Genetics Chapters 16-19, 21

Front 1

A mutation in gamete-forming tissue.
An individual with the “new” germinal mutation will not show the phenotype but the mutation can be
transmitted to progeny.

Back 1

Germinal Mutation

Front 2

A mutation in any non-gamete producing tissue.
The individual with the “new” somatic mutation may exhibit a mutant phenotype but the mutation can't be
transmitted to progeny.

Back 2

Somatic Mutation

Front 3

The allele only expresses the mutant phenotype under certain environmental conditions.
(e.g.temperature sensitive)
The protein product functions at the permissive temperature but is non-functional at the
restrictive (non-permissive) temperature.

Back 3

Conditional Mutation

Front 4

The individual must be supplied with certain nutrients (amino acids, nucleotides, vitamins).
Commonly used when studying microorganisms.
2. WT is___________ (nutritionally self-sufficient).

Back 4

1.Auxotrophic Mutation
2.prototrophic

Front 5

Confers the ability to grow in the presence of an inhibitor. (e.g., antibiotic or virus/phage)
Antibiotic resistance mutations can arise from a mutation in the gene encoding the target of the drug or a
mutation in the gene encoding a transporter of the drug.

Back 5

Resistance Mutation

Front 6

Germinal mutations are detected by the sudden appearance of the abnormal phenotype in a pedigree with
no previous record of abnormality.
Dominant mutations are relatively easy to detect.
Recessive mutations can go unnoticed for several generations.
X-linked recessive mutations are easier to detect than autosomal.

Back 6

Human Genetics

Front 7

Experiments designed to isolate mutants that affect a specific biological function.

Back 7

Mutant Hunts

Front 8

Techniques designed to separate rare mutant individuals from WT.
Need a selectable phenotype.
Only the appropriate mutants survive.

Back 8

Genetic Selections

Front 9

Strategies designed to identify desired mutant individuals (mutant phenotype) from a large number of
individuals.

Back 9

Genetic Screens
(Figure 16-4)

Front 10

Used to increase mutation rates.
(e.g., chemicals or UV radiation)

Back 10

Mutagens

Front 11

Analyzing heritable mutant phenotypes at the genetic level before performing molecular analyses of the
isolated mutants.
Requires methods to identify the mutant genes (e.g., gene mapping)

Back 11

Forward Genetics

Front 12

This approach starts with a WT molecule (typically a cloned gene or a purified protein).
The WT gene is then mutated to identify the mutant phenotype.

Back 12

Reverse Genetics

Front 13

1) Grow as single cells in liquid culture or as colonies.
2) Easy to examine millions of individuals.
3) Isolated single cells generate a clonal population (colony) of genetically identical cells.
4) Mutants are easily identified (dominant or recessive).

Back 13

Single Cell Haploid Organisms-Bacteria and Fungi (advantages)

Front 14

1) Grow mutant culture in minimal medium + supplement.
2) Plate cells on minimal medium without supplement.
3) Survivors are prototrophs.
The use of a mutagen in step 1 above will increase the mutation rate.

Back 14

Forward Genetic Selections

Detection of Reverse Mutations (auxotrophprototroph)

Front 15

1. The survivors can be a true reversion (WT genotype).
2. The survivors may be second site suppressors that result in the complete or partial
phenotypic reversion to WT.
a. The second site suppressor can be in a new gene.
Often identifies interacting proteins.
b. The second site suppressor can be in the same gene.
Often identifies a critical contact within the protein that is required for function.

Back 15

Survivors
second site suppreseors

Front 16

Auxotrophic selection in bacteria.
Penicillin kills actively growing cells by interfering with cell wall synthesis.
1) Grow cells in rich medium (with or without a mutagen).
2) Transfer to minimal medium.
3) Add penicillin (prototrophs die, auxotrophs survive).
4) Plate cells on rich medium.
Screen cells on minimal medium + various supplements.
Identifies mutants that now require a specific supplement for growth.

Back 16

Penicillin Enrichment

Front 17

Resistance Mutations
Bacteria
1) Grow cells in liquid culture (with or without a mutagen).
2) Plate cells on selective medium (antibiotic or phage).

Back 17

Resistance Mutations

Front 18

Filter enrichment.
Prototrophs grow as fuzzy balls, auxotrophs do not grow.
1) Grow cells in rich medium (with or without a mutagen).
2) Transfer to minimal medium.
3) Filter cells and save the filtrate containing the auxotrophs.
(The prototrophic fuzzy balls are retained on the filter).
4) Plate cells on rich medium.
Screen cells on minimal medium + various supplements.
Identifies mutants that now require a specific supplement for growth.

