Wednesday, November 11, 2009

Practice Chart for the Citric Acid Cycle

Here is a blank chart to practice on. I have not included various enzyme names, etc. It will be set up the same way as glycolysis for the test.

Friday, November 6, 2009

Study Guide for Test 3

Study Guide for Test 3

Chapter 7

1. Know the ultimate source of energy for almost all living organisms.
2. Know the following terms: energy (both kinetic, potential, and chemical), mass, entropy, enthalpy, catabolism, anabolism, metabolism, dynamic equilibrium, endogonic reactions, and exogonic rections.
3. Know the 1st and 2nd Laws of Thermodynamics and their applications.
4. Know how a cell maintains a high degree of order.
5. Know how endogonic and exogonic reactions are coupled. Know how ATP stores energy.
6. Know how electrons/proton transfer transfers energy in redox reactions. Know the molecules that are involved in energy transfer in redox reactions.
7. Know how enzymes function as catalysts, including induced fit, active sits, allosteric sites, allosteric regulation, cofactors, and apoenzymes.

Chapter 8

8. Know, in detail, glycolysis and citric acid cycle. This includes the molecules involved, the names of the molecules, the names of enzymes involved, and energy molecules involved. Be able to fill in each on a blank chart.
9. For extra credit, know in detail the formation of acteyl CoA from pyruvate.
10. Know in detail the electron transport chain of mitochondria. Know the function and process of chemiosmosis.

Chapter 9

11. Know the order of electromagnetic radation, from shortest wavelength to longest.
12. Know how a photon can energize an electron. Know how chloroplasts use this to gain energized electrons for their electron transport chain.
13. Know the composition of chlorophyll, particularly the inorganic atom at its center.
14. Know the parts of the chloroplast and their functions.
15. Know the parts of the leaf and their functions.
16. Know the overall reaction of photosynthesis and which molecules are reduced and which are oxidized.
17. Know, in detail, the light dependant and carbon-fixation reactions. Know where each reaction takes place. Know the reactants and products of both sets of reactions.
18. Know the difference between C3, C4, and CAM plants. Know how each type of carbon-fixation reaction differs.
19. Know the various trophic groups and examples of each.

