PeeEVOLUTION UNIT 1 TARGETS Content-­Based Targets

I. I will apply the concept of evolution by natural selection to examples and scenarios.

variations in selection via mutations, recombinations of DNA during mitosis or meiosis.

more offspring are produced than the environment can support or there are population density

dependent factors that limit the offspring/individuals that survive

limiting factors cause certain variations in the population to survive and reproduce which leads to

adaptations in the population (“selected for”) if the variations are not selected for than the

animals die out

over a LONG LONG period of time, natural selection creates a population that is well adapted

(not individuals)

ex. bunnies hopping at different speeds away from dogs who are hunting the bunnies

selective pressures: food, predators

II. I will state and explain the five assumptions of the Hardy-­Weinberg principle. 1. The population is very large (the effect of chance on changes in allele frequencies is greatly reduced.) 2. Mating is random (no preference) 3. There are no net changes in the gene pool due to mutations. 4. There is no migration of individuals into or out of the population. 5. There is no selection. (all genotypes are equal in reproductive success.) III. I will describe the two main causes of microevolution: genetic drift (bottleneck effect

& founder effect) and natural selection. (Figures 23.8 and 23.9) genetic drift sampling error caused by a small population resulting in offspring genotypes

that are NOT representative of original population;; it can lead to a loss of genetic variation within a population (fluctuation can lead to the elimination of an allele) and can also cause harmful alleles to become fixed

bottleneck original population is dies down randomly founder effect: a small amount of a population sets up a new population with more

or less of a random gene IV. I will use evidence from embryology, anatomy (homologous structures & vestigial

structures), molecular biology, paleontology, and physiology to support the theory of evolution by natural selection and classification by common ancestry (descent with modification). embryology : compare how embryos are similar in the womb anatomy

homologous structures: common structures that are inherited from a common ancestor

vestigial structures: structures used less often, but remain part of the anatomy of an organism and just are not used (ex. wings on ostriches, leg bones on whale)

molecular biology: studies the similarities and differences in DNA or proteins (amino acid sequences) among different species to determine evolutionary relationships (more closely 2 organisms are related, the less time that has elapsed since they shared common ancestor)

paleontology: looks at relative or actual age of fossils to locate common ancestors done with radioactive dating or relative position in the stratigraphy

Evolutionary physiology is the study of physiological evolution, which is to say, the manner in which the functional characteristics of individuals in a population of organisms have responded to selection across multiple generations during the history of the population

V. I will write definitions for the terms species and population.

“A specie is a single organism that is capable of reproduction (only with others of the same species) while a population is a group of species in the same geographical area.”

VI. I will describe the mechanisms for macroevolution (modes of speciation) and generate several examples of: allopatric speciation (adaptive radiation) and sympatric speciation. macroevolution requires an immense amount of time with lots of successive events of

natural selection for numerous advantageous traits, resulting in a population so different from the original that it can’t breed with the original-­-­>population=new species

allopatric speciation: 2 populations of the same species gets physically separated and become 2 new species as a result of living in different environments (ex. river separating a species of trees)

sympatric speciation: when a new species starts in the same location as the “parental” species, due to polyploidism or hybridization (ex. navel oranges)

VII. I will state several examples of prezygotic barriers and postzygotic barriers as causes of reproductive isolation between species. prezygotic barriers: habitat isolation, temporal isolation (different breeding seasons),

behavioral isolation, mechanical isolation (mating attempted but morphological differences prevent success), gametic isolation (sperm not able to fertilize egg)

postzygotic barriers: reduced hybrid viability (parent genes don’t mix well for survival), reduced hybrid fertility(sterile), hybrid breakdown (offspring are sterile or feeble)

: VIII. I will distinguish between and provide examples of the modes of (natural) selection:

disruptive, directional, stabilizing, diversifying, frequency-­dependent, and sexual. disruptive selection: conditions favor both extremes of phenotypic range over

intermediate range stabilizing: acts against extreme phenotypes and favors intermediate phenotypes directional: conditions favor one extreme of phenotypic rance, shifting frequency range in

one direction or other

sexual selection: individuals with certain characteristics are more likely than other individuals to obtain mates

diversifying: same as disruptive selection frequency-­dependent selection: fitness of phenotype declines if it becomes too common

lets add some more evolution diagrams to throughout unit 1 to fill up this space, needs work

BIOCHEMISTRY UNIT 2 TARGETS I. Compare and contrast chemical bonds (ionic and covalent) with intermolecular forces (hydrogen bonds, Van der Waals forces). Rank them in order of strength and describe the behavior of valence electrons.

strong to weak: ionic, hydrogen bonds, disulfide bridges, van der waals. chemical bonds

intermolecular forces big difference between hydrogen bonds and van der waals. covalent: share electrons, some have partial charges

polar(hydrophilic) when: has a charge nonpolar(hydrophobic) when: no charge at al

ionic compounds: transfer electrons always polar since they are always charged

intermolecular forces hydrogen bond: between hydrogen and sulfur/oxygen/nitrogen. Strongest force. van der waals: between 2 nonpolar molecules. Weakest force disulfide bridges: Opposite charged regions of a polar molecule. Stronger than