Back 18

Selection of Fungal Auxotrophs
(Figure 16-6)

Front 19

1) Grow cells in rich medium (with or without a mutagen).
2) Plate cells and visually screen for abnormal colonies.
Identifies mutants that have defects in hyphal tip growth and branching

Back 19

Morphogenesis Mutants in Fungi
(Figure 16-8)

Front 20

1.Mutant screens of yeast are responsible for much of our understanding about the cell cycle.
2. _____________ were first obtained in a screen looking for mutants that blocked
the mitotic cell cycle at specific points.

3._____________has shown that these same genes function in a similar manner in humans.
Many of these genes are defective in cancers.

Back 20

1.Yeast Cell Cycle
2.cdc mutants(cell-division cycle) (Figure 16-9)
3.Comparitive genomics

Front 21

Nobel Prize for yeast cell cycle

Back 21

Leland Hartwell and Paul Nurse

Front 22

A. Random Mutagenesis
Combine new mutants with a known mutation of the gene of interest.

Back 22

Reverse Genetics

Front 23

1.
Mutagenize the WT parent.
a+a+b+b+ X aabb

a+a b+b a+abb aab+b
mutant 1 mutant 2

Back 23

Diploid organism (germinal)

Front 24

Mutagenize the individual.
Look for sectoring in a heterozygote.

Back 24

Diploid organism (somatic)

Front 25

See notes for exam 3.
Replace the wild type gene with a gene that has been disrupted by a drug resistance gene.

Back 25

Gene-specific Mutagenesis
Gene Replacement (Inactivation)
(Figure 16-15)

Front 26

See notes from exam 3.
Introduce point mutations, deletions or insertions in a gene of interest using mutagenic
oligonucleotides.

Back 26

Gene-specific Mutagenesis
Site-directed Mutagenesis
(Figure 16-16)

Front 27

Use conditions in which PCR exhibits reduced fidelity of a coding region.
Then use a genetic selection or screen to identify interesting mutants.

Back 27

Gene-specific Mutagenesis

Error-prone PCR

Front 28

Applying mutagenic and selective techniques to animal and plant cell cultures.
Often only identify dominant mutations because the organism is diploid.

Back 28

Somatic Cell Genetics

Front 29

Once several mutants are identified in a selection or screen, it is important to identify the mutant genes.
There could be one or several different genes giving rise to a particular phenotype.
(e.g., the inability to grow in the absence of an amino acid)

Back 29

Analysis of Recovered Mutations

Front 30

A. __________
1. Use conjugation and transduction mapping techniques to localize the mutant gene.
2. Clone by complementation.
B.__________
Use the complementation test.

Back 30

Analysis of Recovered Mutation
A.Prokaryotes
B. Eukaryotes

Front 31

Mechanisms exist to ensure that cell numbers remain balanced.
Cancer is a genetic disease of somatic cells that is caused by mutations that result in the failure of cell
cycle control and/or the failure of apoptosis (programmed cell death).

Back 31

Cell Cycle Control and Apoptosis

Front 32

G1 (Gap 1)Time between mitosis and DNA replication.
SDNA synthesis.
G2 (Gap 2) Time between DNA replication and mitosis.
MMitosis
G0Optional "resting" phase.
Eacly embryonic cells (no G0)
Differentiated cells (continous G0)
Stem cells (fluctuate between G0 and the cell division cycle)
Rates of cell division are regulated to ensure sufficient cells to replace dying ones, and to prevent
production of excess cells.

Back 32

The Cell Cycle (G1-S-G2-M)

Front 33

Progression of one stage of the cell cycle to the next depends on protein complexes
consisting of a cyclin and a cyclin-dependent protein kinase (CDK).
Protein kinases phosphorylate specific proteins.
Cyclins only expressed at specific cell cycle stages.

Cyclins tether the CDK to a specific target protein so that the target protein can be phosphorylated,
thereby changing the activity of the target protein.
The timing of gene expression of different cyclins results in phosphorylation of different target proteins at
different times.
Phosphorylation of the target initiates a chain of events leading to the activation of transcription factors
that promote transcription of genes required for the next stage of the cell cycle.
Sequential activation of different CDK-cyclins leads to sequential activation of transcription factors and,
in turn, progression of the cell cycle.
Various checkpoints serve as monitors of the status of DNA replication, spindle apparatus formation, etc.