Monday, November 2, 2009

Lectures 11 - 13, 18 - 19

Chapter 11
Outline

I. Mendel’s principles of inheritance
A. Gregor Mendel
1. Monk in Austria, published work in 1866, not recognized until 1900.
2. Performed experiments with garden pea plants, results of which became the foundation of genetics.
B. Pea plant crosses
1. Before Mendel, breeders of animals and plants knew about true-breeding stock and hybrids.
a. True-breeding stock produce off-spring that are the same, and so on
All have the same phenotype (physical appearance of an organism)
Genotype - genetic composition of an organisms - all genes may not be expressed
b. Hybrids are the result of a cross of true-breeding stock. Hybrids are similar but their off-spring are different, with mixed traits.
2. Other researchers had picked models that were not good for beginning:
a. Humans and calico cats – too complicated, generations are too long.
3. Peas are simpler -
a. Easy to grow, easy to control pollenation/breeding, several varieties commercially available, short generations.
b. Mendel selected plants with seven characteristics, two variations only.
c. Alleles - alternative forms of a gene - informational unit.
d. Flower color - white & purple; seed color – yellow & green; seed shape – smooth & wrinkled; stem height – tall & short, etc.
4. Mendel purchased seeds that were true-breeding
a. Raised them for several generations to be sure.
b. Called the P or parental generation.
5. Performed various crosses of the P generation.
a. F1 generation - first cross (1st filial [Latin, sons & daughters] generation).
b. F2 generation - cross of the F1 plants.
6. Previous thought had been that inheritance involved blending of traits.
a. Breed a tall plant with a short plant and you will get a medium-sized plant.
b. Mendel’s pea plants showed a different result - F1 plants all tall, F2 plants, 2/3 tall, 1/3 short.
II. Prediction of monohybrid crosses
A. Monohybrid crosses
1. Punnett square
2. Tall (T) & short (t) stems
3. Purple (P) & white (p) flowers
4. 3:1 ratio
5. Terms
a. Monohybrid cross - involves individuals with different alleles of a given locus
b. Locus - designates the position of a gene on the chromosome (Fig 11-5 on page 239).
c. Dominate - factor/gene expressed in F1
d. Recessive - factor/gene masked by the dominate gene when both are present.
e. Homozygous - have the same allele on homologous chromosomes (either dominate or recessive).
f. Heterozygous - have different alleles on homologous chromosomes
B. Principles of segregation and independent assortment
1. Principle of segregation - before sexual reproduction occurs, the two alleles by an individual parent must become separated (that is segregated).
a. During meiosis, homologous chromosomes (and genes on them) separate.
b. Alleles go not mix or destroy each other.
c. Recessive genes are not lost can reappear in later generations
2. Principle of independent assortment - members of any gene pair segregate from one another independently of the members of the other gene pairs.
a. Each gamete contains one allele for each locus.
b. Alleles of different loci are assorted at random with respect to each other in the gametes.
B. Test crosses
1. Guinea pigs (Fig 11-7) - want to know if a black guinea pig is homozygous or heterozygous
2. Breed with a homozygous recessive guinea pig and check the results.
3. ½ black and ½ brown - heterozygous
4. All black - probably homozygous
Why probably? Because of heterozygous have an equal change of producing gametes carrying the black or brown allele - by change, all the off-spring may have only received the dominate gene.
Example of black haired, brown-eyed Italian father; blonde, blue-eyed mother with two blonde, blue-eyed children.
C. Dihybrid crosses
1. Guinea pigs (Figure 11-8 on page 242).
2. Black, short-haired dominate; brown, long-haired recessive
3. F1 - all black, short-haired
4. F2 - ratio 9:3:3:1
Hair color and hair length separate independently
9/16 will be black, short-haired
3/16 will be black, long-haired
3/16 will be brown, short-haired
1/16 will be brown, long-haired
5. Blood type example - ABO and Rh factor
III. Gene linkage
A. Work with the fruit fly, Drosophila melanogaster.
1. Did not follow classic Mendelian genetics
2. Figure 11-11 on page 246
3. Of 2300 off-spring, expected 1/4 of off-spring to be from each independent assortment phenotype.
4. In the actual results, 83% belonged to each of the two parental classes, and 17% belonged to the two recombinant classes.
5. Determined that wing shape and body color are linked, or on the same chromosome.
6. Because they are on the same chromosome, they do not assort independently.
B. Frequency of crossing-over
1. If on the same chromosome, how do we get any recombinant phenotypes?
2. Crossing-over during meiosis
Homologous chromosomes line up - synapsis.
Enzymes break the chromosomes and reattach the pieces on the homologous chromosome.
3. Use frequency of crossing-over to determine linear order of linked genes.
Use data to determine percentage of crossing-over.
Divide number of individuals in the two recombinant classes of offspring by total number of offspring and multiply by 100
Fruit fly example: (391/2300)*100 = 17%
V locus and B locus have 17% recombination between them.