Van der Waals but weaker than Hydrogen Bond valence electrons are the electrons in the outer shell and want the shell to be full, will gain

or lose electrons to do this. H.O.N.C. if you have valence electrons 1,2,3,4 II. Differentiate between nonpolar covalent bonds and polar covalent bonds. Describe how electronegativity differences between atoms affect the polarity of a molecule. electronegativity is how much an atom wants electrons to fill valence electrons to be stable so there are two types of bonds: covalent -­ slight difference in charge and shares electrons [nonpolar with equal sharing;; polar with atoms of higher electronegativity]-­usually greater than 1.7 in difference ionic -­ trade electrons with a full charge difference [polar] III. Identify and discuss the 4 unique properties of water (cohesion, moderation of temperature, expansion upon freezing, and versatility as a solvent). Explain how water’s unique polar structure and its ability to hydrogen bond is essential for living organisms and their environments. COHESION-­Hydrogen bonds give water molecules its cohesive property because it holds and makes it stick to itself. ( i.e water falling in droplets-­rain)( ie. water move along a pathway-­circulatory system in the human body) Moderation of Temperature-­High specific heat(amount of a substance to change 1 C) and a high heat of vaporization( energy required to change from liquid to gas) Expansion upon freezing-­solid expands results in floating rather than sinking. When water freezes hydrogen bonding aligns itself in a tetrahedral form -­which leaves some extra space in between the molecules. Versatility as a solvent-­ Water is known to be the universal solvent. IT can break up molecules by surrounding the molecule and dissolving the molecule. IV. Describe the properties of acids, bases, and buffers. Explain how these affect the pH

scale (0-­14) by discussing the mathematical relationship between H+ and OH-­ concentration.

Acid: turns litmus paper red and is a proton donor. Have a pH less than 7 Base turns litmus paper blue and is a proton acceptor. Have a pH more than 7 Buffers: buffer zone, only small pH change when an acid or base is added pH=-­log(H+) ( ← this is only most widely used) & pH=-­log(OH-­) H2O(l) H+(aq) + OH-­(aq) Kw = [H+] * [OH-­] = 1 X 10-­14

V. Describe how Carbon’s special valence electron arrangement can lead to the

formation of structural isomers, geometric isomers, and enantiomers (chiral compounds). Compare and contrast each type of isomer.

Isomers have same molecular formula, but different molecular structure.

Structural Isomers: Same Chemical Formula but atoms are bonded in different orders. Geometric Isomers: Different arrangement of the groups around a double bond. Same

direction=”cis” …... opposite direction=”trans” Enantiomers: Mirror images of each other (ie. left-­right hands)

VI. Identify the 7 functional groups (hydroxyl, carbonyl, carboxyl, amino, sulfhydryl, phosphate, and methyl) when presented with a structural formula (drawing) of their structures. For each, describe the types of molecules they form, their functions, and their polarity.

Hydroxyl: (-­OH) usually alcohols (ie. ethanol) POLAR Carbonyl: (-­CO) either Ketones or Aldehydes. KETONES are located within the carbon

skeleton, while ALDEHYDES are located at the end of the skeleton. Sugars Carboxyl: (-­COOH) it is an organic acid. POLAR Amino: (-­NH2) one,two, three covalent bonds Sulfhydryl: (-­SH) “Thiols”-­name of compound Phosphate:(-­[PO4]3-­) Methyl: (-­CH3) In fatty acids NONPOLAR

VII. Describe the processes of dehydration synthesis and hydrolysis in Carbohydrates, Proteins, and Lipids. Identify the monomers, polymers, and bond names for each macromolecule.

carbs, lipids, and proteins all created by dehydration synthesis and destroyed by hydrolysis

Carbs: monomer=monosaccharide, polymer=disaccharide and polysaccharides, bond

names=glycosidic linkage Lipids: monomer=fatty acids and glycerol;; polymer=steroids, cholesterol, phospholipids,

etc;; bond names= ester linkage Proteins: monomer=amino acids;; polymer=protein (is a polymer), enzymes, structural

proteins;; bond names=peptide bonds Dehydration synthesis involves monomers combining to form larger molecules

(monosaccharides forming disachharides or polysachharides) Produces water in the process

Hydrolysis involves breaking the macromolecules into their monomers (breaking a protein down to its amino acids)

While hydrolysis needs water to proceed

VIII. Identify the elemental composition, general formula, structural formulas (drawings),

types, and functions of Carbohydrates. Carbohydrates are made up of anywhere between 5-­7 carbon atoms with H and OH groups attached in a chain for monosaccharides. The elements are C H & O. Carbs functions are to fuel, they help make cellulose, which helps structure the cells (strengthens cell wall) Saccharides have a general formula of (CH20)n n=3-­9, the most common being 5,6 glucose and fructose monosaccharides.

← structural formula Dehyradtion Synthesis removes water from the monosaccharides and forms disaccharides.

A chain of these disaccharides make up the polysaccharides( ie. starch) IX. Identify the elemental composition, general formula, structural formulas (drawings), types, and functions of Lipids.

