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= 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

Blood Types

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 unitLinked genes

Linked genes don’t produce typical Mendelian results in breeding experimentsWhen 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 andvestigial 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 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?