Crossing-over more likely to occur if loci are further apart on the chromosome.
Map units - 1% recombination between two loci equals a distance of 1 map unit apart.
V and B loci are 17 map units apart.
4. Fig 11-13 on page 248 - Gene mapping
IV. Sex chromosomes
A. Inheritance of sex
1. Sex determined by different factors in different species.
2. Alligators - temperature
3. Most species, genes are the most important factor.
4. Some species, specific sex chromosomes - rest called autosomes.
a. In mammals, the Y chromosome determines male sex.
b. Several genes on Y chromosome contribute to male developement.
c. Sex reversal on Y (SRY) gen is the major male-determining gene on the Y chromosome, acts as a ‘genetic switch’ that causes testes to develop in the fetus.
d. Testes produce testosterone, which causes other male characteristics to develop.
e. Some genes on X chromosome, some on autosomes.
f. Evidence suggests that X & Y chromosomes were originally a homologous pair, but the Y chromosome lost most of its genetic material. Functional genes preserved on the X chromosome.
B. X-linked genes
1. Because X & Y can’t line up for synapsis, cross-over is very rare/nonexistent.
2. Genes located on the X-chromosome have unusual heiritance patterns.
3. Used to be called ‘sex-linked genes’, because they are on a sex chromosome, but now called ‘X-linked genes’ which is more appropriate. Genes on X-chromosome are not necessarily linked to sex/sex development.
a. Color perception
b. Blood clotting
4. Figure 11-16 on page 250 - X-linked red-green color blindness
a. Normal father with carrier mother - half sons will be red-green color blind and half daughters will be carriers.
b. Color-blind father with carrier mother - half children (male and female) will be color blind and all normal daughters will be carriers.
C. Dosage compensation
1. Female mammals have potentially double-dose of X-genes from both homologous X-chromosomes.
2. Dosage compensation is a mechanism that makes equivalent the two doses in the female and the single dose in the male.
3. Dosage compensation in humans and other mammals involves the random inactivation of one/most of one of the X-chromosomes.
a. During interphase, a dark spot of chromatin is visible at the edge of the nucleus of each female mammalian cell when stained and observed under the microscope.
b. Barr body - dense metabolically inactive X chromosome.
c. 25% of the genes are expressed to some degree.
4. Calico cats - X-linked genes for black and yellow/orange fur (white different gene.
5. Sweat gland expression in humans.
6. Color-blindness - patches of color-blind cells in the retina, normal patches make up the difference.
V. Extensions of Mendelian Genetics
A. Pleiotropy - multiple affects from one gene.
1. Homozygous for recessive allele that causes cystic fibrosis produce abnormally thick mucus in many areas of the body, including respiratory, digestive, and reproductive systems.
a. Two CF parents will have CF children.
b. One CF parent and a normal parent will have all carrier children, unless the normal parent is a carrier, then 50% CF, 25% carrier, 25% non-carrier.
2. Dwarfism - dominate allele for abnormal growth of bones
a. Achondroplaisa - normal sized truck, short limbs, and slightly enlarged head.
b. 70% of all dwarfism
c. May have problems with apnea (central & obstructive) and hydrocephalus
d. Two achondroplaisic parents have 25% chance of a non-dwarf child, 50% of a dwarf child, and 25% of a double-dominate child (fatal)
B. Dominance is not always complete
1. Incomplete dominance - instances in which the heterozygote is intermediate in phenotype
a. Four o’clocks - red + white = pink in F1
b. Not ‘blending’ - red and white show up in F2, genes not ‘lost’
c. Red pigment ‘dosage’ dependant - two genes produce red, one gene produces pink, white in gene that doesn’t code for pigment.
2. Codominance - instances in which the heterozygote simultaneously expresses the pheotypes of both types of homozygotes
a. ABO blood groups
C. Multiple alleles
1. Some genes have multiple alleles
2. Rabbit coat color - sequential dominance
a. Fig 11- 19 on page 254
b. C>cch>ch>C
c. Predict cross between different rabbit coat colors
3. Some alleles of different loci may interact to produce a phenotype.
a. Chickens - two genes on different chromosomes (unlinked).
b. Rose comb [R] vs single comb [r]
c. Pea comb (P) vs single comb (p)
d. Fig 11-20 on page 254
4. Epistasis (standing on) - common type of gene interaction in which the presence of certain alleles of one locus can prevent or mask the expression of alleles of a different locus and express their own phenotype instead.
a. No novel phenotypes are produced.
b. Labrador retrievers coat color - gene for pigment and gene for depositing color in the coat.
c. Black coat (B) vs brown coat (b)
d. Expression of color/depositing color (E) vs blocking color (e, eistatic).
e. e is recessive and blocks expression of B/b.
f. Fig 11-12 on page 255.
5. Polygenic inheritance - when multiple independent pairs of genes have similar and additive effects on the same character.
a. Human skin pigment - as many as 60 loci found so far.
b. Incompletely dominate - more capital letters, darker the skin.
c. For simplicity, limit to three independent loci, A/a, B/b, C/c
d. Fig 11-22 on page 256.