Ratio of Hydrogen to Oxygen in lipids is greater than 2:1. It is made up of Carbon, Hyrdrogen, and oxygen. Lipids include waxes, steroids, phosopholipids, and fats. Lipids function as energy storage, and are part of membranes/cell walls

X. Identify the elemental composition, general formula, structural formulas (drawings), levels of structure (primary, secondary, tertiary, quaternary), and functions of Proteins. elemental composition: C,H,O,N

Primary Structure of a protein is its unique sequence of amino acids. Each amino acid in the diagram comes from one of 20 amino acids.

primary structure is the order of letters in a very long

word precise primary structure is determined by genetic info

Secondary structure (pic below on left);; segments of polypetide chains repeatedly coiled (alpha helix) or folded

coils and folds are the result of hydrogen bonds between repeating polypeptide backbone

the α Helix is a delicate coil held together by hydrogen bonding btwn every fourth amino acid, shown below

the β pleated sheet is the other main type of secondary structure. In this structure two or more regions of the polypeptide chain lying side by side are connected by hydrogen bonds btwn parts of the two parallel polypeptide backbones.

Tertiary Structure (pic above on right) while secondary structure involves interactions btwn backbone constituents,

tertiary structure is the overall shape of a polypeptide resulting from interactions btwn the sidechains (R groups)of the various amino acids

folding of beta pleated sheets and alpha helixes

Quaternary structure multiple tertiary structures combined certain proteins consist of two or more peptide

chains aggregated into one functional macromolecule

Each structure is part of the next structure

Proteins store amino acids and protect against disease Amino Acids-­ organic molecules possessing both carboxyl and amino groups

XI. Describe the relationship between enzymes, coenzymes, cofactors, substrates, catalysts, and activation energy. Identify and explain the factors that affect enzyme specificity, optimization, and activity.

enzymes: Proteins that help speed up chemical reaction in a body;; can be used over and over again and they have a very specific function.

Activation energy: Amount of energy needed to start the chemical reaction. initial amount of energy required to create the reaction. Graph depicts the energy over time for a chemical reaction

Catalyst: Reduces the amount of activation energy required to start the chemical reaction. Makes the initial “hump” in graph above lower. Also, it can be reused over and over again.

Substrates: Fits into the active site of the enzyme to begin the chemical reaction. Broken down by an enzyme.

Coenzymes: Bind to the active site of the enzyme and participate in cataylsis...more needed?

Cofactors: Inorganic material that increase the rate of catalysis XII. Explain the concept of denaturation and how it applies to the different levels of protein structure.

temperature: too high of temps: fried egg pH: high H+ or OH-­ Salt concentration 99% of the time is irreversible secondary, tertiary, and quaternary can be denatured

active site of protein does not match with substrate anymore XIII. Explain the relationship between enzyme activity, competitive inhibitors, allosteric activators/inhibitors, and metabolism (catabolism and anabolism).

enzyme activity: enzymes are natural catalysts that allow chemical reactions happen at biologically significant rates

competitive inhibitors: block active site so substrate cannot bind to enzyme allosteric-­ enzymes oscillate between 1 or 2 possible shapes regulated by feedback

activators: bind to allosteric site near/on one subunit & change enzyme shape to allow binding of substrate (ON switch)

inhibitors: bind to allosteric site near/on one subunit and change entire enzyme shape including active site so no substrate can bind to enzyme (OFF switch)

metabolism -­physical and chemical processes in an organism by which its material substance is produced, maintained, and destroyed

catabolism:reactions that break macromolecules into monomers [hydrolysis] anabolism: reactions that build macromolecules (dehydration synthesis/

condensation)

Unit 3-­ Cells, Membranes, & Transport

Target I Determine the surface area to volume ratio (size) of various cells to predict the rate of diffusion for monomers entering & exiting the cell;; relate calculations to cell shape.

calculations: SA=6s^2, Volume=s^3 cell size: 0.50 cm side → SA:Volume ratio=12:1

cell size 1.00 cm side → SA:Volume ratio=6:1 cell size 2.00 cm side → SA:Volume ratio=3:1 rate of diffusion technically is the same BUT surface area exposed to monomer

entering/exiting the cell is different so cells that are flat with a high surface area are advantageous b/c small

molecules can pass quickly through their membranes via diffusion In a given time, the materials and nutrients will reach smaller cells at a greater proportion of volume. Target II Visually identify & explain the function of the following cell organelles: ribosome, endoplasmic reticulum (rough & smooth), Golgi complex, mitochondrion, vacuole, & chloroplast.

pic of whole animal cell and plant cell is on page 100-­101 FIND THESE PICS AND PASTE HERE, needs work

Target III Describe how organelles interact to accomplish a task within a cell, tissue, organ, or organism. (look at target packet answer key from unit 3) Though, Protein Synthesis is one example.

paste page from target packet answer key from last year, needs work

Target IV Compare & contrast prokaryotic & eukaryotic cells. Prokaryotic & Eukaryotic cells similarity: both have cell membrane and enclosed with

cytosol, cytoplasm, ribosomes, and contain DNA (both have chromosomes), can have flagella, eukaryotic plant cells and some prokaryotic cells (not bacteria/archaea)

