extending mendelian genetics
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Extending Mendelian Genetics. Chapter 12 (YES that is the correct chapter). Simple Mendelian genetics. Makes a couple of assumptions Organisms are genetically simple Traits are controlled by a single gene Each gene has only 2 alleles One allele is completely dominant over the other - PowerPoint PPT PresentationTRANSCRIPT
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Extending Mendelian Genetics
Chapter 12 (YES that is the correct chapter)
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Simple Mendelian genetics
Makes a couple of assumptions Organisms are genetically simple Traits are controlled by a single gene Each gene has only 2 alleles One allele is completely dominant over
the other It is rarely this simple
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Incomplete dominance Heterozygotes show an
intermediate phenotype RR= red flowers rr= white flowers Rr= pink flowers
Blending theory? Why or why not?
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Incomplete dominance
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Co-dominance
In heterozygotes, both alleles appear phenotypically as individual distinct traits White Black Roan
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Co-dominance in blood types
ABO blood groups 3 alleles= IA, IB, and i
IA and IB are dominant over I IA and IB are co-dominant to each other Determines the presence of specific
oligosaccharides on the surface of red blood cells
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Blood Types
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Blood Types
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Blood compatibility
Matching compatible blood groups is critical for blood transfusions
A person produces antibodies against foreign oligosaccharides If donor’s blood has an A or B oligosaccharide
that is foreign to the recipient, antibodies in the recipient’s blood will bind to the foreign oligosaccharides
Causes the donated blood cells to clump together and can result in death for the recipient
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Blood donation
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Pleiotropy
One gene affects more than one phenotypic character
MOST are pleiotropic Wide ranging
effects due to a single gene
Dwarfism Gigantism
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Acromegaly: Andre the Giant
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Pleiotropy
Pleiotropic allele may be dominant with respect to one trait and recessive with respect to another
Effects often hard to predict Many inherited disorders in
humans Sickle cell anemia Timothy disease
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Epistasis
One gene masks another Products of genes interact with each
other Coat color in mice=2 genes
C=pigment c=no pigment B=black pigment b= brown pigment If cc is present, then albino (no coat color)
regardless of BB, Bb or bb
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Epistasis
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Epistasis in Labrador retrievers EE or Ee= dark pigment (black coat color) ee= no pigment (yellow coat color)
BB= dark pigment deposition (black coat color) bb= light pigment deposition (chocolate coat
color)
Doesn’t affect just coat color eebb=yellow dogs with brown on nose, lips and eye
rims eeB_=yellow dogs with black pigment on nose, lips
and eye rims
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Epistasis
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Polygenic inheritance Phenotype is determined by an additive
effect of 2 or more genes acting on a single character Phenotypes on a continuum Human polygenic traits
Skin color Height Weight Eye color Intelligence behaviors
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Nature vs. Nurture
The interaction of genes and environment control phenotype
Hotly contested debate
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Examples of nature v. nurture
A tree can have leaves that vary in shape, size and color depending on exposure to wind and sun
In humans nutrition influences height Exercise alters build Sun exposure darkens skin Experience improves performance on
intelligence tests Even identical twins (genetic equals)
accumulate phenotypic differences due to unique experiences
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Chromosome Theory of Inheritance
Walter S. Sutton, Theodore Boveri and others (~1902) Mendelian genes
have specific loci on chromosomes
The chromosomes segregate and undergo independent assortment
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Thomas Hunt Morgan
First to associate a specific gene with a specific chromosome Drosophilia Melanogaster
Common fruit fly Fly room Why flies?
Prolific breeders New generation every 2 weeks 4 pairs of chromosomes
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Fruit Flies
Took Morgan a year of breeding flies to get a mutant Male with white eyes Wild type is red
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Discovery of Sex linkage
Mated P1=White eyed male X red eye female
F1=all red eyes (red is dominant) P2=Red eye F1
F2=3:1 phenotypic ratio Only the males had white eyes
½ red, ½ white All females were red eyes
Eye color is linked to X
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Sex linked genes
Morgan’s evidence of sex linkages added credibility to the chromosome theory of inheritance
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Linked genes
Far more genes than chromosomes Each chromosome has 100’s of genes Genes on the same chromosome tend
to be inherited as a unitLinked genes
Linked genes don’t produce typical Mendelian results in breeding experimentsWhen linked they are unable to assort
independently
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Complete the following cross
In flies Wild type flies have gray bodies (b+) and normal wings (vg+) the mutant flies have black body (b) and vestigial wings (vg)(much smaller than normal). What is the genotype for all the flies in the F1 generation?