Chapter 12
Outline
DNA Replication
I. General structure of DNA molecule
A. Contributors and ‘pieces’ discovered
1. Erwin Chargaff - Chargaff’s rules.
a. Reports relationships among DNA bases that provide a clue to the structure of DNA.
b. Ratios of purines (adenine, guanine) to pyrimidines (thymine, cytosine) and also adenine to thymine and guanine to cytosine were not far from 1.
2. Franklin & Wilkins
a. Rosalind Franklin - performed the X-ray diffraction on DNA crystals that showed DNA has a helical structure.
b. Maurice Wilkins was the head of the lab, received the Nobel Prize
3. Watson and Crick
a. Used information from other researchers to put together a model that reflected the data.
b. Used Franklin’s X-ray data for a helical/double helical structure and distance between the nucleotide bases and turns of the helix.
c. Used Chargaff’s data of the 1:1 ration of adenine:thymine, guanine:cystosine to determine that pairing keeps the distance between the two helixes the same. H-bonding is favored, too.
d. Determined the two helixes would need to be antiparallel to each other.
3. Meselson and Stahl
a. Demonstrated semiconservative replication.
b. Fig 12-7, page 268.
c. 15N labeled DNA in E. coli. 15N labeled DNA heavier than normal 14N DNA.
d. Isolated DNA by density gradient centrifugation for a heavy base reading.
e. Grew E. coli in normal media for one generation (20 minutes).
f. Isolated DNA - intermediate density of all 14N labeled DNA - supported semiconservative (one helix is the template for new strand) and dispersive (parental and new strands are randomly mixed during replication) replication. Conservative (both parent strands remain together, as would the new strands) would have had two bands.
g. Second generation had two bands of DNA, one intermediate and one light - supported semiconservative. Dispersive would have had one band again, but less dense.
B. Final picture
1. Double helix
2. Pairing of adenine and thymine; guanine and cystosine.
3. Replication by semiconservative replication.
II. DNA replication
A. General Process
1. Helix ‘unzipped’.
2. New strand formed by pairing with nucleotide bases.
3. Duplicated DNA has one parental strand and one new strand - two helixes total.
B. Detailed Process (Fig 12-11, page 272; Fig. 12-12, page 273)
1. Major enzyme groups (Table 12-3, page 270)
2. Helicases ‘unzip’ the helix by breaking H-bonds between nucleotide bases.
3. Single-strand binding (SSB) proteins bind to single DNA to:
a. Prevent strands from reannealing
b. Prevent hydrolysis of strands by nucleases
4. Topoisomerases produce breaks in the DNA to unwind the helix - relieve supercoiling produced by the helicases.
5. DNA primase
a. Produce a RNA primer (5 - 14 nucleotides) for the start of replication.
b. Replication proceeds from 5' to 3' of an existing polynucleotide strand.
6. DNA polymerases
a. Add nucleotides to 3' of the growing strand.
b. Nucleotides with three phosphate groups brought in.
c. Nucleotide paired with base of the template strand.
d. As nucleotide is linked, two phosphates leave - strongly exergonic reaction.
e. Polynucleotide chain is elongated by linkage of the 5' phosphate group to the 3' hydroxyl group of the sugar at the end of the existing strand - always synthesis from 5' to 3'.
7. Leading strand vs lagging strand
a. Replication happening on both sides of the replication fork.
b. One strand is replicated continuously (5' - 3' direction toward replication fork) and is called the leading strand.
c. Other strand is replicated discontinuously (5' - 3' direction away from replication fork) and is called the lagging strand.
d. Lagging strand is replicated in ‘pieces’ called Okazaki fragments.
e. Okazaki fragments replicated until come to RNA primer, DNA polymerase degrades and replaces the RNA primer with DNA
f. Starts again further up the strand toward the replication fork.
8. DNA ligase
a. Joins the DNA fragments by linking the 3' hydroxyl group with the 5' phosphate fo the DNA next to it, forming a phosphodiester linkage.
b. Also joins breaks in non-replicating DNA.
9. Number of replication forks
a. Multiple replication forks in eukaryotic cells - replication precedes faster.
b. Prokaryotic cells have circular DNA strands - one origin of replication, fewer bases.
c. Either way, forks meet and merge.
10. Telomerases cap eukaryotic chromosome ends.
a. Because eukaryotic chromosomes are linear, some DNA is lost on the end.
b. Not a problem because ends are composed of telomeres, stretches of short, simple, noncoding DNA sequences the repeat many times.
c. Human gametes 5'—TTAGGG—3'
d. Cell can divide many times before start losing essential genetic information.
e. Telomerases lengthen telomeric DNA by adding repetitive nucleotide sequences to the ends of eukaryotic chromosomes.
Active in rapidly dividing cells - germ cell lines, blood cells, skin cells, etc.
Telomere shortening implicated in cell aging and apoptosis - some cells only divide a certain number of times then die.
Cancer cells - telomeres shorten to critical lengths, telomerases kick in rather than the cells dying.
Balance between controlled immortality and cancer.