Back 33

How does the cell “know” when to divide?
(Figure 17-2)
(Figure 17-3)

Front 34

inhibit the kinase activity of the CDK until the cell is ready to progress
into the next stage of the cell cycle.

Back 34

CDK-cyclin-binding proteins

Front 35

E2F is a transcription factor that turns on genes encoding enzymes for DNA synthesis (i.e., replication).
Rb (retinoblastoma) protein binds to and inhibits E2F function.
cyclin A-Cdk2 complex phosphorylates Rb (i.e., Rb is the target protein).
Phosphorylated Rb can't bind to E2F.
p21 is the cyclin A-Cdk2 binding protein, which prevents Rb phosphorylation.
p53 activate p21 expression in the presence of DNA damage.
Thus, enzymes for DNA synthesis are not turned on until the DNA damage is repaired.
The key is the negative regulators that inhibit the kinase activity of the CDK-cyclin complexes.
(e.g., p21 in the example above)

Back 35

G1 to S Transition Checkpoint
(Figures 17-4 and 17-5)

Front 36

Elimination of damaged (potentially harmful) cells through a self-destruct and disposal mechanism.
Activation of the self-destruct mechanism leads to fragmentation of the chromosomes, disruption of
organelle structure, and loss of normal cell shape.
Eventually the cells are fragmented and eaten by scavenger cells.

Back 36

Apoptosis (Programmed Cell Death)
(Figure 17-6)

Front 37

The engines of self-destruction.
A group of enzymes that cleave other proteins (proteases).
In normal cells, each executioner caspase is present in an inactive state called the zymogen form.
The zymogen is converted to an active caspase by proteolytic removal of a portion of the polypeptide.

The active caspase then cleaves its target proteins.

Back 37

Caspases
(Figure 17-7)

Front 38

1. Other zymogens
2. A protein responsible for inactivating a DNA endonuclease.
Leads to activation of the endonuclease and chromosomal fragmentation.
3. Actin (a component of the cytoskeleton)-leads to abnormal cell shape.
4. others
In addition to its role in cell cycle control, p53 is an activator of apoptosis.
Thus, p53 function is critical in controlling cell number and the elimination of abnormal cells.

Back 38

Target proteins may include:

Front 39

Cells communicate with each other via signal transduction pathways.

Back 39

Intercellular Communication

Front 40

A small molecule (ligand) is released from one cell and interacts with a membrane bound receptor of
another cell.

Back 40

Signal Transduction
(Figure 17-11)

Front 41

Cancerous cells are uncoupled from the regulatory mechanisms that keep cell proliferation in check.
Caused by multiple mutations in a single cell that causes it to:
1) Proliferate out of control.
2) Decrease the susceptibility to apoptosis.
3) Increase the general mutation rate of the cell so that proliferation or apoptotic mutation is
more likely to occur.
Some of these mutations are inherited, while others originate in the somatic cell lineage.

Back 41

CANCER

Front 42

Mutations resulting in proteins that are activated when they shouldn’t be.
Typically these proteins are components of intracellular communication pathways such that the cell
always behaves as if it is receiving a signal to proliferate. (e.g., Ras) (Figure 17-16)
Ras (G-protein) mutations result in Ras always being in the GTP bound (active) form. Results in the
continuous propagation of the signal that promotes cell proliferation.

Back 42

Dominant Oncogene Mutations

Front 43

Ligand bound receptors dimerize.
Dimerized receptors autophosphorylate the cytoplasmic domain.
The phosphate is then transferred to another protein in the signaling pathway.
Often the next step in propagating the signal is to activate a G-protein
(Figure 17-12).
G-proteins cycle between GDP bound (inactive) and GTP bound (active) forms.
The active G-protein eventually leads to phosphorylation of transcription factors,
and thereby changes in gene expression.
(Often occurs through phosphorylated intermediates)

Back 43

Signal Transduction
(Figure 17-13)

Front 44

Mutations in genes whose proteins normally contribute to the inhibition of cell proliferation.
A. Proteins involved in inhibiting progression of the cell cycle (i.e., inhibitor protein is inactive).
B. Proteins involved in the repair of DNA damage (e.g., p53).
C. Proteins that promote apoptosis (e.g., p53).
p53 is a DNA-binding regulatory protein that is activated by DNA damage.
p53 prevents progression of the cell cycle until the DNA is repaired.
.
If p53 is non-functional, cell division proceeds in the absence of DNA repair, which leads to the
accumulation of additional mutations.
This increases the chance of mutations in other genes involved in controlling the cell cycle and/or
apoptosis leading to uncontrolled cell growth and cancer.
p53 is non-functional in more than 50% of human cancers!!