Eukaryotic cells: plant/animal cells, nucleus (location of linear DNA), many organelles, evolved from bacteria (endosymbiont hypothesis summary), larger than prokaryotes

Prokaryotic cells: no organelles, usually smaller than eukaryotic cells, nucleoid (location of circular DNA)

Prokaryotic Cells Prokaryotic & Eukaryotic Cells Eukaryotic Cells

no membrane-­bound

organelles usually smaller than

eukaryotic cells can have flagella nucleoid (location of

circular chromosomes)

both have cell

membranes Eukaryotic plant cells as

well as some Prokaryotic cells (not bacteria/ archaea) have cells walls.

contain ribosomes and cytoplasm.

have organelles

(membrane-­ bound)(including a nucleus)

larger than prokaryotes can have flagella nucleus (location of

linear DNA)

Target V Infer similarities and differences about the three domains of life (Bacteria, Eukarya, & Archaea) from a cladogram.

-­LUCA= Last Universal Common Ancestor -­Which domains share the most recent common ancestor?

Archaea and Eukarya -­Which domain have cells that contain a nucleus？

Eurkarya Which domain contain multicellular organism? -­Eukarya Domain Bacteria and Domain

Archaea= prokaryotic Domain Eukarya= eukaryotic Target VI Explain the Miller-­Urey experiment, RNA world hypothesis, formation of protobionts, and endosymbiont hypothesis in the context of the early Earth & the origin of life on Earth.

Miller-­Urey experiment: used methane, ammonia, hydrogen, and water vapor. runs continuous electric currents to simulate lightning (common on early earth) and showed organic compounds like amino acids (essential to life) could be made under conditions present on early earth

RNA world hypothesis:

Formation of Protobionts forms abiotically and spontaneously into a collection of

organic molecules, this exhibits reproduction and metabolism early form of life RNA error during duplicating leads to more stability and

better replication (natural selection → survivors have more reproduction → creates more stable RNA cells)

Endosymbiont Hypothesis: process of prokaryotic cell transforming into eukaryotic cells;; resulted in eukaryotic cells by forming organelles

Target VII Identify the component structures of the plasma membrane: phospholipid, integral protein, peripheral protein, cholesterol, hydrophilic head (polar), hydrophobic tails (non-­polar), glycolipids, & glycoproteins.

-­Fluid Mosaic Model: proteins and stuff floating in the membrane -­Integral proteins: transmembrane （material can move across membrane), act as channels to transport substances across the membrane, pumps -­Peripheral proteins (extracellular matrix): cell recognition, joining of cells, cell signals, stabalizes -­Peripheral proteins （cytoskeleton): stabalizes cytoskeleton, not embeded in lipid bilayer, -­Cholesterol: makes membrane less fluid by restraining phospholipid movement

-­temperature buffer glycoproteins: covalently bonded to proteins glycolipids: covalently bonded to lipids Target VIII Use representations, models, and or data to analyze situations in which molecules move passively by diffusion, osmosis, or facilitated diffusion.

passive transport: movement of molecules down concentration gradient from high to low membrane must be permeable & this requires no energy

facilitated diffusion: help polar and charged molecules across membrane and can use transport proteins [examples: aquaporins & gated channels]

diffusion: gases just passing through membrane, are nonpolar and small （Phosphate lipid: phosphate head= polar hydrophylic;; fatty acid tails= nonpolar, hydrophobic)

osmosis: diffusion of water across membranes from hypotonic solutions to hypertonic solutions

Target IX Use representations, models, and or data to analyze situations in which molecules move actively to establish concentration gradients across a membrane or

move large molecules into or out of a cell (endocytosis & exocytosis).

UNIT 4 TARGETS DNA Replication and the Cell Cycle

I. Explain the historical significance of the following scientists’ contributions to the study of DNA and utilize their data to support the claim that DNA is the source of heritable information: Griffith,Avery/McLeod/McCarthy, Hershey/Chase, Watson/Crick, Franklin/Wilkins Griffith Experiment:

Conclusion made: If bacteria is dead, then live bacteria (II-­‐r) can still use genetic

information from dead bacteria(III-­‐s)to transform into a bacteria that is virulent.

“transformation” involved: type II-­‐s (dead) coupled with II-­‐r (live) transformed

into III-­‐r

Hershey and Chase Experiment:

Hershey and Chase tracked the location of DNA and protein through radiation

concentrations and discovered that DNA, not protein, is the source of genetic material

Avery’s Experiment:

Avery’s group broke open heat-­‐killed pathogenic bacteria and extracted the cellular

contents. They then used specific treatments that inactivated DNA, RNA, and protein. He

tested each treated sample for its ability to transform live nonpathogenic bacteria; only

when DNA was allowed to remain active did transformation occur.

