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  It is rarely this simple

Incomplete dominance  Heterozygotes show an intermediate phenotype  RR= red flowers  rr= white flowers  Rr= pink flowers  Blending theory? Why or why not?

Incomplete dominance

Co-dominance  In heterozygotes, both alleles appear phenotypically as individual distinct traits White Black Roan

Co-dominance in blood types  ABO blood groups  3 alleles= I A, I B, and i I A and I B are dominant over I I A and I B are co-dominant to each other Determines the presence of specific oligosaccharides on the surface of red blood cells

Blood Types

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

Blood donation

Pleiotropy  One gene affects more than one phenotypic character  MOST are pleiotropic Wide ranging effects due to a single gene  Dwarfism  Gigantism

Acromegaly: Andre the Giant

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

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

Epistasis

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

Epistasis

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

Nature vs. Nurture  The interaction of genes and environment control phenotype  Hotly contested debate

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

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

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

Fruit Flies  Took Morgan a year of breeding flies to get a mutant Male with white eyes Wild type is red

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

Sex linked genes  Morgan’s evidence of sex linkages added credibility to the chromosome theory of inheritance

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 unit Linked genes  Linked genes don’t produce typical Mendelian results in breeding experiments When linked they are unable to assort independently

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?

Resulting F1 flies were all hybrids presenting the wild type (gray bodies and vestigial wings)

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

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

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

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

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?

Crossing over  Lets watch the animation Lets watch the animation  Percentage of recombinants (recombination frequency) is related to the distance between linked genes

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

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 %

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)

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

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

In fact..  Two of Mendel’s genes (seed color and flower color) are on the same chromosome but still assort independently

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

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

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.

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.

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?