implies that the differences are not
due to chance alone, but instead may be
caused by other factors at work.
• The formula to the right is for a
chi-squared test
O = observed frequencies
E = expected frequencies
X2 =
10.2 U.5 Chi-squared tests are used to determine whether the
difference between an observed and expected frequency distribution is
statistically significant.
10.2 U.5 Chi-squared tests are used to determine whether the
difference between an observed and expected frequency distribution is
statistically significant.
10.2 U.5 Chi-squared tests are used to determine whether the difference between an observed
and expected frequency distribution is statistically significant. AND 10.2.S3 Use of a chi-squared
test on data from dihybrid crosses.
Testing dihybrid crosses – using the chi-squared test
Now try extending your understanding of dihybrid crosses using data.
Follow the links below and complete the examples and problems:
CORN GENETICS CHI SQUARE
ANALYSIS by Biology Corner
Answer key:
http://www.biologycorner.com/worksheets/corn_chi_key.
html
http://www.biologycorner.com/worksheets
/corn_chi.html
Chi-square in biology: Testing
for a dihybrid ratio
http://mathbench.umd.edu/modules/s
tatistical-
tests_chisquare_intro/page15.htm
3.5 Genetic modification and biotechnology
Essential idea: Biologists have developed techniques for
artificial manipulation of DNA, cells and organisms.
http://www.nacentralohio.com/wp-content/uploads/2013/01/WW_0113_GMO_AppleOrange.jpg
3.5 U.2 PCR can be used to amplify small amounts of DNA.
• PCR (polymerase chain reaction) is a
laboratory technique that takes a single
or few copies of DNA and amplifies
them to generate millions or more
copies of a particular DNA sequence.
• When you collect DNA from different
sources such as sperm samples or small
drops of blood, there are usually very
little usable cells to collect DNA.
• Therefore, PCR is used to create enough
DNA to be analyzed for investigations
such as forensics or custody cases.
• Once large quantities of the DNA have
been created, other methods such as gel
electrophoresis are used to analyze the
DNA.
Click4Biology
3.5 U.2 PCR can be used to amplify small amounts of DNA.
PCR occurs in a thermal cycler and
involves a repeat procedure of 3
steps:
1. Denaturation: DNA sample is
heated to separate it into two
strands
2. Annealing: DNA primers attach
to opposite ends of the target
sequence
3. Elongation: A heat-tolerant DNA
polymerase (Taq) copies the strands
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
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Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
Topic Five: Genetics
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Topic Five: Genetics
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Topic Five: Genetics

  • 2. 3.1 Genes • Essential idea: Every living organism inherits a blueprint for life from its parents. http://pixabay.com/static/uploads/photo/2014/05/11/14/25/elephants-341981_640.jpg
  • 3. Chromosome Histones DNA double helix 3.1 U.1 A gene is a heritable factor that consists of a length of DNA and influences a specific characteristic Chromosome is a length of DNA with associated proteins (histones).
  • 4. 3.1 U.1 A gene is a heritable factor that consists of a length of DNA and influences a specific characteristic Eukaryotic Chromosomes • The Photograph shows a human chromosome, which are made of DNA and protein. The picture show two chromatids newly replicated DNA (uncoiled chromosomes) joined at the centromere. Each Eukaryotic chromosomes is composed of genes which control specific characteristics (alleles are different forms of a specific gene)
  • 5. 3.1 U.2 A gene occupies a specific position on a chromosome. http://www.daviddarlin g.info/images/gene.jpg •Gene is a heritable factor that controls a specific characteristic. There are over 20,000 genes that make up a human. One gene codes for one polypeptide.
  • 6. http://physrev.physiology.org/content/physrev/91/1/151/F11.large.jpg A gene can exist in many different forms, called alleles. An example hair color: black hair, brown hair, auburn hair, red hair, blond hair, etc. You inherit one allele for each gene from your mother and one from your father. Each allele maybe a dominant characteristic (represented with a upper case letter) or weak or recessive (represented with a lower case letter). You need only one dominant allele to express a dominant trait. You need two recessive alleles to show a recessive form of a trait. See the heredity diagram for tongue rolling to see how dominant and recessive alleles work. No tongue rolling is the recessive form of the gene.
  • 8. Allele is one specific form of a gene, differing from other alleles by one or a few bases only and occupying the same gene locus as other alleles of the gene. 3.1 U.3 The various specific forms of a gene are alleles.
  • 9. •Gene loci is the position of the gene on a chromosome. Alleles have the same gene loci.
  • 10. 3.2 U.1 Prokaryotes have one chromosome consisting of a circular DNA molecule. Prokaryotes have two types of DNA: a single chromosome and plasmids (not associated with any of the cells life processes • Prokaryotic DNA is circular and is not associated with any histone proteins (“naked DNA) • There is one copy of each gene except when the cell and its DNA are replicating • The single chromosome is coiled up and concentrated in the nucleoid region. Bacterial DNA Plasmids
  • 11. Nucleosome Structure: consist of 8 histones. They help create super coiling of chromatin, which creates a chromosome during cell replication 3.2 U.3 Eukaryote chromosomes are linear DNA molecules associated with histone proteins. http://micro.magnet.fsu.edu/cells/nucleus/images/ chromatinstructurefigure1.jpg
  • 12. • Eukaryotic chromosomes are linear and are made up of DNA and histone proteins, important in gene expression and replication. • Histones are globular shaped protein in which the DNA is wrapped around. • DNA wrapped around 8 histone proteins is called a nucleosome. • The DNA wraps twice around the histone protein core. • Another histone protein is attached to the outside of the DNA strand. This helps maintain the colloidal structure of the nucleosome. • DNA, because of its negative charge is attracted to the positive charge on the amino acids of the histone proteins. 3.2 U.3 Eukaryote chromosomes are linear DNA molecules associated with histone proteins. http://upload.wikimedia.org/wikipedia/commons/ 4/45/Nucleosome_organization.png
  • 13. 3.2 U.1 Prokaryotes have one chromosome consisting of a circular DNA molecule. https://classconnection.s3.amazonaws.com/730/flashcards/127 6730/jpg/cell_types1330806303645.jpg
  • 14. 3.2 U.2 Some prokaryotes also have plasmids but eukaryotes do not. http://upload.wikimedia.org/wikipedia/en/6/6c/PBR322_pl asmid_showing_restriction_sites_and_resistance_genes.jpg Features of Plasmids: • Naked DNA - not associated with histone proteins • Small circular rings of DNA • Can be passed between prokaryotes • Plasmids are small separate (usually circular) DNA molecules located in some prokaryotic cells • Plasmids are also naked (not associated with proteins) and are not needed for daily life processes in the cell. • The genes in plasmids are often associated with antibiotic resistant and can be transferred from one bacterial cell to another. • Plasmids are readily used by scientists to artificially transfer genes from one species to another (ie. Gene for human insulin)
  • 15. 3.2 U.4 In a eukaryote species there are different chromosomes that carry different genes. https://s-media-cache- ak0.pinimg.com/originals/4d/d0/7 a/4dd07a61c384abefef3c521bcab 6bad8.jpg http://dxline.info/img/new_ail/chro mosome-4.jpg • Chromosomes are linear, varying in length and in position of the centromere that holds the sister chromatids together. • In humans there are 23 types of chromosomes. Each chromosome carries a specific sequence of genes along the linear DNA molecule. The position where the gene is located is called the locus. • Alleles are different forms of a gene on the chromosome. Chromosome 1 Chromosome 4 2,000 Genes 1,000 Genes
  • 16. https://public.ornl.gov/site/gallery/originals/ Eukaryotes possess multiple chromosomes. All individuals of a species possess the same chromosomes, with the same gene loci. For example all humans have twenty three pairs. Chromosomes can vary by: • Length – the number of base pairs in the DNA molecule • Position of the centromere • Genes occur at a specific locus (location), i.e. it is always found at the same position on the same chromosome (the locus and genes possessed vary between species) 3.2 U.4 In a eukaryote species there are different chromosomes that carry different genes.
  • 17. 3.1 U.6 The genome is the whole of the genetic information of an organism. Allows for: • Physical mapping of chromosomes • Used to screen for genetic diseases • Lead to a better understanding of genetic diseases. • May lead to the development of better drugs to fight diseases. • Maybe be used for comparison of genomes with other species.
  • 18. 3.2 A.2 Comparison of genome size in T2 phage, Escherichia coli, Drosophila melanogaster, Homo sapiens and Paris japonica. [Genome size is the total length of DNA in an organism. The examples of genome and chromosome number have been selected to allow points of interest to be raised Name Genome Length (million base pairs) Number of Genes T2 phage (Virus) 0.18 300 Escherichia coli (Bacteria) 5 4,377 Drosophila melanogaster (Fruit Fly) 140 17,000 Paris japonica (Woodland Plant) 150,000 Unknown Homo sapiens (Human) 3,000 19-23,000
  • 19. 3.1 A.2 Comparison of the number of genes in humans with other species. [The number of genes in a species should not be referred to as genome size as this term is used for the total amount of DNA. At least one plant and one bacterium should be included in the comparison and at least one species with more genes and one with fewer genes than a human.]