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Resulting F1 flies were all hybrids presenting the wild
type (gray bodies andvestigial wings)
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F2
What is the expected ratio of phenotypes when you mate one of the hybrids from the F1 with an individual that is homozygous recessive for both traits?
This is a testcross
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Linkage
These results indicated that body type and wing shape are inherited together
So how do you explain the black, normal and the gray, vestigial
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Test cross
Peas Heterozygous for yellow round crossed
with homozygous for green wrinkled YyRr X yyrr ½ of phenotypes match original P
phenotypes Either yellow round or green wrinkled
½ have a new combination of traits Green round or Yellow wrinkled
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Unlinked genes
Show a 50% recombination Indicates separate chromosomes for
these genes Not Linked genes This form of recombination is the
result of independent assortment of chromosomes and therefore alleles
Seed color gene has no bearing on the seed shape gene
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Recombination between linked genes
Refer back to Morgan’s dihybrid test cross
Assuming linkage between body and wing, expected 1:1:0:0
How do you explain the alternate phenotypes?
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Crossing over
Lets watch the animation Percentage of recombinants
(recombination frequency) is related to the distance between linked genes
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Long story short
A genetic map of loci can be made using percentage of crossing over between linked genes The more often genes cross over, the
farther apart they are The closer two genes are together, the
less they cross over Linkage map=a genetic map based
on recombination frequencies
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Figuring out a linkage map
Three gene loci Body color (b) Wing size (vg) Cinnabar (affects eye color) (cn)
Recombination frequency Between cn and b =9% Between cn and vg = 9.5 % Between b and vg = ~17 %
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So…
cn is about midway between b and vg
Distance between gene is measured in map units 1 map unit is equal to 1%
recombination frequency (centimorgan-in honor of Thomas Morgan)
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Did you notice… The three recombination frequencies in
the example given don’t quite add up 9% (b-cn) + 9.5% (cn-vg)>17% (b-vg) This results from multiple crossing over
events A second crossing over event “cancels
out” the first and reduces the observed number of recombinant offspring
Genes farther apart (b-vg) are more likely to experience multiple crossing over events
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Genes that are very far apart
If genes are far apart on a chromosome, then recombination is almost a certainty
Maximum recombination frequency between genes is 50%
This is indistinguishable from genes on different chromosomes
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In fact..
Two of Mendel’s genes (seed color and flower color) are on the same chromosome but still assort independently
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Misc. Genes located far apart on a
chromosome are mapped by adding the recombination frequencies between the distant genes and intervening genes
A linkage map is an imperfect picture of a chromosome
Map units indicate relative distance and order, NOT precise location Frequency of crossing over is not actually
uniform over the length of a chromosome
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Combined with other methods
Chromosomal banding Cytogenetic maps can be produced Indicate positions of genes with respect
to chromosomal features Recent techniques show physical
distances between gene loci in DNA nucleotides
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Think about it:
Determine the sequence of genes along a chromosome based on the following
recombination frequencies: A –B , 8 map units; A –C , 28 map units; A –
D, 25 map units; B –C, 20 map units; B –D, 33 map units.
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Think about it:
Genes A, B, C, and D are located on the same chromosome. After
calculating recombination frequencies, a student
determines that these genes are separated by the following map
units: C–D: 25 map units; A–B: 12 map units; B–D: 20 map units;
and A–C: 17 map units.
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Think about it:
Genes A, B, and C are located on the same chromosome. Testcrosses show that the recombination frequency between A and B is 28% and between A and C is 12%. Can you determine the linear order of
these genes?