Chapter 13
Outline
Gene Expression
I. Overview of RNA
A. Difference between RNA & DNA
1. Draw structures
2. Ribose vs deoxyribose - OH group on the 2' carbon atom
3. Uracil vs thymine
4. Both joined 5' to 3'
5. Both have purines H-bonding with prymidines
B. Three Main Types of RNA
1. mRNA - single strand of RNA that carries the information for making a protein.
2. tRNA
a. Single strand of RNA that folds bac on itself to form a specific shape.
b. Each kind of tRNA bonds with only one kind of amino acid and carries it to the ribosome.
c. Because there are more kinds of tRNA molecules than there are amino acids, many amino acids are carried by two or more kinds of tRNA molecules.
3. rRNA
a. Is in a globular form.
b. Is an important part of the structure of ribosomes.
c. Has catalytic functions needed during protein synthesis.
d. rRNA made in nucleolar organizer in the nucleolus. Proteins needed for ribosomes synthesized in the cytoplasm, imported into the nucleolus, where the ribosomes are assembled.
4. Rest of the major groups are in Table 13-1, page 296. We’ll cover some of them as we go along.
II. From DNA to proteins
A. DNA genetic code
1. Triplet code
a. DNA has four bases.
b. If each base coded for one amino acid, it could only could for four amino acids.
c. Twenty amino acids commonly found in cells.
d. If each base served as a letter in a four-letter "alphabet", could code for more amino acids.
e. If each ‘word’ consisted of three ‘letters’, could ‘spell’ 64 ‘words’, more than enough to code for all the naturally occurring amino acids.
2. Codons
a. Each three base ‘word’ is called a codon.
b. Figure 13-5 on page 284.
c. More than one codon specifies certain amino acids - the code is redundant - ‘wobble’ at 3' end of the triplet.
d. Code is nearly universal - same in bacteria, yeast, and humans.
e. Some codons do not specify an amino acid, but ‘punctuation’, such as ‘start’ and ‘stop.’
3. Using the chart, ‘spell’ out examples of proteins.
a. Example one
Non-template DNA strand
5' - ATGTTTGGTGGTTTA - 5'
Template DNA strand
3' - TACAAAGGTCCAAAT - 5'
mRNA strand
5' - AUGUUUCCAGGUUUA - 3'
Codons indicated
AUG/UUU/CCA/GGU/UUA
Find codons on the chart for amino acids/instructions coded
Met (start)/Phe/Pro/Gly/Stop
b. Example two
Non-Template DNA strand
5' - ATGGTGTCACGAUGA - 3'
Template DNA strand
3' - TACCTCTGTGGTACT - 5'
mRNA strand
5' - AUGGUGUCACGAUGA - 3'
Codons indicated
AUG/GUG/UCA/CGA/UGA
Have students find codons on the chart
Met (start)/Val/Ser/Arg/Stop
c. Example three
Non-Template DNA strand
5' - ATGGGGCAGAGCGTGATTTAA - 3'
Template DNA strand
3' - TACCCCGTCTCGCACTAAATT - 5'
mRNA strand
5' - AUGGGGCAGAGCGUGAUUUAA - 3'
Codons indicated
AUG/GGG/CAG/AGC/GUG/AUU/UAA
Find codons on the chart
Met (start)/Gly/Gln/Ser/Val/Ile/Stop
B. Transcription
1. Unwinding of the DNA molecule from the histones.
a. Like in replication, must unwind and ‘unzip’ to be ‘seen.’
b. Part of gene regulation is modifying the DNA molecule so it can be exposed or not.
2. DNA-dependent RNA polymerases
a. Types of RNA polymerases
RNA polymerase I catalyzes the synthesis of several kinds of rRNA molecules that are components of the ribosome.
RNA polymerase II catalyzes the production of the protein-coding mRNA
RNA polymerase III catalyzes the synthesis of tRNA and one of the rRNA molecules.
b. Behavior
Require DNA as a template (‘DNA-dependent’ portion of the name).
Carry out synthesis in the 5' 6 3' direction - begin at the 5' end of the RNA molecule being synthesized then continue to add nucleotides at the 3' end until the molecule is complete.
Use nucleotides with three phosphate groups (like ATP, GTP), two phosphate groups come off after polymerization.
Doesn’t require a RNA primer.
3. Synthesis of mRNA
a. Initiation
Promoter - nucleotide sequence in DNA to which RNA polymerase and associated proteins initially bind.
Promoter is not transcribed (no mRNA made from it) – the RNA polymerase moves past it before beginning transcription.
Promoters slightly different for each gene - another method of control.
Promoters upstream (toward 5' end of the mRNA sequence, toward the 3' end of the template DNA) of the point where transcription will begin.
Once RNA polymerase recognizes the promoter, unwinds DNA molecule and initiates transcription.
b. Elongation
First RNA nucleotide retains its triphosphate group on the 5' end.
During elongation, each RNA nucleotide added to the 3' end loses two of the phosphate groups in an exergonic reaction.
Remaining phosphate group is incorporated into the phosphate sugar backbone.
Terminal 3' end has an exposed OH group.
c. Termination
Elongation continues until the RNA polymerase recognizes a termination squence.
This signal leads to a separation of RNA polymerase from the template DNA and the newly synthesized RNA.
Prokaryotes - termination signal causes the RNA polymerase to separate from the DNA template strand and the newly synthesized mRNA strand.
Eukaryotes - after termination signal, RNA polymerase adds nucleotides (about 10 - 35) before seperating.
4. Use examples from II.A. for transcription.
5. Mutations (Figure 13 - 20, page 299)
a. Base substitution
Involves a change in only one pair of nucleotides.
Usually result from errors in base pairing during replication.
b. Missense
Base substitutions that result in replacement of one amino acid by another.
Missense mutations have a wide range of effects.
Occurs at or near enzyme active site with a replacement by a very different amino acid (acid replacing a basic amino acid) - enzyme activity may be decreased or completely inactivated.
Replacement of a very different amino acid occurs further away from active site - enzyme may be unaffected, if folding is unaffected.
Replacement with a very similar amino acid (acid for acid), may not change anything - a silent mutation.
c. Nonsense
Base substitutions that covert an amino acid-specifying codon to a stop codon.