  • 20. 3.1 U.6 The genome is the whole of the genetic information of an organism. Genome Size and Number of Genes
  • 21. 3.2 A.2 Comparison of genome size in T2 phage, Escherichia coli, Drosophila melanogaster, Homo sapiens and Paris japonica. [Genome size is the total length of DNA in an organism. The examples of genome and chromosome number have been selected to allow points of interest to be raised Paris japonica Largest Known Genome
  • 22. 3.2 A.3 Comparison of diploid chromosome numbers of Homo sapiens, Pan troglodytes, Canis familiaris, Oryza sativa, Parascaris equorum. Homo sapiens (46)
  • 23. 3.2 A.3 Comparison of diploid chromosome numbers of Homo sapiens, Pan troglodytes, Canis familiaris, Oryza sativa, Parascaris equorum. Pan troglodytes (48) http://static1.squarespace.com/static/51b89f0ce4b0000660671282/t/540785e0e4b00aec95acf4cb/1409779168825/floandfigan.jpg
  • 24. 3.2 A.3 Comparison of diploid chromosome numbers of Homo sapiens, Pan troglodytes, Canis familiaris, Oryza sativa, Parascaris equorum. Canis familiaris 78 https://c1.staticflickr.com/5/4007/5076413065_6a42962771_b.jpg
  • 25. 3.2 A.3 Comparison of diploid chromosome numbers of Homo sapiens, Pan troglodytes, Canis familiaris, Oryza sativa, Parascaris equorum. Species Name Number of Chromosomes Homo sapiens 46 Pan troglodytes 48 Canis familiaris 78 Oryza sativa 24 Parascaris equorum 2 Rice: Oryza sativa http://upload.wikimedia.org/wikipedia /commons/6/65/Oryza_sativa_Kyoto_J PN_001.JPG Round Worm: Parascaris equorum http://www.vetnext.com/fotos/quizc2a4.jpg
  • 26. Chromosome 16 3.1 U.6 The genome is the whole of the genetic information of an organism. http://physrev.physiology.org/content/physrev/91/1/151/F11.large.jpg
  • 27. 3.1 U.6 The genome is the whole of the genetic information of an organism. Advantages from the project Advantage 1 • Screening can also be done for adults, to see if they may be carriers of potential genetic conditions. Certain Jewish and Canadian populations regularly obtain voluntary screening for Tay- Sachs disease, a known child-killer. This information has been used to help make decisions about future marriage partners. Advantage 2 Perhaps the greatest benefit will come from what is called gene-based therapy. Understanding the molecular workings of genes and the proteins they encode will lead to more precise drug treatments. The more precise the drug treatment, the fewer and milder will be the side effects.
  • 28. 3.1 U.7 The entire base sequence of human genes was sequenced in the Human Genome Project. • The Human Genome Project (HGP) was an international scientific research project with the goal of determining the sequence of chemical base pairs which make up human DNA, and of identifying and mapping all of the genes of the human genome from both a physical and functional standpoint. It remains the world's largest collaborative biological project • Key findings of the draft (2001) and complete (2004) genome sequences include:  There are approximately 20,500 genes in human beings, the same range as in mice.  Significantly more segmental duplications (nearly identical, repeated sections of DNA) than had been previously suspected.  At the time of publishing fewer than 7% of protein families appeared to be vertebrate specific.
  • 29. 3.1 U.7 The entire base sequence of human genes was sequenced in the Human Genome Project. http://www.bioscience.heacademy.ac.uk/imagebank/search/Fullimage.aspx?IDvalues=6513 Key advances in technology: • Biotechnology techniques such as PCR are used to prepare samples: the DNA needs to be copied to prepare a sufficiently large pure samples to sequence • Computers automate the sequencing process • Fluorescent labeling techniques enable all four nucleotides to be analyzed together • Lasers are used to fluoresce the dye markers • Digital camera technology reads the dye markers • Computers are used to assemble the base sequence
  • 30. 3.1.S1 Use of a database to determine differences in the base sequence of a gene in two species. One use of aligning base sequences is to determine the differences between species: this can be used to help determine evolutionary relationships. http://www.ncbi.nlm .nih.gov/genbank GenBank http://bitesizebio.s3.amazonaws.com/wp-content/uploads/2012/10/header-image-copy18.jpg Your task is to analyze the differences between three or more species (the skill asks for two species, but the online Clustal tool works better with a minimum of three). For each chosen species retrieve the base sequence: • Go to GenBank website http://www.ncbi.nlm.nih.gov/genbank • Select ‘Gene’ from the search bar • Enter the name of a gene (e.g. AMY1A for salivary amylase 1A or COX1 for cytochrome oxidase 1) AND the organism (use the binomial) and press ‘Search’ n.b. if you are comparing species the gene chosen needs to be the same for each species • Select the ‘Name/Gene ID’ to get a detailed view • Scroll down to the ‘Genomic regions, transcripts, and products’ section and click on ‘FASTA’ • Copy the entire sequence from ‘>’ onwards • Save the sequence – you will need to align with the other species next
  • 31. 3.1.S1 Use of a database to determine differences in the base sequence of a gene in two species. http://www.ebi.ac.uk/Tools/msa/clustalo/ Analysis: • ‘Alignments’ allows you to visually check the results – this is easier if the chosen gene has a short base sequence • Under ‘Results Summary’ use the ‘Percent Identity Matrix’ to quantify the overall similarity (0 = no similarity, 100 = identical) • Under ‘Phylogenic Tree’ chose the ‘Real’ option for the Phylogram to get a visual representation of how similar the species are (based on the chosen gene). http://bitesizebio.s3.amazonaws.com/wp-content/uploads/2012/10/header-image-copy18.jpg To align the sequences: • Go to the Clustal Omega website http://www.ebi.ac.uk/Tools/msa/clustalo/ • In STEP 1 Select ‘DNA’ under ‘a set of’ • Paste the chosen sequences into the box (each sequence must start on a new line) • Press ‘Submit’ (and wait – depending on the size of the sequences you may have to wait for a couple of minutes)
  • 32. 3.3 Meiosis Essential Question: Alleles segregate during meiosis allowing new combinations to be formed by the fusion of gametes. http://i.huffpost.com/gen/1123387/original.jpg Grandmother Granddaughter
  • 33. 10.1 Meiosis Essential idea: Meiosis leads to independent assortment of chromosomes and unique composition of alleles in daughter cells. http://upload.wikimedia.org/wikipedia/c ommons/4/46/Dividing_Cell_Fluorescenc e.jpg
  • 34. Review: 1.6.U.1 Mitosis is division of the nucleus into two genetically identical daughter nuclei. http://highered.mheducation.com/sites/0072495 855/student_view0/chapter2/animation__mitosis _and_cytokinesis.html Use the animated tutorials to learn about mitosis http://www.johnkyrk.com/mitosis.html http://www.sumanasinc.com/webcontent/animations/content /mitosis.html http://outreach.mcb.harvard.edu/animations/cellcycle. swf
  • 35. 3.2 U.7 Haploid nuclei have one chromosome of each pair. [The two DNA molecules formed by DNA replication prior to cell division are considered to be sister chromatids until the splitting of the centromere at the start of anaphase. After this, they are individual chromosomes.] • Haploid nuclei have one copy of each chromosome or one full set of the chromosomes (Sex cells) in that particular species eg. Human 23 chromosomes (or n) • These are called gametes, which are sperm and egg • Human sperm and eggs each contain 23 chromosomes (when the fuse together the become diploid with 46 chromosomes (or 2n, a somatic body cell developing into any type of cell (ex. Blood cell, liver cell heart cell)
  • 36. 3.2 U.7 Haploid nuclei have one chromosome of each pair. [The two DNA molecules formed by DNA replication prior to cell division are considered to be sister chromatids until the splitting of the centromere at the start of anaphase. After this, they are individual chromosomes.] • Chromosome to the right splits at the centromere. • This occurs during Anaphase • The chromatids move to opposite poles of the cell to become chromosomes in a newly created cell. Chromosome With two Chromatids Become Two Chromosome
  • 37. 3.3 U.1 One diploid nucleus divides by meiosis to produce four haploid nuclei. http://cis.payap.ac.th/wp-content/uploads/2011/11/142.png • Cells divide twice • Result: 4 daughter cells, each with half as many chromosomes as parent cell
  • 38. • What if a complex multicellular organism (like us) wants to reproduce? – joining of egg + sperm • Do we make egg & sperm by mitosis? 46 46+ 92 egg sperm zygote What if we did, then…. Doesn’t work! No! 3.3 U.2 The halving of the chromosome number allows a sexual life cycle with fusion of gametes.
  • 39. 3.3 U.1 One diploid nucleus divides by meiosis to produce four haploid nuclei. Edited from: https://commons.wikimedia.org/wiki/File:Diagram_of_meiosis.svg Meiosis is a reduction division of the nucleus to form haploid gametes One diploid (2N) body cells contain a homologous pair of each chromosome (except for sex cells) Four haploid (N) gametes contain one of each chromosome Second division of the nucleus First division of the nucleus Chromosomes are replicated to form sister chromatids
  • 40. 3.3 U.1 One diploid nucleus divides by meiosis to produce four haploid nuclei. http://www.sumanasinc.com/webcontent/animations/c ontent/meiosis.html http://highered.mheducation.com/sites/dl /free/0072437316/120074/bio19.swf http://www.stolaf.edu/people/giannini/fla shanimat/celldivision/meiosis.swf http://www.biostudio.com/d_%20Meiosis.htm The animations are a great way to visualize the process – watch and take notes. Meiosis is a reduction division of the nucleus to form haploid gametes
  • 41. 3.3 U.2 The halving of the chromosome number allows a sexual life cycle with fusion of gametes. http://lh6.ggpht.com/n2T0xB_8BXE/UY4bhlV52uI/AAAAAAAABxg/7ZjWF QMBfx4/s1600-h/Humanlifecycle9.jpg Sexual life cycle • Meiosis: reduces # of chromosomes creating gametes (n) • Fertilization: combine gametes (sperm + egg) creating a fertilized egg or zygote (2n)
  • 42. 3.3 U.2 The halving of the chromosome number allows a sexual life cycle with fusion of gametes. Stage of the sexual life cycle Chromosome number (N/2N) Haploid or Diploid Adults 2n = 46 D Gametes (egg and sperm cells) n = 23 H Zygote 2n = 46 D Juvenile 2n = 46 D
  • 43. Duplication of Chromosomes During S stage of Interphase Single chromosome (called chromatin during Interphase) Two identical sister chromatids joined by a centromere DNA replication during S stage of Interphase 3.3 U.3 DNA is replicated before meiosis so that all chromosomes consist of two sister chromatids. .