Back 44

Recessive Tumor Suppresser Genes (Figure 17-5)

Front 45

Autosomal recessive predisposition to skin cancer caused by a mutation in a gene involved in repairing
UV damaged DNA.

Back 45

Xeroderma Pigmentosum

Front 46

A. Hereditary Predisposition (Family history)
1) Inherited as RB rb (recessive).
2) rb rb is generated by rare mitotic X-overs.
Two eyes are often affected because only one mitotic X-over is required in each retina.
B. Sporadic (No previous family history)
1) Inherited as RB RB
2) RB rb is generated from a new somatic mutation.
3) rb rb is then generated by a rare mitotic X-overs.
Only one eye affected.
Two affected eyes would require two independent mutations (one in each eye).

Back 46

Retinoblastoma (Retinal cancer)
(Figure 17-20)

Front 47

The study of the events that occur during the transfiguration of a single cell (fertilized egg) to an adult
organism that is composed of thousands, millions or trillions of cells organized into tissues and organs.

Back 47

DEVELOPMENTAL BIOLOGY

Front 48

Cells adopt specific fates or the capacity to differentiate into specific types of cells (gradual process).
Periodic decisions are made in each cell lineage to more exactly specify the fates of the daughter cells.
In general, the same basic sets of regulatory proteins govern the major developmental events in many, if
not all, higher animals.
Many highly differentiated organisms can regenerate new organs and tissues.
(e.g., starfish arms, damaged human liver, gecko tails).

Back 48

Cell determination

Front 49

Establishing the body plan.

Back 49

PATTERN FORMATION

Front 50

In addition to their role in determining cell shape, Microtubules and Microfilaments serve as molecular
highways in the cell.
Proteins, and vesicles are transported throughout the cell by molecular motors.
(e.g., kinesin and microtubules)
Proteins are targeted to various geographic regions within a cell because they contain a "molecular
address".
Localization of RNA to geographic regions can occur via interaction with a protein that is already
localized to its address.
The cytoskeleton plays a critical role in early pattern formation of developing animals.

Back 50

Role of the Cytoskeleton in Pattern Formation
(Figure 18-5)

Front 51

(roundworm)
The adult nematode is composed of only a few thousand cells.
The cell lineage has been traced from fertilized egg to adult.
(Figure 18-6)
P granules become restricted to one side of the egg upon fertilization and give rise to the germ line of the
worm (gamete producing cells).
Early cell divisions are asymmetric giving rise to two distinct cells.
The dorsal-ventral (top-bottom) (D/V) and anterior-posterior (head-tail) (A/P) axes of the nematode are
established very early during development.
Cell position is also important because neighboring cells communicate to each other via signal
transduction.
The potential fate of a cell becomes progressively restricted as cell divisions continue.

Back 51

C. elegans

Front 52

Generation of the egg cell.
Stem cellprimary oocyte16 cells, one of which becomes the oocyte itself.
The other 15 cells are nurse cells that dump their cytoplasmic contents into the oocyte.

Back 52

Oogenesis

Front 53

1._______________ (RNA & protein) form at the posterior pole of the oocyte (tethered by the cytoskeleton).

2. Nuclear division without complete cell division forms a________________

3.______________ form at the posterior end, which form the entire germ line of the fly.

The other cells in the syncitium give rise to the soma (all other cell types).

Back 53

1.Polar granules
(Figure 18-7)
2.syncitium.
3.Pole cells

Front 54

The Drosophila larva is highly differentiated along ___________

Back 54

Formation of the Body Plan
(Figure 18-8)

the A/P & D/V axes.

Front 55

~10 hrs after fertilization 14 body segments are formed along the A/P axis.
(3 head, 3 thoracic, 8 abdominal)
Each segment gives rise to body parts of the adult.

Back 55

Segmentation Pattern

Front 56

The egg contributes localized gene products that establish polarity along the A/P & D/V axes, which
ultimately determines cell fates.
Cell fate is determined during development by the selective local activation of a set of master regulatory
proteins due to the concentration gradient of localized determinants (RNA & protein) established in the
egg (Maternal-effect genes).