DNA was determined to be the sites of genes/genetic material

Chargaff’s Experiments/Rule:

Chargaff used a chromatographic

technique to isolate the different

bases of DNA (A,T,G,C)

Chargaff’s Ratios: in a single

molecule of DNA:

A=T, G=C

Franklin/Wilkins:

Franklin took the picture (below) with X-­‐ray diffraction/X-­‐ray crystallography

Watson concluded from it that DNA was helical in shape and enabled him to approximate

the width of the helix and spacing of nitrogenous bases along it (basically double helix

shape)

II. Compare and contrast the structure of a single nucleotide (phosphate, nitrogen base, sugar) from both DNA and RNA.

PUR(ine) AG Double ring Pure silver will double your

wealth PYR(imidine) CTU Single Ring

Cing TUt lies in a single pyramid

Purines bind to pyrimidines DNA: Adenine & Thymine RNA: Adenine & Uracil (AU = gold) DNA & RNA: Guanine & Cytosine ← look at picture

III. Describe the structure of DNA. Be sure to use the following terms: deoxyribose, nucleotide, hydrogen bond, phosphodiester bond, purine, pyrimidine, 5’, 3’, A, T, C, G, and anti-­parallel. You may be asked to label a diagram.

antiparallel: DNA composed of 2 separate, linear strings of nuleotides that are anitparallel.

-­one strand runs from 5’ to 3’

-­one strand runs from 3’ to 5’

-­5’ comes from phosphate group

-­3’ comes from deoxyribose Deoxyribose: sugar in DNA nucleotide: 1 nucleotide consists of a sugar (deoxyribose), phosphate, and nitrogen base

Phosphodiester bond: bonds sugar and phosphate Hydrogen bond: bonds nitrogen bases of each strand

-­stabalizes the helix -­bond angles form double helixes

Nitrogen bases: Adenine, Thymine, Guanine, Cytosine -­adenine and thymine bond together

-­2 hydrogen bonds -­guanine and cytosine bond

-­3 hydrogen bonds Purine: PUre Silver (AG) doubles your wealth

-­double ring Pyrimidne: Cing Tut lives in a single pyramid IV. Compare and contrast the genome structure and organization of prokaryotes and eukaryotes.

Comparing DNA Packaging

Prokaryotes both Eukaryotes

bacterial chromosome

1 double-­‐stranded

circular DNA

associated w/ small

amounts of protein

E-­‐Coli bacteria = 100x

more DNA than viruses

DNA approx. 1 mm

Nucleoid: dense region

DNA, not membrane

bound

proteins cause

chromosome to coil

& super coil

Composed of DNA +

Protein

1 linear DNA molecule

w/lots of protein

Complex of DNA and

protein called

chromatin

human somatic cell =

1000x more DNA the

E-­‐Coli

DNA approx. 4 cm

Nucleus:

Membrane-­‐bound

organelle with DNA

inside

Packaged by proteins

called histones (H2A,

H2B, H3, H4)

histones bind

to each other

and DNA to

form

nucleosomes

metaphase

chromosome: normal

compaction of

chromatin

euchromatin: less

compaction of

chromatin

heterochromatin: highly

condensed chromatin

^^metaphase, euchromatin

and heterochromatin

influenced by histone

modifications

V. State and explain the function of the following enzymes or structures involved in DNA replication: DNA polymerase (I, II, III), DNA ligase, primase, helicase, topoisomerase, nuclease, and single stranded binding proteins.

The hints should say: Enzymes move 3' to 5' on template Enzymes synthesize 5' to 3' Leading template runs 3' to 5' into fork Lagging template runs 3' to 5' out of fork

Protein Function

Helicase Unwinds parental double helix at replication forks

Single-­Strand Binding Protein Binds to and stabilizes single-­stranded DNA until it can be used as a template

Topisomerase Relieves “overwinding” strain ahead of replication forks by breaking, swiveling, and rejoining DNA strands

Primase Synthesizes an RNA primer at 5’ end of leading strand and of each Okazaki fragment of lagging strand

DNA pol III Using parental DNA as a template, synthesizes new DNA strand by covalently adding nucleotides to the 3’ end of a pre-­existing DNA strand or RNA primer

DNA pol I Removes RNA nucleotides of primer from 5’ end and replaces them with DNA nucleotides

DNA ligase Joins 3’ end of DNA that replaces primer to rest of leading strand and joins Okazaki fragments of lagging strand

VI. Explain the process of replication. Explain the differences between replication of the leading and the lagging strands. Include the terms: Okazaki fragments,

semiconservative, telomeres, telomerase, replication fork. Telomeres: Pretty much, they’re RNA “caps” that prevents the actual DNA from being cut off during replication.

Models of Replication

VII. Describe the process and significance of errors in DNA replication, DNA repair mechanisms, and mutagens.

Tumors usually have short telomeres;; and this is expected from cells that have undergone too many cell divisions -­ however, telomerase stabilizes the telomere length and prevents the self-­destruction of the tumor cells, which is what makes cancer so dangerous.