  • 44. 10.1 U.1 Chromosomes replicate in interphase before meiosis. • During the S phase of the cell cycle, so that each chromosome has a copy of itself and consists of two sister chromatids. • During meiosis I, chromosomes condense and synapse to form bivalents (homologous chromosomes are aligned next to each other). http://www.columbia.edu/cu/biology/co urses/c200507/images/meiosis_19.gif
  • 45. 3.2 A.1 Cairns’ technique for measuring the length of DNA molecules by autoradiography. • John Cairns first supplied the E. Coli bacteria cells with suitable radioactive material (replaces normal hydrogen in thymidine). • Used to selectively label only DNA and will not label RNA. • Intact E. Coli bacterial chromosomes are placed on slides. These slides are then covered by photographic emulsion and stored in dark. • During this storage the particles are emitted the radioactive material, which exposed the film • The photographs show the regions of labelled DNA. Findings • The images showed that E. coli possesses a single circular chromosome which is 1,100 μm long (E. coli cells have a length of only 2 μm) • Cairns images also provided evidence to support the theory of semi-conservative replication
  • 46. 3.2 U.5 Homologous chromosomes carry the same sequence of genes but not necessarily the same alleles of those genes. https://molecularhelix.files.wordpress.com/2011/07/homolg.gif • Homologous chromosomes are chromosomes within each cell that carry the same genes • One chromosome came from an individual’s mother and one from the father • They have the same shape and size • These chromosomes pair up during meiosis • Even though these chromosomes carry the same genes, they could have different alleles (different versions of the same gene) Crossing Over is an exchange of genetic information as two of the same chromosomes move near to each other and exchange information
  • 47. 3.2 U.6 Diploid nuclei have pairs of homologous chromosomes. • Diploid nuclei have two copies of each type of chromosome. One chromosome comes from the mother and one from the father. • Haploid gametes (sperm and egg) fuse during sexual reproduction which produces zygote with a diploid nucleus • This cell will then divide by mitosis to produce numerous cells, all with a diploid nucleus • Each nucleus has two copies of each gene, accept the sex chromosomes
  • 48. 3.2 U.8 The number of chromosomes is a characteristic feature of members of a species. • The chromosome number is a characteristic feature of that species 2 (n) in a diploid cell or n in a haploid cell. • A chromosome number does not indicate how complicated an organism might be • Organisms with different numbers of chromosomes which is why the letter n is used because it is a variable and they might not be able to interbred • Chromosome number tends to remain unchanged over millions of years of evolution; however, sometimes through evolution chromosomes can fuse together or split to change the number of chromosomes an organism contains • During human evolution, two ancestral ape chromosomes fused to produce human chromosome 2
  • 49. First variation created by meiosis: Crossover • Red and Blue chromosomes are a homologous pair. • They have replicated during the interphase and each copy is held together by the centromere (black dot) 3.3 U.4 The early stages of meiosis involve pairing of homologous chromosomes and crossing over followed by condensation. [The process of chiasmata formation need not be explained.]
  • 50. •The chromosomes overlap and exchange lengths of DNA. •The chiasmata are the points at which the red and blue chromosomes overlap. •At a molecular level the exchange has to be exact avoiding loss of bases. Such an error would have profound effect on the genes. 3.3 U.4 The early stages of meiosis involve pairing of homologous chromosomes and crossing over followed by condensation. [The process of chiasmata formation need not be explained.]
  • 51. • At anaphase I the homologous pairs are separated. • Notice that sections of DNA have been exchanged. • These will mean that the alleles on these section have also been exchanged • Crossing over, Independent Assortment, combined with Fertilization: A couple can produce over 64 trillion (8.3 million x 8.3 million) different combinations in a zygotes during fertilization 3.3 U.4 The early stages of meiosis involve pairing of homologous chromosomes and crossing over followed by condensation. [The process of chiasmata formation need not be explained.]
  • 52. 10.1 U.4 Chiasmata formation between non-sister chromatids can result in an exchange of alleles. • Chiasmata are points where two homologous non-sister chromatids exchange genetic material during crossing over in meiosis. • Chromosomes intertwine and break at the exact same positions in non-sister chromatids. • The two chromosomes are now attached at the same corresponding position on the non- sister chromatid. • Once attached the non-attached portions of the chromatids actually repel each other. • Chiasmata refer to the actual break of the phosphodiester bond during crossing over. • The chiasmata are separated during anaphase 1 which can result in an exchange of alleles between the non-sister chromatids from the maternal and paternal chromosomes.
  • 53. 10.1 U.5 Homologous chromosomes separate in meiosis I. • During meiosis I, unlike mitosis homologous chromosomes separate to opposite poles; however, their sister chromatids remain attached to each other • Homologous chromosomes can exchange material in a process called crossing over • Meiosis I is considered reduction division because the chromosome number is reduced by half (2n -> n in humans) • In Meiosis II there is no further reduction of the chromosomes. This stage resembles mitosis. http://i.ytimg.com/vi/RfRtVDkABPI/hqdefault.jpg
  • 54. 10.1 U.7 Independent assortment of genes is due to the random orientation of pairs of homologous chromosomes in meiosis I. Second variation created by meiosis: Independent assortment
  • 55. 10.1 S.1 Drawing diagrams to show chiasmata formed by crossing over. http://www.thestudentroom.co.uk/showthread.php?t=1561992 • The centromere is a section of DNA on the chromosome where the chromatids are closest, it's generally tightly packed, the DNA typically doesn't have a defined sequence and is often satellite DNA, non-coding. It's the point at which the mitotic spindle attaches itself. • Chiasmata are places where the chromosomes cross over, and sometimes exchanges of DNA take place. This is usually looser packed, often functional DNA
  • 56. 3.3 U.5 Orientation of pairs of homologous chromosomes prior to separation is random. http://cnx.org/resources/9d7cd91abd65e9c7fa857cf6c6d133ec/Figure_08_03_06.jpg Independent assortment of chromosomes alleles for a trait separate when gametes are formed. These allele pairs are then randomly united at fertilization. • meiosis introduces genetic variation in several ways • gametes of offspring do not have same combination of genes as gametes from parents
  • 57. 10.1 U.2 Crossing over is the exchange of DNA material between non- sister homologous chromatids.
  • 58. 10.1 U.6 Sister chromatids separate in meiosis II. • During meiosis II sister chromatids separate (some are non-identical sister chromatids due to crossing over • This type of separation is very similar to mitosis as the chromatids are separated from each other
  • 59. 10.1 U.7 Independent assortment of genes is due to the random orientation of pairs of homologous chromosomes in meiosis I. • Independent assortment is an essential component in explaining how chromosomes align themselves during meiosis. • It also explains how unlinked genes are passed on from generation to generation. • When homologues line up along the equatorial plate in metaphase I, the orientation of each pair of is random; meaning the maternal or paternal homologue can orient towards either pole. • The orientation of how one set of homologues line up has no effect on how any of the other homologues line up. • Two young women to the right are non identical twin sisters, who because of independent assortment have different hair and skin coloring http://hellogiggles.com/bi-racial-twins/
  • 60. 10.1 U.3 Crossing over produces new combinations of alleles on the chromosomes of the haploid cells. http://www.indiana.edu/~oso/lessons/Genetics/ figs/sheep/2_fXe_sp2.jpg
  • 61. 3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the formation of four haploid cells. http://upload.wikimedia.org/wikipedia/commons/5/54/Meiosis_diagram.jpg
  • 62. Meiosis 1 • interphase • prophase 1 • metaphase 1 • anaphase 1 • telophase 1 Meiosis 2 • prophase 2 • metaphase 2 • anaphase 2 • telophase 2 2nd division of meiosis separates sister chromatids (1n  1n) * just like mitosis * 1st division of meiosis separates homologous pairs (2n  1n) “reduction division” 3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the formation of four haploid cells.
  • 63. Meiosis: Generates haploid gametes • Reduces the number of chromosomes by half. • Also produces genetic variability, each gamete is different, ensuring that two offspring from the same parents are never identical. • Two divisions: Meiosis I and meiosis II. Chromosomes are duplicated in interphase prior to Meiosis I. 3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the formation of four haploid cells.
  • 64. Meiosis I: Separation of Homologous Chromosomes 1. Prophase I: • Chromatin condenses into chromosomes. • Nuclear membrane disappear. • Centrosomes move to opposite poles of cell and microtubules attach to chromatids. • Synapsis: Homologous chromosomes pair up and form a tetrad of 4 sister chromatids. 3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the formation of four haploid cells.
  • 65. Prophase I: Crossing Over Between Homologous Chromosomes Crossing over combined with Fertilization: A couple can produce over 64 trillion (8.3 million x 8.3 million) different combinations in a zygotes during fertilization. 3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the formation of four haploid cells.
  • 66. Meiosis I: 2. Metaphase I: 3. Anaphase I: 4. Telophase I and Cytokinesis: 3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the formation of four haploid cells.
  • 67. 3.3 U.6 Separation of pairs of homologous chromosomes in the first division of meiosis halves the chromosome number. • Each individual has a pair chromosomes both chromosomes of a pair carry “matching” genes control same inherited characters homologous = same information • To create sex cells the number of chromosomes must be reduce for Humans from 46 chromosomes  to 23. Reduced by half. http://41.media.tumblr.com/4c08ea694bcf30c5a9d8423 187fba2de/tumblr_mx5zd49kWN1qjofuoo1_1280.jpg
  • 68. Random Assortment of Homologous Chromosomes During Meiosis I Generates Many Possible Gametes
  • 69. Meiosis II: Separation of Sister Chromatids During interphase that follows meiosis I, no DNA replication occurs. Interphase may be very brief or absent. Meiosis II is very similar to mitosis. 1. Prophase II: • Very brief, chromosomes reform. • No crossing over or synapsis. • Spindle forms and starts to move chromosomes towards center of the cell. 3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the formation of four haploid cells.
  • 70. Meiosis II: Separation of Sister Chromatids 2. Metaphase II: • Very brief, individual chromosomes line up in the middle of the cell. 3. Anaphase II: • Chromatids separate and move towards opposite ends of the cell. 3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the formation of four haploid cells.