Back 56

How are segmentation patterns established?

Front 57

Concentration gradients of the maternal-effect proteins are established while the soma nuclei are part of
the syncitium (one common cytoplasm).

Back 57

A/P Concentration Gradients
(Figure 18-9)

Front 58

(A maternal-effect gene)
________ mRNA is localized to the anterior pole of the embryo by association with microtubules.
Following translation, BCD protein diffuses forming a concentration gradient from anterior to posterior.
BCD is a transcription factor that is directed to the nuclei.
Nuclei closest to the anterior pole have the highest BCD concentration

Back 58

BCD (bicoid)

Front 59

(A maternal-effect gene)
______ mRNA is localized to the posterior tip of the embryo by association with microtubules.
Following translation, NOS protein diffuses forming a concentration gradient from posterior to anterior.
NOS protein is a translational repressor of hb-m mRNA.

Back 59

NOS (nanos)

Front 60

(A maternal-effect gene)
________mRNA is uniformly distributed throughout the embryo.
NOS represses translation of hb-m mRNA.
Thus, HB-M is most highly expressed in the anterior pole.
HB-M protein is also a transcription factor and is directed to the nuclei such that the concentration is
highest at the anterior pole.
Similar gradients define the D/V axis.
Gradients of protein products establish polarity (different geographic positions) along the A/P & D/V
axes of the Drosophila embryo.

Back 60

HB-M (hunchback)

Front 61

(different geographic positions) along the A/P & D/V
axes of the Drosophila embryo. which determine cell fate

Back 61

polarity

Front 62

As development continues a hierarchy of gene expression establishes the number of body segments, then
subsegments, then segment identity, etc...

Back 62

Hierarchy of Gene Expression

Front 63

BCD and HB-M are transcriptional regulatory proteins that activate and/or repress a set of genes called
cardinal genes (also called gap genes)

Back 63

Segmentation Pattern

Front 64

1.___________Repressed by high [BCD] but activated by low [BCD] and low [HB-M].
2.
Activated by low [HB-M] and repressed by BCD.
Thus, kni is expressed more posteriorly than kr.
This differential gene expression is caused by BCD and HB-M concentration gradients that was originally
established in the syncitium.
Gap genes encode the next layer of regulatory proteins (transcription factors).
Cellularization of the syncitium at this point traps the transcription factors in the resulting cells at varius
concentrations depending on their geographic location.

Back 64

Gap Genes (Figure 18-23a)
1.kr (kruppel)
2.kni (knirps)(Figure 18-20)

Front 65

1__________regulate expression of the 2___________.
3____________ genes encode transcription factors that regulate expression of 4_________ genes.
Some of the 5.___________genes also encode regulatory proteins while others encode different classes
of proteins.
BCD and HB-M--->gap--->pair-rule--->segment-polarity.

Back 65

1.Gap genes
2.pair-rule genes(Figure 18-23b)
3.Pair-rule
4.segment-polarity
5.segment-polarity

Front 66

Parallel cascades establish ___________

Back 66

Segement Identity

Front 67

Mutations in homeotic genes change the segmental identity into that of another.
(i.e., same number of segments but a duplication of one segment with another segment missing).
All homeotic genes encode transcription factors.
Gap gene proteins activate homeotic genes.
Thus, the number and identity of segments are determined in the early embryo.

The antp mutation (antennapedia) results in legs instead of antenna in the head.
(Chapter 18-cover figure)
Legs are the default.
Antp is required for antenna formation.
Another mutation doubles the number of wings (bithorax).

Wings are the default and the Bithorax complex is required for haltere formation.
(Transparency figures of Drosophila mutants)

Back 67

Homeotic Genes
(Figure 18-26)
(Figure 18-25a-c)

Front 68

Homeotic (segment identity) genes exist in humans and mice, etc... (i.e.) homologous genes
Developmental strategies in animals are ancient.
Animals as divergent as Drosophila and humans develop using the same regulatory switches.

Back 68

Applications to Higher Animals

Front 69

Organisms don't live as isolated individuals; they live in populations.
__________________tries to understand the genetic composition of a population and the forces that
determine and change that composition.
Genetic variation within and between populations arises from the existence of various alleles at different
genetic loci.
Population geneticists want to determine the allele frequency at any given gene locus.
The allele frequency in a population can be changed by mutation, natural selection, migration, nonrandom
mating, and genetic drift.