DNA repair tries to find the errors in DNA (frequent

DNA repair is required), but if the DNA repair mechanisms are damaged then when the DNA replicates it can lead to mutations in the cell

nuclease: DNA-­cutting enzyme that cuts out the damage in DNA & fills it with nucleotides using undamaged parts as a template;; DNA poly & DNA ligase involved

Mutagens: Reactive emissions, reactive chemicals, x-­rays, UV light, certain molecules in cigarette smoke & more

VIII. Describe the importance of mitosis in single-­celled and multi-­cellular organisms in terms of function, chromosome number of daughter cells and location.

functions: repair, growth/development, and reproduction for single-­celled organisms (prokaryotic cells), cell division is also known as binary

fission “The method by which the circular DNA molecule is replicated;; then the cell splits

into two identical cells, each containing an exact copy of the original cell's DNA.” for multi-­cellular organisms cell division is known as mitosis

“Mitosis is the process of forming identical daughter cells by replicating and dividing the original chromosomes”

BASICALLY 2 new cells are formed when a cell goes through the cell cycle and the chromosome number doubles overall for both binary fission and mitosis

IX. Describe the events occurring during the three phases of interphase: G1, S and G2.

X. Explain the relationship between Interphase, mitosis (prophase, metaphase, anaphase and telophase) and the G0 phase

(G0 is when the cell cycle temporarily/permanently stops;; nothing is happening;; cells that don’t divide ex.) brain cells;; never reach mitosis again unless some chemical event puts them back into G1. Interphase: consists of G2, S, G1;; prepares the cell for mitosis;; cell grows and single chromosomes duplicate in S phase Prophase: single chromosomes condense into visible double chromosomes (with 2 sister chromatids held together at the centromere. -­ mitotic spindles form Prometaphase (not on target but...): important thing is the nuclear envelope framents Metaphase: longest stage of mitosis;; centrosomes go to opposite poles of the cell -­ chromosomes allign on the metaphase plate (invisible “equator”) Anaphase: shortest stage of mitosis;; two sister chromatids from a double chromosome are pulled apart -­the two sister chromatids become their own single chromosomes -­single chromosomes move to the opposite ends of the cell as the kinetichore mircrotubles shorten Telophase/ Cytokines: (Telophase): still a single cell but with two daughter nuclei forming in the cell -­nuclear envelope arises from the nuclear envelope fragments

-­chromosomes become less condensed -­clevage furrow is seen (Cytokinesis): division of cytoplasm -­ cell splits into 2 separate cells XI. Define checkpoint in the cell cycle. Describe the differences between normal cells and cancer cells that reflect disruptions in cell cycle controls.

events in cell cycle directed by cell cycle control system, a set of molecules in the cell that triggers and coordinates key parts in the cell cycle.

checkpoint is a control point where stop and go-­ahead signals can regulate the cycle: Out of G1, G2, and M checkpoints, G1 is usually the most important.

cancerous cells may have telomerase present which causes telomeres never to break down (fray), so DNA replication keeps continuing

cancerous cells are also immortal and don’t follow checkpoints, they make their own signals

other characteristics of cancer cells: metastasis, avoidance of G0, oncogene (mutated genes that cause cell growth acceleration) over-­activation of cyclin, extra growth factor receptors

XII. Describe the relative concentrations of MPF, Cdk, kinases, and cyclin and how the interact to control the cell cycle. Regulatory molecules: 1) Protein kinases: enzymes that activate or inactivate other proteins by phosphorylating them. 2) Cyclins: Protein that has a cyclically fluctuating concentration in the cell. Some kinases depend on cyclin to be activated: these are called “cyclin-­dependent kinases.” MPF (Maturation/Mitosis Promoting Factor) initiates mitosis -­ when cyclin that accumulates during G2 combine with Cdk molecules to create MPF, the resulting MPF phosphorylates proteins necessary to initiate mitosis. XIII. Explain the process of gel electrophoresis. Use the following terms: Negatively charged DNA, positive end, negative end, wells, gel, buffer, long fragments, short fragments, distance traveled, electric field, restriction enzyme.

DNA is negatively charged due to phosphate so it goes towards the positive end of the cathode which is made by an electric field

short fragments travel farther in gel than long fragments

gel: Buffer & agarose restriction enzyme:

enzyme that cuts DNA at specific recognition nucleotide sequences known as restriction sites

picture explains process cathode= negative end

of chamber anode= positive end of

chamber

UNIT 5 Ecology TARGETS

I. Identify and define the different levels of ecological organization. Address the following categories: Biosphere, Biome, Ecosystem, Community, Population, Individual, Cell.

Biosphere: global ecosystem Biome:and of the world’s major ecosystems, often

classified according to predominant vegetation and characteristics by adaptations of organisms

Ecosystem: community of organisms in an area and the physical factors of how these organisms interact

Community: group of populations of different species in an area Population: group of individuals of same species living in same area Individual: member of a population Cell: smallest unit of life

II. Describe the flow of energy and materials through trophic levels in ecosystems. Be able to analyze and explain the following graphical representations: Food Webs, Food Chains, Pyramids (Energy, Biomass, Numbers). Incorporate the following terms in your explanations: biological magnification, producer, primary consumer, secondary consumer, tertiary consumer, herbivore, carnivore, omnivore, 10 % rule 10% rule: the next trophic level up can only take 10% of the energy, biomass, or numbers from the previous level Biological magnification: Chemicals (toxins, etc) increase in concentration as the levels increase, as consumers eat more of the lower levels.( Tertiary consumers have most concentration of toxins) Food webs are food chains but more complicated.. Herbivore: eats plants Carnivore: eats animals Omnivore: eats plants and animals Producer: produces own food via photosynthesis;; autotrophs;; plants Primary consumers: herbivores;; will eat only producers Secondary consumers: next trophic level about primary consumer;; eats primary consumer Tertiary consumers: next trophic level about secondary consumer, eats secondary conumer III. Analyze and explain survivorship curves (Type I, II, III). Explain how r-­selected and K-­selected species relate to specific curve types.