  • 71. Meiosis II: Separation of Sister Chromatids 4. Telophase II: • Nuclei form at opposite ends of the cell. • Cytokinesis occurs. Product of meiosis: Four (4) haploid gametes, each genetically different from the other. 3.3 S.1 Drawing diagrams to show the stages of meiosis resulting in the formation of four haploid cells.
  • 72. Outline the differences between the behavior of chromosomes in Mitosis and Meiosis 5 marks Mitosis Meiosis One division Two divisions Diploid cells produced Haploid gametes produced No crossing-over in prophase Crossing-over in prophase I No chiasmata formation Chiasmata form Homologous pairs do not associate and line up at the equator in metaphase Homologous pairs do line up at the equator in metaphase I Sister chromatids separate in anaphase Homologous pairs separate in anaphase I Sister chromatids separate in anaphase II
  • 73. • Mendel’s Law of Independent assortment of chromosomes – Sexual reproduction creates genetic variation – gametes of offspring do not have same combination of genes as gametes from parents • Note: Not know during Mendel’s time, random assortment in humans produces 223 (8,388,608) different combinations in gametes from Dadfrom Mom offspring new gametes made by offspring 3.3 U.7 Crossing over and random orientation promotes genetic variation.
  • 74. Fertilization Sources of Genetic Variation – Human Sperm + Human Egg come together to produce 8 million X 8 million = 64 trillion combinations! 3.3 U.8 Fusion of gametes from different parents promotes genetic variation. http://news.bbcimg.co.uk/media/images/63005000/jpg/_63005 026_p6480029-sperm_fertilizing_egg-spl.jpg http://worms.zoology.wisc.edu/dd2/echino/fert/files/page19_2.gif
  • 75. • Sperm + Egg = ? – any 2 parents will produce a zygote with over 64 trillion (223 x 223) possible diploid combinations http://worms.zoology.wisc.edu/dd2/e chino/fert/files/page19_2.gif 3.3 U.8 Fusion of gametes from different parents promotes genetic variation.
  • 76. Sexual reproduction allows us to maintain both genetic similarity & differences. Baldwin brothers Jonas Brothers http://upload.wikimedia.org/wikipedia/commons/f/ff/Jonas_Brothers_2009.jpg Martin & Charlie Sheen, Emilio Estevez 3.3 U.8 Fusion of gametes from different parents promotes genetic variation.
  • 77. 3.3 A.1 Non-disjunction can cause Down syndrome and other chromosome abnormalities. Accidents During Meiosis Can Cause Chromosomal Abnormalities – Nondisjunction: Chromosomes fail to separate. – Gametes (and zygotes) will have an extra chromosome, others will be missing a chromosome. • Trisomy: Individuals with one extra chromosome, three instead of pair. Have 47 chromosomes in cells. • Monosomy: Missing a chromosome, one instead of pair. Have 45 chromosomes in cells. http://upload.wikimedia.org/wikipedia/commons/3/31/Mitotic_nondisjunction.png
  • 78. 3.2 U.9 A karyogram shows the chromosomes of an organism in homologous pairs of decreasing length. https://commons.wikimedia.org/wiki/File:NHGRI_human_male_karyotype.png
  • 79. 3.2 U.9 A karyogram shows the chromosomes of an organism in homologous pairs of decreasing length. http://learn.genetics.utah.edu/content/chromosomes/karyotype/ Karyotype is a property of the cell described by the number and type of chromosomes present in the nucleus (of a eukaryote cell). Karyogram is a diagram or photograph of the chromosomes present in a nucleus (of a eukaryote cell) arranged in homologous pairs of decreasing length. a Karyogram is a diagram that shows, or can be used to determine, the karyotype.
  • 80. Normal Male 3.2 A.4 Use of karyograms to deduce sex and diagnose Down syndrome in humans. http://commons.wikimedia.org/wiki/File:NHGRI_human_male_karyotype.png
  • 81. Normal Female 3.2 A.4 Use of karyograms to deduce sex and diagnose Down syndrome in humans.
  • 82. X and y Chromosomes y X Females have a sex chromosome arrangement of XX, both the same. Males have a sex chromosome arrangement Of Xy (about 200 genes
  • 83. Male with Down Syndrome: Trisomy 21; mental deficiencies , as well as physical abnormalities 3.2 A.4 Use of karyograms to deduce sex and diagnose Down syndrome in humans. http://upload.wikimedia.org/wikipedia/commons/6/6f/Trisomie_21_Genom-Schema.gif
  • 84. 3.3 A.3 Description of methods used to obtain cells for karyotype analysis e.g. chorionic villus sampling and amniocentesis and the associated risks. http://www.hopkinsmedicine.org/healthlibrary/GetImage.aspx?ImageId=161385 • Amniocentesis medical procedure used in prenatal diagnosis of chromosomal abnormalities and also used for sex determination. • A small amount of amniotic fluid, which contains fetal tissues, is sampled from the amniotic sac surrounding a developing fetus, and the fetal DNA is examined for genetic abnormalities.
  • 85. 3.3.A3 Description of methods used to obtain cells for karyotype analysis e.g. chorionic villus sampling and amniocentesis and the associated risks. Can be carried out in the 16th week of the pregnancy with around a 1% chance of a miscarriage http://www.medindia.net/animation/amniocentesis.asp
  • 86. 3.3.A3 Description of methods used to obtain cells for karyotype analysis e.g. chorionic villus sampling and amniocentesis and the associated risks. Can be carried out in the 11th week of the pregnancy with around a 2% chance of a miscarriage
  • 87. 3.3 A.2 Studies showing age of parents influences chances of non- disjunction. http://knowgenetics.org/wp-content/uploads/2012/12/DownSyndromeRisk.png • Weakening of cohesive ties holding together chromosomes in sex cells may contribute to maternal age- related errors. • The loss of cohesion may contribute to incorrect microtubule attachment during meiotic divisions
  • 88. 3.4 Inheritance Essential Question: The inheritance of genes follows patterns. http://upload.wikimedia.org/wikipedia/commons/1/11/Peas_in_pods_-_Studio.jpg
  • 89. Gregor Mendel • Austrian monk who published results of garden pea plants inheritance in 1865 • Used artificial pollination in a series of experiments by using a small brush to place the pollen on the reproductive parts of the flowers 3.4 U.1 Mendel discovered the principles of inheritance with experiments in which large numbers of pea plants were crossed.
  • 90. Key terminology 1. Genotype – symbolic representation of pair of alleles possessed by an organism, typically represented by two letters • Ex: Bb, GG, tt 2. Phenotype – characteristics or traits of an organism • Ex: five fingers on each hand, color blindness, type O blood 3. Dominant allele – an allele that has the same effect on the phenotype whether it is paired with the same allele or a different one; always expressed in phenotype • Ex: Aa give dominant trait A because the a allele is masked; the a allele is not transcribed and translated during protein synthesis 3.4 U.1 Mendel discovered the principles of inheritance with experiments in which large numbers of pea plants were crossed.
  • 91. 4. Recessive allele – an allele that has an effect on the phenotype only when present in the homozygous state • Ex: aa gives rise to the recessive trait because no dominant allele is there to mask it 5. Codominant allele – pairs of alleles that both affect the phenotype when present in a heterozygote • Ex: parent with curly hair and parent with straight hair can have children with different degrees of curliness as both alleles influence hair condition when both are present in the genotype 6. Locus – particular position on homologous chromosomes of a gene 3.4 U.1 Mendel discovered the principles of inheritance with experiments in which large numbers of pea plants were crossed.
  • 92. 7. Homozygous – having two identical alleles of a gene Ex: AA is a genotype which is homozygous dominant whereas aa is the genotype which is homozygous recessive 8. Heterozygous – having two different alleles of a gene Ex: Aa is a heterozygous genotype 9. Carrier – an individual who has a recessive allele of a gene that does not have an effect on their phenotype 3.4 U.1 Mendel discovered the principles of inheritance with experiments in which large numbers of pea plants were crossed.
  • 93. 10. Test cross – testing a suspected heterozygote plant or animal by crossing it with a known homozygous recessive (aa). Since a recessive allele can be masked, it is often impossible to tell if an organism is AA or Aa until they produce offspring which have the recessive trait. 3.4 U.1 Mendel discovered the principles of inheritance with experiments in which large numbers of pea plants were crossed.
  • 94. 3.4 U.1 Mendel discovered the principles of inheritance with experiments in which large numbers of pea plants were crossed.
  • 95. 3.4 U.1 Mendel discovered the principles of inheritance with experiments in which large numbers of pea plants were crossed.
  • 96. 3.4 U.2 Gametes are haploid so contain only one allele of each gene. • Gametes are sex cell. • Sex cells contain one chromosome of each type, as an example Humans have 23 types. • Parents pass information in the form of genes in gametes (sex cell) • These cell will fuse together with the cell of the opposite sex to create a zygote. Review
  • 97. Meiosis = reduction division • Cells divide twice • Result: 4 daughter cells, each with half as many chromosomes as parent cell 3.4 U.3 The two alleles of each gene separate into different haploid daughter nuclei during meiosis. Review
  • 98. Life cycle: reproductive history of organism, from conception  production of own offspring • Fertilization and meiosis alternate in sexual life cycles • Meiosis: cell division that reduces # of chromosomes (2n  n), creates gametes • Fertilization: combine gametes (sperm + egg) – Fertilized egg = zygote (2n) • Zygote divides by mitosis to make multicellular diploid organism 3.4 U.4 Fusion of gametes results in diploid zygotes with two alleles of each gene that may be the same allele or different alleles. . Review
  • 99. Human Life Cycle 3.4 U.4 Fusion of gametes results in diploid zygotes with two alleles of each gene that may be the same allele or different alleles. . Review
  • 100. 3.4 U.4 Fusion of gametes results in diploid zygotes with two alleles of each gene that may be the same allele or different alleles. . Review
  • 101. Mendel’s Law’s Three parts • Law of SegregationFor each characteristic, an organism inherits two alleles, one from each parent. • Law of Dominance Alternative versions of genes dominant or recessive that account for variations in inherited characteristics. Organized into three possible combinations. Homozygous dominant, Homozygous recessive or Heterozygous dominant. • Law of Independent Assortment The two alleles for each characteristic segregate during gamete production 3.4 S.1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
  • 102. Principle of Segregation: Each Parent or Gamete Contributes One Allele to Offspring 3.4 S.1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
  • 103. Punnet Square: Used to determine the outcome of a cross between two individuals. In the example we have two parents that are heterozygous dominant for a trait Offspring: Genotype: 1/4 PP, 1/2 Pp, and 1/4 pp Phenotype: 3/4 Purple and 1/4 white 3.4 S.1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
  • 104. 3.4 S.2 Comparison of predicted and actual outcomes of genetic crosses using real data.