Back 69

Population genetics

Front 70

Frequency Distribution of a Genotype
(e.g., MN blood types) (Table 19-1)
More typically the allele frequencies are used.
Determine the allele frequency by counting the homozygotes and half the heterozygotes.
A/A = 0.36; A/a = 0.48, a/a = 0.16
Allele frequency of A = 0.36 + 0.24 = 0.60
p + q = 1
If p(A) = 0.60, then q(a) = 1 – 0.60 = 0.40

Back 70

Allele Frequency

Front 71

The occurrence in a population of several phenotypic forms associated with alleles of a particular gene.

Back 71

Polymorphism

Front 72

Can easily see with the naked eye.
Doesn't tell you anything about what is actually changed.

Back 72

Morphologic Variation

Front 73

1.
ABO Blood groups

2. ___________
(e.g., change in number of charged amino acids)
Only detects a few of the total number of changes.
Only examines protein-coding sequences.

Back 73

Protein Polymorphism
(Figures 19-2, 19-3)
1.Immunologic polymorphism
(Table 19-2)
2.Amino acid sequence polymorphism

Front 74

1. ________________
Only detects some of the changes.
Don't know where they are located.
Green sea turtle RFLP study.
2._________________
Detects all of the changes, including regulatory changes.

Back 74

DNA Sequence Polymorphism

1. RFLP Mapping
2. Complete Sequence

Front 75

A/A A/a a/a
p2 + 2 pq + q2 = 1
Random mating results in an equilibrium distribution of genotypes after only one generation.
Sexual reproduction does not cause a reduction in genetic variation.
Genetic variation is maintained.

Back 75

Hardy-Weinberg Equilibrium

Front 76

1 0.3 0.0 0.7
2 0.2 0.2 0.6
3 0.1 0.4 0.5
1 p = f (A/A) 1/2 f (A/a) = 0.3 + 1/2(0) = 0.3 + 0.0 = 0.3
2 p = 0.2 + 1/2 (0.2) = 0.2 + 0.1 = 0.3
3 p = 0.1 + 1/2 (0.4) = 0.1 + 0.2 = 0.3
In each case: q = 1 – p = 0.7
Despite the different genotypic compositions, they all have the same allele frequency.
After one generation of random mating, each of the 3 populations will have the same genotypic
frequency.
A/A A/a a/a
(0.3)2 = 0.09 2(0.3)(0.7) = 0.42 (0.7)2 = 0.49
They will remain identical through each generation unless something perturbs the equilibrium.

Back 76

Hardy-Weinberg Equilibrium
A/A A/a a/a
p2 + 2 pq + q2 = 1

Front 77

The total frequency of heterozygotes for a given gene.
Heterozygosity can be calculated using the Hardy-Weinberg equilibrium (2pq).
Heterozygosity is greatest when several alleles of a gene exist in equal frequency.

Back 77

Heterozygosity
(Figure 19-6)

Front 78

Evolution is dependent on renewed variation.
A.________________
rates are very slow and they take an extremely long time to be fixed into the population.
B. _______________
The generation of recombinants via intrachromosomal recombination (X-overs)
contributes to variation.
Requires preexisting mutations.
C. _______________
Migration of genes into a population from another population.
Requires preexisting mutations.
The ultimate source must be mutation.

Back 78

Sources of Variation
A. Mutation
B. Recombination
C. Immigration

Front 79

Hardy-Weinberg assumes random mating but this is not always the case.
A. ____________
When mating between relatives is more common than what would occur purely by chance.
B. _____________
When mating between relatives is less common than what would occur purely bychance.
C. _______________
A bias towards mating because of a resemblance caused by a particular locus.
D. _________________
Mating with unlike partners.

Inbreeding and positive assortative mating lead to decreased heterozygosity.
Putbreeding and negative assortative mating lead to increased heterozygosity.

Back 79

Non-random Mating
A. Inbreeding
B. Outbreeding
C. Positive Assortative Mating
D. Negative assortative mating

Front 80

Differential rates of survival and reproduction of particular genotypes .

Back 80

SELECTION
(Natural Selection)

Front 81

Relative probability of survival and rate of reproduction of a phenotype or genotype.
Fitness is a consequence of the interaction of the phenotype with its environment.
The fitness of a particular phenotype (genotype) will differ in different environments.