Chapter 18
Outline
Introduction of Darwinian Evolution
I. Pre-Darwinian Ideas About Evolution
A. Aristotle - movement toward ‘perfection.’
B. Jean Baptiste de Lamarck
1. Giraffe example - each generation stretches to reach higher leaves, passes traits onto offspring.
2. Problems with this:
a. Human example - tribe in ? where long necks in women are considered beautiful.
b. Rings put around neck of girls, slowly stretching the necks.
c. When this went out of fashion, women didn’t have any longer necks than before.
d. Can pass on the genes for long necks, can’t pass on traits gained from exercise or other modification of the body.
II. Ideas That Influenced Darwin
A. Combined previous ideas with this own observations made on the 5 year journey of the HMS Beagle.
B. Principles of Geology by Charles Lyell
1. Mountains, valleys, and other physical features of Earth’s surface did not originate in their present forms.
2. Developed slowly over long periods of time.
3. Earth is extremely old.
C. Artificial selection
1. Breeders can cause extreme changes in a species in just a few generations by selecting for a desired trait and breeding (artificial selection of mates) for it.
2. Dog example - find some good pictures on the internet.
3. Broccoli example:
a. All one species, Brassica oleracea.
b. Selectively breeding the wild cabbage (colewort) produced seven vegetables:
Enlarged terminal bud - cabbage
Enlarged, eatable flowers - broccoli and cauliflower
Enlarged axillary buds - brussels sprouts
Enlarged stems - kohlrabi
Eatable leaves - collard and kale
D. Thomas Malthus, a British clergyman and economist
1. Human population growth can be geometrical (2 64 68).
2. Food supply only grows arithmetically (1 62 63, or, you can’t eat your seed corn).
3. Competition for limited resources causes war, famine, disease, etc., all of which puts a break on population growth.
4. Strong and constant check on human population growth.
III. Darwinian Evolution
A. Terms
1. Evolution - beginning definition is the accumulation of inherited changes within populations over time.
2. Population - group of individuals of one species that live in the same geographic areas at the same time.
3. Species - a group of organisms, with similar structure, function, and behavior that are capable of interbreeding with one another.
a. Working definition.
b. Some arguments over what ‘species’ means, especially when defining an endangered or threatened species.
c. The definition given is good enough for the present.
4. Natural selection - better-adapted organisms are more likely to survive and become the parents of the next generation.
a. Dog example
b. Release the average chichaia into the Provo ‘wilds,’ meaning a backyard, and it will survive anywhere from ten minutes to three months, depending upon the weather. Would do OK in a warmer climate such as California. This dog breed is the result of artificial selection.
c. Release the average non-working breed of dog into the mountains of the Washach Front and, The Incredible Journey not withstanding, it’s going to starve.
d. This will be the case for most domesticated animals and plants - product of artificial selection - traits selected by humans to meet needs of humans, not the needs of the organisms for the environment.
e. Some wild species, on the other hand, are doing quite well in their changing environment - deer, coyotes, domesticated mice and rats, roaches, all do well in contact with humans.
f. Wild species that do not do well in contact with humans are becoming extinct.
B. Darwin proposed that evolution occurs by natural selection.
1. Variation
a. Individuals in a population exhibit variation.
b. Each individual has a unique combination of traits, such as color, size, ability to resist infection, etc.
c. Some traits either improve an individual’s chances of survival or do not.
d. Variation necessary for evolution by natural selection must be inherited.
2. Overproduction
a. The reproductive ability of each species has the potential to cause the population to geometrically increase over time.
b. More offspring are produced than typically survive to reproduce.
c. Female frog lays ~10,000 eggs, two of which will hatch and survive to reproductive age.
3. Limits on population growth, or a struggle for existence.
a. There is only so much food, water, light, growing space and other resources available to a population.
b. Organisms compete with one another for these limited resources.
c. More individuals than the environment can support, not all survive to reproduce.
d. Other limits on population growth are predators, disease organisms, and unfavorable weather conditions.
4. Differential reproductive success
a. Those individuals that have the most favorable combination of characteristics (those that make individuals better adapted to their environment) are more likely to survive and reproduce.
b. Offspring tend to resemble their parents - inherited traits.
c. Successful reproduction is key - best-adapted individuals produce more offspring, less-adapted individuals die prematurely or produce fewer or inferior offspring.
C. Modern synthesis/synthetic theory of evolution
1. Darwin didn’t know how traits were inherited or why there was variation in a population.
2. Once genetics introduced, able to explain.
3. Combined the principles of Mendelian inheritance with Darwin’s theory of natural selection to form a unified explanation of evolution known as the modern synthesis or the synthetic theory of evolution.
IV. Evidence Supporting Darwinian Evolution
A. Fossil record
1. Remains of life preserved in sedimentary rock
2. Remains also found in bogs, tar, amber, and ice – anything that will slow decay and protect from weathering.
3. Fossils show a progression from unicellular organisms to multicellular organisms, demonstrating that life evolved through time.
a. Able to determine what’s older.
b. Younger layers of sediment usually on top.
c. Index fossils - lots of them in only one age of rock.
d. Radioactive decay
Potassium-40 has a half-life of 1.3 billion years and is used to date rock - decays in to argon-40 - magma solidifies with potassium-40 - gas escapes liquid rock but is trapped in solid rock
Uranium-235 has a half-life of 704 million years and is used to date rock
Carbon-14 has a half-life of 5730 years - used to date the carbon remains of anything that was once living.
Try to use more than one radioisotope for dating - more accurate.
4. Fossils form only under select conditions that slow or prevent the decay process:
a. Buried in a reduced oxygen medium.
b. Remains covered quickly by a sediment of fine soil particles suspended in water - bogs, mudflats, sandbars, deltas, floodplains.
c. Remains covered quickly by windblown sand - deserts.
d. Trapped in tree sap - forested areas.
e. Volcanic eruptions - volcanic ash kills and buries all in one swipe.
f. Quick frozen - arctic areas - relatively recent, because freeze-thaw is a good way to destroy a specimen.
5. Fossil record not a random sample of past life.
a. Biased toward aquatic organisms.
b. Biased toward terrestrial organisms living in few areas conducive to fossil formation - river plains, by lakes, volcanic areas, etc.