  • 105. 3.4 S.2 Comparison of predicted and actual outcomes of genetic crosses using real data.
  • 106. F0 F1 Genotype: R ? r r Phenotypes: All red Test Cross Used to determine the genotype of an unknown individual. The unknown is crossed with a known homozygous recessive. Homozygous recessiveunknown Phenotype: Key to alleles: R = Red flower r = white Some white, some red Unknown parent = RR Unknown parent = Rr Possible outcomes: gametes r r R Rr Rr R Rr Rr gametes r r R Rr Rr r rr rr 3.4 S.2 Comparison of predicted and actual outcomes of genetic crosses using real data.
  • 107. 3.4 U.5 Dominant alleles mask the effects of recessive alleles but co- dominant alleles have joint effects. . http://gestblog.scientopia.org/wp-content/uploads/sites/35/2012/07/10-04.gif
  • 108. Problems with predictions I. Codominance II. Multiple alleles III. Sex linked genes
  • 109. 3.4 U.5 Dominant alleles mask the effects of recessive alleles but co-dominant alleles have joint effects. 3.4 A.1 Inheritance of ABO blood groups. [The expected notation for ABO blood group alleles: O = i, A=IA, B = IB.] There are 4: A, B, AB (codominant example)and O. A & B refer to 2 genetically inherited A and B antigens on the surface of red blood cells. IA – codes for A IB – codes for B i - codes for no antigen = type O blood
  • 110. http://www.anatomybox.com/tag/erythrocytes/ 3.4.U5 Dominant alleles mask the effects of recessive alleles but co-dominant alleles have joint effects. Dominant alleles have the same effect on the phenotype whether it is present in the homozygous or heterozygous state IAi Type “O” allele present and blood type is not O therefore the type “O” allele is recessive to type “A” Type “A” allele present and blood type is A therefore the type “A” allele is dominant to type “O” IAIB Type “A” and “B” alleles are present and blood type is AB therefore type “A”and “B” alleles are codominant Codominant alleles are pairs of different alleles that both affect the phenotype when present in a heterozygote Recessive alleles only have an effect on the phenotype when present in the homozygous state
  • 111. http://www.anatomybox.com/tag/erythrocytes/ 3.4.U5 Dominant alleles mask the effects of recessive alleles but co-dominant alleles have joint effects. Dominant alleles have the same effect on the phenotype whether it is present in the homozygous or heterozygous state IAi Type “O” allele present and blood type is not O therefore the type “O” allele is recessive to type “A” Type “A” allele present and blood type is A therefore the type “A” allele is dominant to type “O” IAIB Type “A” and “B” alleles are present and blood type is AB therefore type “A”and “B” alleles are codominant Codominant alleles are pairs of different alleles that both affect the phenotype when present in a heterozygote Recessive alleles only have an effect on the phenotype when present in the homozygous state
  • 112. Multiple Alleles: There are three alleles That control blood type, type A, type B and Type O This produce 4 different Blood Groups Blood type O: Universal donor. Blood type AB: Universal acceptor 3.4 U.5 Dominant alleles mask the effects of recessive alleles but co-dominant alleles have joint effects. 3.4 A.1 Inheritance of ABO blood groups. [The expected notation for ABO blood group alleles: O = i, A=IA, B = IB.]
  • 113. 3.4 A.1 Inheritance of ABO blood groups. [The expected notation for ABO blood group alleles: O = i, A=IA, B = IB.]
  • 114. http://image.slidesharecdn.com/incompletecodominancemultiplealleles-120127095123-phpapp02/95/incomplete-codominance-multiplealleles-7-728.jpg 3.4 U.6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases are due to dominant or co-dominant alleles
  • 115. 3.4 U.6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases are due to dominant or co-dominant alleles Inheritance characterized by full expression of both alleles in the heterozygote. Seen in: • Roan Cattle • Tay Sacs disease • Blood Types You have a brown Bull and a white Cow. You cross them and get a mix of the two colors. BB = Brown Bull/Cow WW = white Bull/Cow BW = Mixture of the two colors
  • 116. 3.4 U.6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases are due to dominant or co-dominant alleles https://jeanurquharthighlandsandislandsmsp.files.wordpress.com/2014/06/cavans-bourbon-orkney-by-robert_scarth-on-flickr.jpg
  • 117. Sickle Cell Anemia: (example of a codomainant gene mutation and its consequences through protein synthesis) The Genetics of Sickle Cell Anemia •HBA HBA Suceptible to malaria without anemia • HBA HBs Increase resistance to malaria with mild anemia • HBs HBs Sickle cell shaped cell suceptible to malaria with severe anemia 3.4 U.6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases are due to dominant or co-dominant alleles
  • 118. This may cause: •Change in the base sequence of the gene •Change in the mRNA made in transcription •Change in the amino acids sequence / primary structure of the protein •Change in the secondary/ tertiary or quaternary structure of the protein •Change the shape of the protein •Change the function of the protein. http://evolution.berkeley.edu/evolibrary/images/interviews/pax_mouse_fly.gif 3.1 U.5 New alleles are formed by mutation. Review
  • 119. Examples of the mutations to the gene Deletion Duplication Inversion Translocation 3.1 U.5 New alleles are formed by mutation. [Deletions, insertions and frameshift mutations do not need to be included.] http://www.goldiesroom.org/Multimedia/Bio_Images/19%20Applie d%20Genetics/05%20Chromosome%20Mutations.jpg
  • 120. The Disease: •Sickle cell anemia is an inherited disorder that affects hemoglobin, a protein that enables red blood cells to carry oxygen to all parts of the body. •The disorder caused by a base substiution mutation, produces abnormal hemoglobin, which along with the abnormalities of the cells membrane cause damage from be trapped and freed in the capillaries. This shortens the blood cells life to as little as 4 days. Disease Symptoms: •chronic anemia, •acute chest syndrome, •stroke, •spleen and renal dysfunction, •pain crises and susceptibility to bacterial infections, particularly in children. •Sickle cell disease is also associated with significant mortality. 3.1 A.1 The causes of sickle cell anemia, including a base substitution mutation, a change to the base sequence of mRNA transcribed from it and a change to the sequence of a polypeptide in hemoglobin
  • 121. 3.1 U.4 Allleles differ from each other by one or only a few bases. Review
  • 122. 3.1 A.1 The causes of sickle cell anemia •The HBA gene carries the instructions for the manufacture of a protein that is a component of hemoglobin. •Hemoglobin is the protein containing a sequence GAG which codes for the amino acid glutamic acid. •In a base substitution mutation of Hemoglobin proteins in Sickle Cell Anemia. GAG has been changed into GTG which codes for the amino acid valine. It is given the symbol HBs for the gene. http://upload.wikimedia.org/wikipedia/commons/a/ac/Sickle_cell_01.jpg
  • 123. Alleles, genotypes and Phenotypes of Sickle Cell: HbA Normal allele producing normal protein using the amino acid glutamic acid Hbs Sickle allele which is the abnormal mutation of the normal allele. Is produced using the replacement amino acid valine. 3.1 A.1 The causes of sickle cell anemia http://www.zo.utexas.edu/faculty/sjasper/images/17.23.gif
  • 124.
  • 125. 3.1.A1 The causes of sickle cell anemia, including a base substitution mutation, a change to the base sequence of mRNA transcribed from it and a change to the sequence of a polypeptide in hemoglobin.
  • 126. •In these regions sickle cell trait ( HbAHbs ) are resistant to the malaria parasite. This is because the Hbs allele makes it difficult for the parasite to live inside the red cells. Sickle cell trait (carriers) therefore survive malaria infection. •HbAHbA normal hemoglobin people are susceptible to malaria infection and do not survive well. •HbsHbs have sickle disease and do not survive well. HbAHbs no or reduce sickle cell symptoms and resistant to malaria •The sickle allele survives well in malaria regions accounting for its high frequency in these regions. •With migration the allele has spread around the world. Compare the frequency distribution of sickle allele with the distribution of malaria.
  • 127. 3.4 U.6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases are due to dominant or co-dominant alleles
  • 128. 3.4 U.6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases are due to dominant or co-dominant alleles
  • 129. World Sickle allele Distribution: Compare the frequency distribution of sickle allele with the distribution of malaria. •Notice that the high frequency of sickle coincides with malaria regions
  • 130. 3.4 U.6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases are due to dominant or co-dominant alleles
  • 131. 3.4 U.6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases are due to dominant or co-dominant alleles
  • 132. 3.4 U.6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases are due to dominant or co-dominant alleles
  • 133. 3.4 U.8 Many genetic diseases have been identified in humans but most are very rare. • Medical research has identified over 4,000 genetic diseases, however many individuals do not suffer from one. • Most genetic diseases are caused by rare recessive alleles. Making the chance of inheritance very small. • Genetic sequencing of the human genome current estimates are that there maybe as little as 75-200 genes out of over 20,000 genes in the genome that contain these traits.
  • 134. 3.2 U.10 Sex is determined by sex chromosomes and autosomes are chromosomes that do not determine sex.