Type I: Humans -­very flat for a long time;; most survive to an old age and die off as they approach max life span -­few offspring;; a lot of care -­large mammals Type II: Squirrels -­equal chance of dying everyday -­fewer offspring;; some care -­rodents, small animals Type III: Oysters -­100% survivorship at birth, but then die out rapidly due to minimal care and predation;; about .3% of survivors make it to 25% of max life span -­those that survive past 50% of max life will usually survive until 100% of max life span -­produce a lot of offspring;; little or no care -­invertebrates, fish R-­Selected K-­Selected

Relative size of organism

small body size larger body size

Relative number of offspring

many not many

Relative life expectancy short life expectancy, die quickly

long life expectancy

Survivorship curve type Type III Type I

Number of times organisms reproduce

many few

Relative age to sexual maturity

short, mature quickly

long, takes time to mature

Examples Oyster, Turtle Human, Panda R-­selected: lives in the “r” (growth rate), quickly grow, short life, produce a lot of offspring, little or

no care K-­ selected: lives in the “K” (carrying capacity), long life,produce few offspring, a lot of care, IV. Analyze and explain logistic and exponential growth graphs. Apply the following terms to your discussion: limiting factors, carrying capacity, and equilibrium. Logistic or S-­curve graphs contain organisms that are usually long lived. There will be a steady increase of the population until a limiting factor (climate change/food/predators-­food and predators would become an issue when the carrying capacity of the population is reached) causes the population rise to cease, then the population will reach an equilibrium and not have a huge increase/decrease in numbers. Exponential or J-­curve graphs contain organisms that are usually short lived. There are no limiting factors until the carrying capacity is reached, no food is available for all the organisms and the population size will eventually decrease drastically. V. Discuss how biotic and abiotic limiting factors (density dependent and density independent) affect population size. For biotic factors, address predator/prey relationships and competition.

Biotic factors-­Density Dependent: disease, predation [inverse relationship], competition [for food resources], change in population size

Abiotic factors-­Density Independent: Sunlight, climate, soil, pollution, natural disasters VI. Explain how the different relationships among organisms can affect populations. Address the following: mimicry (Mullerian and Batesian), symbiosis (Mutualism, Parasitism, Commensalism), niche, habitat, competitive exclusion, camouflage, adaptation

Mimicry-­ Mullerian: Both are harmful (when 2 species are both undesirable and fend off predators;; together they can represent a larger population that is more obvious to be avoided)

Mimicry-­Batesian: one is harmful and one is not harmful (when 1 species has a similar appearance to a poisonous or dangerous species and thus leads predators to avoid it)

symbiosis -­ mutualism: both organisms benefit from this relationship Parasitism: one organism benefits from the relationship and one is hurt from the

relationship Commensalism: one organism benefits from the relationship and the other is not affected niche-­ an organism’s role in its habitat

VII. Define, explain, and identify examples of organism behavior. Address the following: taxis, kinesis, sign stimulus, fixed action patterns, imprinting

Taxis is a directional response in which an organism moves toward or away from a stimulus

Kinesis is a movement that is random and doesn't result in orientation with respect to a stimulus (a non-­directional response).

Fixed action patterns are when you have a series of behaviors that are initiated by some stimulus

ex. parent bird lands in nest (sign stimulus), bird opens its mouth and cheeps (fixed action pattern)

Imprinting involves both learned (behaviors that change as a result of experience) and innate (uniform within a population regardless of different environmental conditions or different experiences) components, and can only be acquired during the short sensitive period of an organism's life (usually initiated by a stimulus).

VIII. Compare and contrast the biogeochemical cycles: nitrogen, phosphorus, carbon. biogeochemical cycles: pathway of molecule or chemical element that moves through

atmosphere and biosphere

carbon cycle: starts in air, then used by plants for photosynthesis, they get stored in carbon compounds, either they return to the atmosphere through animal consumption of plants: respiration or decaying of the plant into coal over time: burned as fuel so CO2 is back into the air

nitrogen cycle: 80% of atmosphere is nitrogen, nitrogen has to be fixed by bacteria before it can be used by organisms. Bacteria chemically combine nitrogen in air to make nitrate or ammonia which plants use to make proteins. Animals eat plants, ingest the proteins and excrete the ammonia. Bacteria change ammonia/nitrogen back into nitrogen.

phosphorus cycle: largest reservoir of phosphorus is in sedimentary rock. through weathering it gets deposited into water and soil, so plants take up the phosphate ions which animals eat. the phosphorus can return to soil through urine, feces and through decomposition of when plant and animals die