Back 81

Darwinian Fitness

Front 82

The fitness of the individual does not depend on the composition of the population.
No competition.
(e.g., plants in a desert depend on depth of roots to obtain water)

Back 82

Frequency Independent Selection

Front 83

The fitness of the individual depends on the composition of the population.
Competition!
(e.g., carnivores)
The allele with the highest average fitness increases in the population.

Back 83

Frequency Dependent Selection

Front 84

All populations are finite in size.
If two parents only have a small number of offspring, even in the absence of selective forces, the
frequency of a gene will not be exactly reproduced in the next generation (sampling error).
New mutations can be fixed even if they are not favored by natural selection due to random genetic drift.
New favorable mutations can also be lost.

Back 84

Random Genetic Drift

Front 85

Occurs when a small group breaks off from a larger population to found a new colony.
Probably responsible for the virtual lack of blood type B in Native Americans.

Back 85

Founder effect

Front 86

1. Populations of a given species includes individuals with varying characteristics
(i.e., different phenotypes and genotypes)
2. The population of the next generation will contain a higher frequency of those types that are most
successful at surviving and reproducing.
(i.e., natural selection)
3. The frequencies of the various types within the species will change over time.

99.9% of all species that ever existed are extinct.
Yet, the number of species has increased during the past billion years.
Thus, evolution gives rise to new species.

Back 86

Darwin's Theory of Evolution
(Figure 21-2)

Front 87

13 different finch species with variation in form and function.

Evolution occurs within populations and not between individuals.
(i.e., the gene pool)

Back 87

Darwin's Finches in the Galapagos Islands
(Figure 21-4)

Front 88

A group of organisms which are capable of exchanging genes within the group but are genetically unable
to exchange genes in nature with other groups.
New species form as a result of geographic isolation.
(e.g., continental drift, different islands)
Populations that are geographically isolated will diverge from one another genetically as a consequence
of:
1) unique mutations
2) natural selection
3) genetic drift
Migration interferes with evolution.
In the absence of migration, genetic differences between populations become so great that the formation
of hybrids becomes impossible.
These biologically isolated populations are new species.

Back 88

Species

Front 89

1.The failure to form zygotes.

2.The failure of fertilized zygotes contribute gametes to future generations.
(e.g., sterility of the hybrids)
horse + donkey = sterile mule (horses and donkeys are different species)

Back 89

1.Prezygotic Isolation
2.Postzygotic Isolation

Front 90

Evolution consists of more than substitution of one allele for another.
In some cases, continuous transformation leads to new form and function without totally new genes.
(e.g., development of the mammalian inner ear from reptilian jaw bones)
New genes and proteins are necessary in many instances.
(e.g., photosynthesis, hemoglobin, immune system)

Back 90

Origin of New Genes

Front 91

A. ______________
Duplication of the entire genome.
The duplicated genes can diverge and take on altered or new function.
Common occurrence in plants.
When n (haploid number) > 12, most plants have an even number of chromosomes.
B. ______________
Following __________ of a chromosome region,
the ___________ genes can diverge and take on altered or new function.

Back 91

Where does the DNA for new genes come from?

A. Polyploidy (Figure 21-9)
B. Duplications

Front 92

1) _______________
(α2β2) α and β are 50% identical.
α on chromosome 16, β on chromosome 11

2)______________(α2γ2) γ and β are 75% identical.
γ and β are adjacent to one another on chromosome 11.

From embryos to adults, the relative makeup of hemoglobin changes in the order of the genes on
chromosomes 11 and 16.

Back 92

Human Hemoglobin
(Figure 21-10)(Figure 21-11)

1) Adult hemoglobin
2) Fetal hemoglobin

Front 93

1) Mitochondria and chloroplasts
Bacterial engulfmentsymbiosisorganelle
2) Horizontal transfer
Transposable elements and plasmids can transfer DNA from one species to another.

Back 93

Imported DNA

Front 94

Mutations can have three consequences.
1) Decrease fitness
2) Increase fitness
3) No effect on fitness (neutral)

Back 94

Molecular Evolution

Front 95

Evolution of a gene proceeds according to a molecular clock that is dictated by the mutation frequency.
Synonymous substitutions occur at a faster rate than non-synonymous substitutions.

Different proteins have different molecular clocks.

Back 95

Molecular Clock
(Figure 21-13)
(Figure 21-14)

Front 96

in Evolution
All organisms are descended from a single common ancestor.
(Figure 21-15)

Back 96

Common Ancestry