  • 135. 3.2 U.10 Sex is determined by sex chromosomes and autosomes are chromosomes that do not determine sex. • The X and Y chromosome determine the sex of an individual • If an individual has two homologous X chromosomes they will be a female and if they have an two non homologous X and a Y chromosome they will be a male • All other chromosomes are called autosomes and do not affect the sex of an individual • The X chromosome has many genes located on it essential to human development, while the Y chromosome has a small number of genes (some of these are shared with the X chromosome). The rest of the genes on the Y chromosome are only necessary for male development. • Color Blindness and hemophilia are two examples of sex linked traits http://images.zeit.de/wissen/gesundheit/2014 -01/y-chromosom/y-chromosom-540x304.jpg
  • 136. Sex Linkage Examples: Hemophilia • Hemophilia is an example of a sex linkage condition. • The hemophilia allele is recessive to the normal allele. • The gene is located on the non- homologous region of the X chromosome. • The disease is associated with an inability to produce a clotting factor in blood. • Internal bleeding takes longer to stop. 3.4 A.2 Red-green color blindness and hemophilia as examples of sex- linked inheritance. http://blog.nz-online.de/lieb/wp-content/uploads/sites/8/2010/07/blut.jpg
  • 137. Female carriers of sex linked alleles • Female heterozygote's for sex linked alleles e.g. Hemophilia XHXh or Color Blindness XBXb are carriers of the allele. • They are unaffected by the condition. • They do pass on the allele which may result in a homozygous female or a male with the sex linked recessive allele. 3.4 A.2 Red-green color blindness and hemophilia as examples of sex- linked inheritance.
  • 138. Sex Chromosomes 3.4 U.7 Some genetic diseases are sex-linked. The pattern of inheritance is different with sex-linked genes due to their location on sex chromosomes. [Alleles carried on X chromosomes should be shown as superscript letters on an upper case X, such as Xh.] • The X chromosome in humans spans more than 153 million base pairs (the building material of DNA). It represents about 1400 out of 20,000 - 25,000 genes. • The Y chromosome containing 200genes, out of the estimated 20,000 to 25,000 total genes in the human genome. Genetic disorders that are due to mutations in genes on the X chromosome are described as X linked. • Because males have so many fewer genes, they are much more likely to express sex linked disorders then females
  • 139. 3.4 U.7 Some genetic diseases are sex-linked. The pattern of inheritance is different with sex-linked genes due to their location on sex chromosomes. [Alleles carried on X chromosomes should be shown as superscript letters on an upper case X, such as Xh.] Male sex chromosomes • There are non-homologous region males in which there is only one allele per gene and that is inherited from the female on the X- chromosome • In the homologous region the male inherited two copies of an allele per gene.
  • 140. Female sex chromosomes • All regions of the X chromosome are homologous. • There are two alleles per gene as with all other genes on all other chromosomes This difference in x and y chromosomes plays a large role in determining rates of genetic inherited defects 3.4 U.7 Some genetic diseases are sex-linked. The pattern of inheritance is different with sex-linked genes due to their location on sex chromosomes. [Alleles carried on X chromosomes should be shown as superscript letters on an upper case X, such as Xh.]
  • 141. Sex Linkage Alleles on the non-homologous region of the X chromosome are more common in females than in males • A gene with two alleles where one is dominant and one is recessive. • Female has three possible genotypes and one is the homozygous recessive. • In a population the chance of being homozygous recessive is 33.3 %. • Males have two possible genotypes. • There is a 50% chance of the homozygous recessive condition in the population. • In sex linked conditions the recessive condition is more common in males than females. 3.4 U.7 Some genetic diseases are sex-linked. The pattern of inheritance is different with sex-linked genes due to their location on sex chromosomes. [Alleles carried on X chromosomes should be shown as superscript letters on an upper case X, such as Xh.]
  • 142. 3.4 U.7 Some genetic diseases are sex-linked. The pattern of inheritance is different with sex-linked genes due to their location on sex chromosomes. [Alleles carried on X chromosomes should be shown as superscript letters on an upper case X, such as Xh.]
  • 143. • Red Green Color Blindness is an example of a sex linked condition. • Red Green Color blindness is a recessive condition. • The color blind allele is recessive to the normal allele. • Female homozygous recessives XbXb are color blind. • Males with the genotype XbY are color blind. • Notice that in a population the probability of having a Red Green color blind genotype in males is higher. 3.4 A.2 Red-green color blindness and hemophilia as examples of sex- linked inheritance. http://en.wikipedia.org/wiki/Color_blindness#/media/File:Ishihara_9.png Above is a color test plate.[The numeral "74" should be clearly visible to viewers with normal color vision.
  • 144. 3.4 A.2 Red-green color blindness and hemophilia as examples of sex- linked inheritance. http://upload.wikimedia.org/wikipedia/commons/a/a3/XlinkRecessive.jpg Normal color vision Red/green color blindness
  • 145. • The homozygous genotype(*) in females has a high mortality. • The genotype XnY in males has a high mortality. 3.4 A.2 Red-green color blindness and hemophilia as examples of sex- linked inheritance. Hemophilia Hemophilia
  • 146. Cystic fibrosis (CF) Non Sex link recessive genetic trait found on Chromosome 7. Cystic fibrosis is a genetic disorder that affects mostly the lungs, Long-term issues include difficulty breathing and coughing up mucus as a result of frequent lung infections. Example: Cross The couple below are heterozygous for CF 3.4 A.3 Inheritance of cystic fibrosis and Huntington’s disease. http://www.bbc.co.uk/staticarchive/088e5fc50b3c51cfb49ebc4b6eaf203b18b93bbc.gif
  • 147. Huntington’s Disease Non Sex link dominant genetic trait which causes a decline in thinking and reasoning skills, including memory, concentration, judgment, and ability to plan and organize The couple two couples below are examples • Couple 1: 1 heterozygous (has trait) with 1 homozygous (without the trait) • Couple 2: Both parents are heterozygous with Huntington's 3.4 A.3 Inheritance of cystic fibrosis and Huntington’s disease. http://vanhornhuntingtonsdisease.weebly.com/uploads/1/3/7/4/13740905/4993818.jpg?1347964948 Couple 1 Couple 2
  • 148. Pedigree Chart • Another way to visualize a monohybrid crosses or determining a genotype is by using a pedigree chart • Knowing the phenotype of individuals in a family will sometimes allow genotypes to be determined. • In genetic counseling this enables probabilities to be determined for the inheritance of characteristics in children. 3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
  • 149. Pedigree Chart • White circle : Normal female • White Square: Normal male • Black Circle: affected female • Black square: affected male • (1) and (2)..Normal Parents • (3) affected female • (4),(5) and (6) normal 3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
  • 150. 3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
  • 151. 3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
  • 152. 1. Phenylketonuria (Pku) • Using the allele key provided state the genotype of parents 1 and 2? • Give the genotype and phenotype of individual 5 ? • Is it possible that the condition is sex linked ? • What is the genotype and phenotype of individuals 7 and 8? • Which two individuals have the incorrect pedigree 3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
  • 153. 2. Muscular Dystrophy • What type of genetic disease is muscular dystrophy? • Give the genotype and phenotype of 1? • Give the genotype and phenotype of 2? • Give the genotype and phenotype of 8 ? • Give the genotype and phenotype of 5 and 6 ? 3.4 S.3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
  • 154. Review: 1.6.U6 Mutagens, oncogenes and metastasis are involved in the development of primary and secondary tumors. mutation in a oncogene If a mutation occurs in an oncogenes it can become cancerous. In normal cells oncogenes control of the cell cycle and cell division. Cancer is uncontrolled cell growth. http://en.wikipedia.org/wiki/Oncogene#mediaviewer/File:Oncogenes_illustration.jpg uncontrolled cell division tumor formation malfunction in the control of the cell cycle
  • 155. 3.4 U.9 Radiation and mutagenic chemicals increase the mutation rate and can cause genetic diseases and cancer. • A mutagen is a physical (radiation), viruses, or ultraviolet light. They can also be chemical agent like Nitrosamines, found in tobacco. These mutagens change the genetic material, usually DNA, of an organism and increases the frequency of mutations above the natural background level. Many mutations cause cancer, mutagens are therefore also likely to be carcinogens.
  • 156. • Radiation-induced cancers do not appear until at least 10 years after exposure (for tumors) or 2 years after exposure (for leukemia). • The risk of cancer after exposure can extend beyond this latent period for the rest of a person’s life for tumors or about 30 years for leukemia. • Risk is calculations are based on: – The type of radiation. • Each type of radiation is different and affects tissues differently. – The energy that it leaves in the body. • More energy means a higher probability of an effect. – Where in the body the energy remains. • Radiation exposure to a non-sensitive area of the body (i.e., wrist) really has no actual effect. Radiation exposure to a sensitive area of the body (i.e., blood-forming organs) can have an effect if the amount of energy left is high enough. 3.4 U.9 Radiation and mutagenic chemicals increase the mutation rate and can cause genetic diseases and cancer.
  • 157. • Indirect damage – Water molecule is ionized, breaks apart, and forms OH free radical. – OH free radical contains an unpaired electron in the outer shell and is highly reactive: Reacts with DNA. – 75 percent of radiation-caused DNA damage is due to OH free radical. • Direct damage – DNA molecule is struck by radiation, ionized, resulting in damage. 3.4 U.9 Radiation and mutagenic chemicals increase the mutation rate and can cause genetic diseases and cancer.
  • 158. Chromosome Damage Formation of a ring and fragments followed by replication of chromosomes. 3.4 U.9 Radiation and mutagenic chemicals increase the mutation rate and can cause genetic diseases and cancer.
  • 159. Chromosome Damage Interchange between two chromosomes forms a chromosome with two centromeres and fragment, followed by replication. 3.4 U.9 Radiation and mutagenic chemicals increase the mutation rate and can cause genetic diseases and cancer.