c. Biased toward organisms with hard body parts, such as bones and shells.
d. Very few fossils found from tropical forest areas - decay quickly on the forest floor from the heat, moisture, and many scavengers.
e. Very few fossils found of organisms will all soft body parts, such as bacteria/unicellular organisms, worms, sharks (only teeth found).
f. Some rocks of certain ages are more accessible to paleontologists than are rocks of other ages - what’s eroded and how much (too little and it’s not exposed, too much as it’s gone), what’s been destroyed by volcanism, etc.
6. Book’s example of whale evolution - excited about because it’s a fairly recent discovery.
B. Biogeography
1. Biogeography - study of the past and present geographic distribution of organisms.
2. The geographic distribution of organisms affects their evolution.
a. Species on ocean islands tend to resemble species of the nearest mainland, even if the environment is different - migration from the nearest mainland.
b. Species on ocean islands do not tend to resemble species on islands with similar environments in other parts of the world - ancestors of the island species separated by distance.
c. Australia has been a separate landmass for millions of years and has distinctive organisms - isolation prevented later placental mammals to compete with monotremes and marsupials.
d. Continental drift - all landmasses were once connected in a supercontinent called Pangaea, which subsequently broke up.
e. Plate tectonics - movement of the crustal plates.
f. Species divergence based on how long ago landmasses separated - old world vs new world monkeys - S. America and Africa both have monkeys (more recently joined than N. America and Eurasia) but they are noticeably different (perihensil tails).
C. Comparative biology
1. Before molecular biology, organisms were grouped according to similarities of anatomy/structure, reproduction, feathers vs scales, etc.
a. Mammals grouped together because they all have hair/fur and feed their young with milk - includes placental, marsupial, and egg-laying species.
b. Birds grouped together because they have feathers, lay eggs, and have hollow bones - includes flightless varieties.
2. Similar features, such as milk production or feathers or forelimbs, indicates a common ancestor.
3. Homologous features - structures derived from the same structure in a common ancestor - condition known as homology.
a. Forelimb in mammals.
b. Leaves in plants - spine in the fishhook cactus and the tendrils of the garden pea are all modified leaves.
4. Convergent evolution - independent evolution of similar structures in distantly related organisms.
a. Wings of birds and bats - because flight requires the same aerodynamics, the wings of different, unrelated species look similar
b. Often the entire animal looks similar if it occupies a similar environmental niche
Heads of mule deer and kangaroos - eyes on sides, ears on top, narrow snout
Heads of T. rex or raptors and wolves - eyes forward, big teeth, prominent nostrils, long limbs
Hind limbs of raptors and emus
c. Homoplastic feature/homoplasy - structurally similar features that are evolved independently in distantly related organisms by convergent evolution.
d. Homoplasy - similarities in different species that are independently acquired.
5. Vestigial structures
a. Organs or parts of organs that are seemingly nonfunctional and degenerate, often undersized or lacking some essential part.
b. Remnants of more developed structures that were present and functional in ancestral organisms.
c. In humans, more than 100 structures are considered vestigial: coccyx (fused tailbones), third molars, muscles that move our ears.
d. Whales and pythons have vestigial hind-limb bones.
e. As structure becomes less essential for survival, can ‘put up with’ mutations cause reduced structure in that limb. Some speculate that lose of one structure can lead to gain of another structure - humans ‘exchanging’ tails and fur for larger brains.
D. Developmental biology
1. Mutations in genes that regulate the orderly sequence of events that occurs during development.
2. Example of snakes - elongation and loss of limbs are from several mutations in the Hox genes that affect expression of body patterns and limb formation.
3. Example of bird beaks - a gene involved with craniofacial skeleton development, BMP4 - when turned on early, the beak is heavier and thicker.
4. Example of vertebrate embryos
All go through a fish stage where it is hard to tell one species from another.
Evolution is a conservative process, building on what is available instead of building from scratch.
Evolution of new features often does not require the evolution of new developmental genes but instead depends on a modification in developmental genes that already exist.
All terrestrial vertebrates are thought to have evolved from fish-like ancestors; therefore, they share some of the early stages of development still found in fishes tody.
E. Comparative molecular biology
1. Genetic code is virtually universal.
2. Proteins and DNA contain a record of evolutionary change.
a. Sequence studies of protein and DNA agree with similarities in structure among living organisms and on fossil data of extinct organisms.
b. DNA sequencing to determine the order of nucleotides - divergence also indicates relatedness.
c. Cytochrome c
A part of the electron transport chain.
Not all amino acids that confer the structural and functional features of cytochrome c are free to change - few spots for silent mutations.
Silent mutations that do arise are indicators of how long two species have diverged - sort of a molecular clock.
3. Sometimes the fossil record and molecular clocks do not agree - whale relatedness
a. Fossil record indicates mesonychains as ancestors
b. Molecular phylogenetic tree indicates artiodactyls (even-toed hoofed mammals) with hippos being the most closely related
c. Mesonychains are not ancient artiodactyls
F. Experimentation
1. Guppy experiments in Venezuela and Trinidad
2. Guppies from high-predation environments mature earlier and are smaller than guppies from low-predation environments.
3. Guppies from low-predation environments grow larger and reach sexual maturity later and produce fewer, larger offspring.
a. High-predation - different kinds and numbers of fishes that prey on guppies.
b. Low-predations - only one kind of predatory fish that occasionally preys on small guppies.
4. Pools separated by waterfalls - populations can’t move upstream from waterfalls.
5. Researchers moved guppies from high-predation pools to low-predation, guppy-less pools and monitored the generations of guppies.
6. 11 years later, the decedents of the high-predation guppies resembled guppies that had always been in low-predation pools.
7. If were me, I’d then introduce lots of predators into these pools to see if the decedents switch back.
8. Other experiments, intentional or otherwise confirm.
a. Antibiotic resistance in bacteria.
b. Pesticide resistance in agricultural pests.
c. Super mice resistant to hemorrhagic and neurological poisons.