  • 160. Commonly Encountered Radiation Doses Effective Dose Radiation Source <= 0.01 rem annual dose living at nuclear power plant panoramic, or full-mouth dental x rays; skull or chest x ray <=0.1 rem single spine x ray; abdominal or pelvic x ray; hip x ray; mammogram <=0.5 rem kidney series of x rays; most barium-related x rays; head CT; any spine x-ray series; annual natural background radiation dose; most nuclear medicine brain, liver, kidney, bone, or lung scans <=1.0 rem barium enema (x rays of the large intestine); chest, abdomen, or pelvic CT <=5.0 rem cardiac catheterization (heart x rays); coronary angiogram (heart x rays); other heart x-ray studies; most nuclear medicine heart scans CT = computerized tomography; a specialized x-ray exam. 3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl.
  • 161. Radiation Doses and Expected Effects (cont.) General radiation doses to the entire body and expected effects: • 100-200 rem received in a short time will cause nausea and fatigue. • 100-200 rem received over a long period will increase a person’s chances of getting cancer. • 200-300 rem received in a short time will cause nausea and vomiting within 24-48 hours. Medical attention should be sought. • 300-500 rem received in a short time will cause nausea, vomiting, and diarrhea within hours. Loss of hair and appetite occurs within a week. Medical attention must be sought for survival; half of the people exposed to radiation at this high level will die if they receive no medical attention. • 500-1,200 rem in a short time will likely lead to death within a few days. • Greater than 10,000 rem in a short time will lead to death within a few hours. 3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl.
  • 162. 3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl. http://inapcache.boston.com/universal/site_graphics/blogs/bigpicture/hiroshima_08_05/h17_04.jpg
  • 163. 3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl. http://www.nucleardarkness.org/include/nucleardarkness/images/cityonfire/hiroshima_after_02_full.jpg Up to one mile from the blast site: 36,000 rem
  • 164. 3.4.A4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl. nuclear bombing of Hiroshima https://upload.wikimedia.org/wikipedia/commons/e/e9/The_patient%27s_skin_is_burned_in_a_pattern_corresponding_to_the_dark_porti ons_of_a_kimono_-_NARA_-_519686.jpg https://upload.wikimedia.org/wikipedia/commons/f/f6/Hiroshima_girl.jpg • Elevated rate of Leukemia (with the greatest impact in children and young adults) • Elevated rates of other cancers • No evidence of stillbirth or mutations in the children of those exposed to radiation http://i.telegraph.co.uk/multimedia/archive/02446/hiroshima-bomb_2446747b.jpg
  • 165. 3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl. http://www.zap-actu.fr/wp-content/uploads/2013/11/pripyat-une-des-villes-fantomes-pres-de-tchernobyl-01.jpg Average thyroid radiation dose observed in Chernobyl evacuees : 49,000 rem
  • 166. 3.4 A.4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl.
  • 167. 3.4.A4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl. accident at Chernobyl nuclear power station https://upload.wikimedia.org/wikipedia/commons/1/16/VOA_Markosian_-_Chernobyl02.jpg • A large area of pine forest downwind of the reactor turned brown and died. • Horses and cattle near the plant died from radiation damage to their thyroid glands. • Bioaccumulation of radioactive caesium in fish (Scandinavia and Germany) and lamb (Wales) - contaminated meat was banned from sale for years afterward. http://i.guim.co.uk/img/static/sys-images/Guardian/Pix/pictures/2014/6/27/1403890449199/933cb303-bf75-4e9a-8b0b-806bbfa6a37b- 2060x1373.jpeg?w=620&q=85&auto=format&sharp=10&s=abe2802021d01fe090859454e9020a44 • Drinking water (and milk) contaminated with radioactive iodine - at least 6,000 thyroid cancer attributed to radioactive iodine. • No clear evidence to support an increase in the rate of leukemia other cancers – in part due to the widely dispersed variable radiation and measures taken in European populations.
  • 168. 10.2 Inheritance Essential idea: Genes may be linked or unlinked and are inherited accordingly. http://upload.wikimedia.org/wikipedia/c ommons/4/46/Dividing_Cell_Fluorescenc e.jpg http://ies.rayuela.mostoles.educa .madrid.org/deptos/dbiogeo/recu rsos/Apuntes/ApuntesBioBach2/i magenes/genetica/droso.png
  • 169. 10.2 U.1 Gene loci are said to be linked if on the same chromosome. • Genes have specific locations (loci) on chromosomes • Chromosomes segregate and assort independently http://web.csulb.edu/~kmacd/361-6-Ch2.htm
  • 170. 10.2 U.2 Unlinked genes segregate independently as a result of meiosis. • Mendel’s law of independent assortment states allele pairs separate independently from other allele pairs during gamete formation (meiosis). • Therefore, traits on different chromosomes are transmitted to the offspring independently of traits on other chromosomes. • An exception to this rule is linked genes https://biologywarakwarak.files.wordpress.com/2011/12/crossingover-rec.jpg
  • 171. 10.2 U.3 Variation can be discrete or continuous. • Discrete Variation is controlled by alleles of a single gene or a small number of genes. The environment has little effect. In this case you either have the characteristic or you don't. Cystic fibrosis is a good example for this; either you have cystic fibrosis or you don’t. • Chi-squared calculations work well when using examples with discrete variation • Continuous Variation is a complete range of phenotypes that can exist from one extreme to the other. Height is an example of continuous variation. Continuous variation is the combined effect of many genes (known as polygenic inheritance) and is often significantly affected by environmental influences. Skin color is another example of continuous variation. http://articles.latimes.com/2013/nov/11/entertainment/la-et- cm-princess-bride-disney-musical-20131111
  • 172. 10.2 A.3 Polygenic traits such as human height may also be influenced by environmental factors. • Polygenic inheritance occurs when two or more genes control the expression of a phenotype. • As the amount of genes that control one trait increase, the number of phenotypes increases. • Each additional gene has an additive affect, increasing the phenotypes. This is called continuous variation. • Example: human height, which varies from person to person within the same race, and varies between identical twin . Below the effect of environmental facts on the height of these identical twin due to diet. http://http://www.dailymail.co.uk/femail/article-2591087/Double-Poignant-portraits-50- year-old-identical-twins-reveal-differences-lifestyle-affected-way-age.html
  • 173. North side, upper branches South side upper branches North side, lower branches South side, lower branches © Horticultural Research International
  • 174. Apple Size • What environmental conditions in the different parts of the tree might have caused the differences in size of the apples ? • The upper branches will receive more sunlight than the lower branches, but on the North side the sunlight would have been less intense • The lower branches will receive less sunlight because of the shade cast by the upper branches. This effect will be more pronounced on the North side Skin color • controlled by 3 or 4 genes and environmental factors leading to a wide range of phenotypes • Exposure to the sun can alter skin color and obscure genotypic differences.
  • 175. Mendel's Second Law of Independent Assortment • This law states that allele pairs separate independently during the formation of gametes. Therefore, traits are transmitted to offspring independently of one another • Polygenic traits are those traits that are controlled by more than one gene. Such traits may even be controlled by genes located on entirely different chromosomes. Human height, eye and hair color are examples of polygenic traits. Skin color is another polygenic trait for humans and a variety of other animals. Multiple alleles are several variant forms of one gene (ex. Genes for blood type A, O, B) 10.2 A.2 Completion and analysis of Punnett squares for dihybrid traits
  • 176. Dihybrid Crosses Monohybrid crosses is one trait. Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea color and pea surface. What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F1 gametesPunnet Grid: F0 Genotype: Phenotype: Heterozygous at both loci Heterozygous at both loci 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 177. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F1 gametesPunnet Grid: F0 Genotype: SsYy SsYy Phenotype: Heterozygous at both loci Heterozygous at both loci Smooth, yellow Smooth, yellow 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 178. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F1 gametes SY Sy sY sy SY SSYY SSYy SsYY SsYy Sy SSYy SSyy SsYy Ssyy sY SsYY SsYy ssYY ssYy sy SsYy Ssyy ssYy ssyy Punnet Grid: F0 Genotype: SsYy SsYy Phenotype: Heterozygous at both loci Heterozygous at both loci Smooth, yellow Smooth, yellow 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 179. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci? F1 gametes SY Sy sY sy SY SSYY SSYy SsYY SsYy Sy SSYy SSyy SsYy Ssyy sY SsYY SsYy ssYY ssYy sy SsYy Ssyy ssYy ssyy Punnet Grid: F0 Genotype: SsYy SsYy Phenotype: Heterozygous at both loci Heterozygous at both loci Smooth, yellow Smooth, yellow Phenotypes: 9 Smooth, yellow : 3 Smooth, green : 3 Rough, yellow : 1 Rough, green 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 180. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Calculate the predicted phenotype ratio for: F1 Punnet Grid: F0 Genotype: Phenotype: Heterozygous at both loci Heterozygous for S, homozygous dominant for Y Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 181. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Calculate the predicted phenotype ratio for: F1 Punnet Grid: F0 Genotype: SsYy SsYY Phenotype: Heterozygous at both loci Heterozygous for S, homozygous dominant for Y Smooth, yellow Smooth, yellow Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 182. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Calculate the predicted phenotype ratio for: F1 gametes SY sY SY Sy sY sy Punnet Grid: F0 Genotype: SsYy SsYY Phenotype: Heterozygous at both loci Heterozygous for S, homozygous dominant for Y Smooth, yellow Smooth, yellow Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 183. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Calculate the predicted phenotype ratio for: F1 gametes SY sY SY SSYY SsYY Sy SSYy SsYy sY SsYY ssYY sy SsYy ssYy Punnet Grid: F0 Genotype: SsYy SsYY Phenotype: Heterozygous at both loci Heterozygous for S, homozygous dominant for Y Smooth, yellow Smooth, yellow Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 184. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. Calculate the predicted phenotype ratio for: F1 gametes SY sY SY SSYY SsYY Sy SSYy SsYy sY SsYY ssYY sy SsYy ssYy Punnet Grid: F0 Genotype: SsYy SsYY Phenotype: Heterozygous at both loci Heterozygous for S, homozygous dominant for Y Smooth, yellow Smooth, yellow Phenotypes: 3 Smooth, yellow : 1 Rough, yellow Present the ratio in the simplest mathematical form. 6 Smooth, yellow : 2 Rough, yellow 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 185. Dihybrid Crosses Common expected ratios of dihybrid crosses. SY Sy sY sy SY SSYY SSYy SsYY SsYy Sy SSYy SSyy SsYy Ssyy sY SsYY SsYy ssYY ssYy sy SsYy Ssyy ssYy ssyy Heterozygous at both loci Heterozygous at both loci SsYySsYy 9 : 3 : 3 : 1 SY sY SY SSYY SsYY Sy SSYy SsYy sY SsYY ssYY sy SsYy ssYy Heterozygous at both loci Heterozygous at one locus, homozygous dominant at the other SsYySsYy 3 : 2 Sy sy SY SSYy SsYy Sy SSyy Ssyy sY SsYy ssYy sy Ssyy ssyy Heterozygous at both loci SsyySsYy 4 : 3 : 1 Heterozygous/ Homozygous recessive ssYYSSyy = All SsYy ssYySsyy = 1 : 1 : 1 : 1 ssyySSYY = all SyYy 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 186. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A rough yellow pea is test crossed to determine its genotype. Punnet Grid: F0 Genotype: Phenotype: Rough, yellow F1 Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 187. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A rough yellow pea is test crossed to determine its genotype. Punnet Grid: F0 Genotype: ssYy Phenotype: Rough, yellow F1 gametes sY sy Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 188. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A rough yellow pea is test crossed to determine its genotype. F1 gametes sY sy sY sYPunnet Grid: F0 Genotype: ssYy or ssYY Phenotype: Rough, yellow Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 189. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A rough yellow pea is test crossed to determine its genotype. F1 gametes sY sy sY sY All sy Punnet Grid: F0 Genotype: ssYy or ssYYssyy Phenotype: Rough, yellow Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 190. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A rough yellow pea is test crossed to determine its genotype. F1 gametes sY sy sY sY All sy ssYy ssyy ssYy ssYy Punnet Grid: F0 Genotype: ssYy or ssYYssyy Phenotype: Rough, yellow Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 191. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A rough yellow pea is test crossed to determine its genotype. F1 gametes sY sy sY sY All sy ssYy ssyy ssYy ssYy Punnet Grid: F0 Genotype: ssYy or ssYYssyy Phenotype: Rough, yellow Phenotypes: Some green peas will be present in the offspring if the unknown parent genotype is ssYy. No green peas will be present in the offspring if the unknown parent genotype is ssYY. 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 192. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F1 Punnet Grid: F0 Genotype: Phenotype: Smooth, green Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 193. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F1 gametes All sy Punnet Grid: F0 Genotype: ssyy Phenotype: Smooth, green Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 194. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F1 gametes Sy Sy All sy Ssyy Ssyy Punnet Grid: F0 Genotype: SSyyssyy Phenotype: Smooth, green Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 195. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F1 gametes Sy Sy Sy sy All sy Ssyy Ssyy Ssyy ssyy Punnet Grid: F0 Genotype: SSyy or Ssyyssyy Phenotype: Smooth, green Phenotypes: 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 196. Dihybrid Crosses Consider two traits, each carried on separate chromsomes (the genes are unlinked). Key to alleles: Y = yellow y = green S = smooth s = rough In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface. A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring. F1 gametes Sy Sy Sy sy All sy Ssyy Ssyy Ssyy ssyy Punnet Grid: F0 Genotype: SSyy or Ssyyssyy Phenotype: Smooth, green Phenotypes: No rough peas will be present in the offspring if the unknown parent genotype is SSyy. The presence of rough green peas in the offspring means that the unknown genotype must be Ssyy. The expected ratio in this cross is 3 smooth green : 1 rough green. This is not the same as the outcome. Remember that each reproduction event is chance and the sample size is very small. With a much larger sample size, the outcome would be closer to the expected ratio, simply due to probability. 10.2.A2 Completion and analysis of Punnett squares for dihybrid traits. AND 10.2.S1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.
  • 197. 10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes Phenotype Key: Phenotypic ratio: – 9 Smooth Yellow: – 3 Smooth green: – 3 Rough Yellow: – 1 Rough Green
  • 198. 10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes
  • 199. 10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes
  • 200. 10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes
  • 201. 10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes
  • 202. 10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes
  • 203. 10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes
  • 204. 10.2 S.1 Calculation of the predicted genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes
  • 205. Example: Human skin color • This trait is controlled by multiple alleles, as many as 6 genes each with its own alleles. • As the number of genes increases the amount of phenotypic variation increases. • The alleles control the production of melanin which is a pigment that colors skin. • In this example the calculation is performed with 3 genes each with 3 alleles. The cross is between two individuals heterozygous at both alleles Allele Key A= add melanin a= no melanin added B= adds melanin b= no melanin added 10.2 U.4 The phenotypes of polygenic characteristics tend to show continuous variation. http://www2.estrellamountain.edu/faculty/farabee/BIOBK/BioBookgeninteract.html
  • 206. 10.2 U.4 The phenotypes of polygenic characteristics tend to show continuous variation.
  • 207. 10.2 U.4 The phenotypes of polygenic characteristics tend to show continuous variation.
  • 208. 10.2 A.1 Morgan’s discovery of non-Mendelian ratios in Drosophila. Thomas Hunt Morgan Drosophila melanogaster – fruit fly Fast breeding, chromosomes (XX/XY) Discovered sex-linked gene, located on X or Y chromosome. These genes characteristic are not expressed in the same ratios as traits located on other chromosomes.
  • 209. 10.2 A.1 Morgan’s discovery of non-Mendelian ratios in Drosophila. • Link genes are located on the same chromosome and do not sort independently • Genes inherited together. Differing from sex linked genes which are genes found on the sex chromosomes. • These genes do not show the expected Mendelian ratios as predicted by the Laws Independent Assortment.
  • 210. Linked Genes 10.2 S.2 Identification of recombinants in crosses involving two linked genes.
  • 211. 10.2 S.2 Identification of recombinants in crosses involving two linked genes.
  • 212. 10.2 S.2 Identification of recombinants in crosses involving two linked genes.
  • 213. 10.2 S.2 Identification of recombinants in crosses involving two linked genes.
  • 214. 10.2 S.2 Identification of recombinants in crosses involving two linked genes.
  • 215. 10.2 S.2 Identification of recombinants in crosses involving two linked genes.
  • 216. 10.2 S.2 Identification of recombinants in crosses involving two linked genes.
  • 217. 10.2 S.3 Use of a chi-squared test on data from dihybrid crosses. • For an example let’s use Mendel’s results from his pea plant crosses • When he did a dihybrid cross between two heterozygotes RrYy x RrYy, the expected phenotypic ratio due to independent assortment would be 9:3:3:1. Look at the chart below to see his actual results.
  • 218. 10.2 S.3 Use of a chi-squared test on data from dihybrid crosses.
  • 219. Chi-Square Analysis Practice • Two true-breeding Drosophila are crossed: a normal-winged, red- eyed female and a miniature-winged, vermillion-eyed male. The F1 offspring all have normal wings and red eyes. When the F1 offspring are crossed with miniature-winged, vermillion-eyed flies, the following offspring resulted: – 233 normal wing, red eye – 247 miniature wing, vermillion eye – 7 normal wing, vermillion eye – 13 miniature wing, red eye • What type of conclusions can you draw from this experiment? Explain your answer. 10.2 S.3 Use of a chi-squared test on data from dihybrid crosses.
  • 220. 10.2 U.5 Chi-squared tests are used to determine whether the difference between an observed and expected frequency distribution is statistically significant. • A chi-square test is a statistical test that can be used to determine whether observed frequencies are significantly different from expected frequencies • These statistical tests enable us to compare observed and expected frequencies empirically and to decide if the results we see are statistically significant. Statistical significance in this case implies that the differences are not due to chance alone, but instead may be caused by other factors at work. • The formula to the right is for a chi-squared test O = observed frequencies E = expected frequencies X2 =
  • 221. 10.2 U.5 Chi-squared tests are used to determine whether the difference between an observed and expected frequency distribution is statistically significant.
  • 222. 10.2 U.5 Chi-squared tests are used to determine whether the difference between an observed and expected frequency distribution is statistically significant.
  • 223. 10.2 U.5 Chi-squared tests are used to determine whether the difference between an observed and expected frequency distribution is statistically significant. AND 10.2.S3 Use of a chi-squared test on data from dihybrid crosses. Testing dihybrid crosses – using the chi-squared test Now try extending your understanding of dihybrid crosses using data. Follow the links below and complete the examples and problems: CORN GENETICS CHI SQUARE ANALYSIS by Biology Corner Answer key: http://www.biologycorner.com/worksheets/corn_chi_key. html http://www.biologycorner.com/worksheets /corn_chi.html Chi-square in biology: Testing for a dihybrid ratio http://mathbench.umd.edu/modules/s tatistical- tests_chisquare_intro/page15.htm
  • 224. 3.5 Genetic modification and biotechnology Essential idea: Biologists have developed techniques for artificial manipulation of DNA, cells and organisms. http://www.nacentralohio.com/wp-content/uploads/2013/01/WW_0113_GMO_AppleOrange.jpg
  • 225. 3.5 U.2 PCR can be used to amplify small amounts of DNA. • PCR (polymerase chain reaction) is a laboratory technique that takes a single or few copies of DNA and amplifies them to generate millions or more copies of a particular DNA sequence. • When you collect DNA from different sources such as sperm samples or small drops of blood, there are usually very little usable cells to collect DNA. • Therefore, PCR is used to create enough DNA to be analyzed for investigations such as forensics or custody cases. • Once large quantities of the DNA have been created, other methods such as gel electrophoresis are used to analyze the DNA. Click4Biology
  • 226. 3.5 U.2 PCR can be used to amplify small amounts of DNA. PCR occurs in a thermal cycler and involves a repeat procedure of 3 steps: 1. Denaturation: DNA sample is heated to separate it into two strands 2. Annealing: DNA primers attach to opposite ends of the target sequence 3. Elongation: A heat-tolerant DNA polymerase (Taq) copies the strands