Chapter 19
Outline
Evolutionary Change in Populations
I. Population Genetics
A. Terms
1. Population - all the individuals of the same species that live in a particular place at the same time.
2. Gene pool - all the alleles for all the loci present in the population.
a. Genetic variation of the individuals of a population indicates that each individual has a different subset of the alleles in the gene pool.
3. Genotype frequency
a. Population of a particular genotype in a population.
b. 1000 individuals sampled
Genotype Number Genotype Frequency
AA 490 0.49
Aa 420 0.42
aa 90 0.09
Total 1000 1.00
4. Phenotype frequency
a. Proportion of a particular phenotype in the population.
b. 1000 individuals sampled
Phenotype Number Phenotype Frequency
Dominate 910 0.91
Recessive 90 0.09
Total 1000 1.00
c. Dominate phenotype is sum of two genotypes, AA & Aa.
5. Allele frequency
a. Proportion of a particular allele in a population.
b. Each individual is diploid, so has two alleles.
Allele Number Allele Frequency
A 1400 0.7
a 600 0.3
Total 2000 1.0
B. Hardy-Weinberg principle
1. Frequencies of alleles and genotypes do not change from generation to generation unless influenced by outside factors.
2. Genetic equilibrium - no net change in allele or genotype frequencies over time - is not undergoing evolutionary change.
3. Hardy-Weinberg principle of genetic equilibrium
a. If population is large, the process of inheritance does not by itself cause changes in allele frequencies.
b. This explains why dominate genes are not more frequent than recessive.
c. Hardy-Weinberg uses phenotype frequencies to calculate the expected genotype frequencies and allele frequencies, assuming a clear understanding of the genetic basis for the character under study.
p2 + 2pq + q2 = 1
Frequency of AA Frequency of Aa Frequency of aa All individuals in pop
d. Any population in which the distribution of genotypes conforms to the relationship p2 + 2pq + q2 = 1, whatever the absolute values for p & q may be, is at genetic equilibrium.
e. Allows biologists to calculate allele frequencies in a given population if we know the genotype frequencies, and vice versa.
f. Use this information to determine if the population is evolving.
4. Conditions for genetic equilibrium:
a. Random mating
Each individual in a population has an equal chance of mating with any individual of the opposite sex.
Mate selection must not be on the basis of genotype or any other factors that result in nonrandom mating.
b. No net mutations
No mutations that convert A into a or vice versa.
A & a frequencies must not change due to mutation
c. Large population size
Allele frequencies are more likely to change in small populations than in large populations.
d. No migration
Doesn’t mean seasonal migrations of the entire population
No exchange of alleles with other populations that might have different allele frequencies.
No migrations of individuals into or out of the populations
e. No natural selection
If natural selection is occurring, some phenotypes are favored over others.
Favored phenotypes (and their genotypes) have greater fitness, which is relative ability to make a genetic contribution to subsequent generations - allele frequencies will change.
5. Human MN blood groups
a. No medical advantage/disadvantage; no phenotype to influence mating, co-dominant (both alleles are expressed phenotypically)
b. Example of a human gene at equilibrium
Genotype Observed Expected
MM 320 313.6
MN 480 492.8
NN 200 193.6
Total 1000 1000.0
II. Genetic Variation in Populations (Microevolution)
A. Nonrandom mating changes genotype frequencies
1. When individuals select mates on the basis of a phenotype, they influence the frequencies of the corresponding genotypes.
2. Inbreeding
a. Neighbors tend to mate with neighbors
b. Increases homozygous genotypes but does not change overall allele frequency
c. Most extreme example of inbreeding is self-fertilization (plants)
Inbreeding not detrimental in some populations, but in others causes inbreeding depression, in which inbred individuals have lower fitness than those not inbred (fertility declines, high juvenile mortality)
3. Assortative mating - individual select mates on the basis of phenotype.
a. Fruit flies with low and high bristle numbers.
b. Low bristle flies were attracted to each other, as where high bristle flies
c. Example of positive assortative mating
d. Negative assortative mating - opposites attract, less common
e. Assortative mating tends to increase homozygosity in the population but does not change overall allele frequency
f. Assortative mating changes genotype frequencies only at the loci involved in mate choice, whereas inbreeding affects the entire genome.
B. Mutation increases variation within a population
1. Mutations are unpredictable - do not arrise with the ‘needs’ of the population.
2. Only involves mutations in reproductive cells - passed onto the next generation.
3. Mutations in somatic cells affect only the individual.
C. Genetic drift
1. Production of random evolutionary changes in a small population through chance.
2. Predators kill the only two individuals with a rare allele and the allele is lost to the population.
3. Larger populations may have the same frequency (percentage) of individuals carrying the rare allele, which results in more numbers of individuals carrying the allele (twenty vs two).
4. Genetic drift decreased genetic variation within a population while it tends to increase genetic differences among different populations.
5. Because of fluctuations in the environment, such as depletion in food supply or an outbreak of disease, a population may rapidly and markedly decrease from time to time - said to go through a bottleneck during which genetic drift can occur in the small population of survivors.
a. As the population recovers and increases its numbers, allele frequencies maybe quite different from allele frequencies before the decline.
b. Black Death of the 1300's - example of human bottleneck - European population decreased by 3/4 is some areas.
c. Cheetah genetic variation was greatly reduced at the end of the last Ice Age - can accept skin grafts from unrelated individuals.
6. Founder effect - when a small number of individuals establish a colony, bringing only a small fraction of the genetic variation present in the original population.
a. Finland settled about 4000 years ago and remained geographically separate from the rest of Europe and Asia.
b. Same thing happened in Iceland - isolation and bottlenecks have made them very genetically similar.
c. Amish of Pennsylvania - 200 founders - Ellis-van Creveld syndrome (six-fingered dwarfism)
D. Natural selection changes allele frequencies in a way that increases adaption
1. Operates on an organism’s phenotype
2. The phenotype represents an interaction between the environment and all the alleles in the organism’s genotype.
a. Rare for one locus to determine a phenotype, such as was seen in Mendel’s peas.
b. More common for phenotype to be polygenicly controlled.
3. Stabilizing selection
a. Associated with a population well adapted to the environment.
b. Selects against extremes - selects for intermediate or average phenotypes.
c. Human birth weight - too little and baby is underdeveloped, too big and the baby (and mother) may not survive birth.
4. Directional selection
a. If an environment changes over time, directional selection may favor phenotypes at one of the extremes
b. Over time, one phenotype gradually replaces the other.
c. Komodo dragons
Monitor lizards that became larger on the island as the larger individuals were favored.
Meanwhile, elephants on the island became smaller and smaller due to limited resources.
d. Poaching elephants
Elephants with big tusks preferentially killed - more money and less risk and less work than killing many elephants with small tusks.
Elephants in highly poached populations are more likely, now, to have small or no tusks.
Elephants in protected populations have larger tusks - aid in fights with other elephants, help in getting food (knocking down trees)
5. Disruptive selection
a. A special type of directional selection in which there is a trend in several directions rather than just one.
b. Selects against the average
c. Rare
d. Finch population on a Galapoagos island
Drought made only two food resources available, wood-boring insects and cactus fruit.
Finches with longer beaks could get at the cactus fruits.
Finches with wider beaks could strip off tree bark to expose insects.
Finches with intermediate beaks died.
3. What forces promote genetic diversity within the gene pool of a population? What forces promote changes in gene pool gene (allele) frequencies?