Unit 6 Targets Plants and Photosynthesis

I. Describe how chlorophyll a, chlorophyll b, and carotenoids interact with different wavelengths of light energy

chlorophylls and carotenoids are used as the light absorbing heads of atoms;; magnesium atom at center. the are the colors that they reflect

chlorophyll a are blue/green and chlorophyll b are olive green they only differ in one of the functional group which allows it to absorb more

wavelengths carotenoids are yellow/green [because absorb violet, green-­blue light]

II. Differentiate between photosystem I and photosystem II using the following terms: light, ATP,

G3P/PGAL, NADPH, NADP+, oxygen, electrons, electron transport chain, water, primary electron receptor, P680, P700

Photosystem I Both Photosystem II

p700 comes earlier in

evolutionary histroy found in primitive

algae and bacteria does not use water

splitting enzyme

produce products for calvin cycle (where G3P is made)

have reaction center complexes

use light to oxidize pigments

electrons reduce primary electron receptor

p680 comes later in

evolutionary history found on all land

plants uses water splitting

enzyme

III. Explain how the first law of thermodynamics relates to sun light energy, chloroplasts, ATP, and glucose (chemical energy)

In photosynthesis, light energy is changed into chemical energy. The first law of thermodynamics is “obeyed” in that the amount of light that is the energy input equals the amount of chemical energy that is the output.

IV. State the reactants and products of the light reactions and differentiate between linear electron flow and cyclic electron flow

The picture shown above is the cyclic electron flow. below is the noncyclic electron flow Cyclic is usually done with algae or other singular cell organisms

it doesn't need a water splitting enzyme to work only makes ATP and NO glucose or sugars PS I

Linear is in multicellular organisms PS II and I working together end of linear cycle, electrons go to NADPH and ATP which is why the electrons

need to be replenished from H20 Both need pigments to be oxidized

V. Identify the structures in a chloroplast and relate structures to the process of photosynthesis.

chloroplasts in all green parts of a plant, but major sites of photosynthesis in leaves chlorophyll-­ the green pigment located

within chloroplasts;; light energy absorbed by this drives the synthesis of organic molecules in the chloroplast

mesophyll-­ tissue in the interior of the leaf;; chloroplasts are mainly found in cells of this

stomata-­ microscopic pores through which

CO2 enters the leaf and Oxygen exits water absorbed by roots delivered to

leaves in veins (which also export sugar) stroma-­ the dense fluid within the chloroplast thylakoids-­ an elaborate system of

interconnected membranous sacs;; segregates stroma from thylakoid space

VI. Explain how paper chromatography can be used to separate pigments and use the terms: molecular mass, solubility

Adhesion: Ability of molecules to stick to the paper

Cohesion: Ability of molecules to stick together Lower molecular mass → will travel further up the paper Greater solubility → will travel further up the paper

VII. Explain the Calvin cycle including: reactants, products, carbon fixation & RubisCO, reduction, how RUBP is regenerated & location in the chloroplast

Calvin cycle-­ second major stage, involving fixation of atmospheric CO2 and reduction of the fixed carbon into carbohydrate

carbon fixation-­ initial incorporation of carbon (CO2 from the air) into organic compounds

then calvin cycle reduces the fixed carbon to carbohydrate by the addition of electrons (requires NADPH)

to convert CO2 to carbohydrate, Calvin cycle requires ATP (this and prev. step requires the chemical energy produced on light reactions

steps in Calvin cycle = dark reactions/light-­independent b/c don’t require light directly;; occur in the stroma

VIII. Explain how C3 plants respond to hot, dry weather and include the terms photorespiration, Calvin cycle, mesophyll cells

the first organic product of carbon fixation for C3 plants is a 3-­phosphoglycerate (3 carbon compound)

photorespiration only occurs when stoma are closed RUBISCO adds O2 instead of CO2 to the calvin cycle CONSUMES ATP DOES NOT PRODUCE IT produces no sugar photorespiration decreases the photosynthesis output photores. drains away ~50% of carbon fixation in plants (sometimes)

carbon fixation is important since heterotrophs need to get the carbon fiaxtion from the plants as well

IX. Explain how C4 plants respond to hot, dry weather and include the following in your explanation: carbon fixation, bundle-­sheath cells, OAA, mesophyll cells, PEP, PEP carboxylase, spatial separation, Calvin cycle

C4 plants start carbon fixation with an alternative carbon fixation (4 carbon compound) Bundle-­sheath cells are arranged into tightly packed sheaths around veins of leafs b/w

bundle sheath and leaf surfaces is made up of loosely arranged mesophyll

cells the calvin cycle is confined to the bundle sheath cells [starts in mesophyll by

incorporating CO2 into organic compounds] 1. PEP carboxylase (enzyme present only in mesophyll cells) adds CO2 to

phosphoenolpyruvate (PEP) forms 4-­carbon oxaloacetate PEP carboxylase (has a higher affinity to CO2 than RUBISCO and no affinity to O2)

PEP carboxylase works more efficiently than rubisco in hot/dry climates with closed stomata

2. C4 plants fixes carbon from CO2, mesophyll cell exports 4-­carbon products (malate in image) to bundle sheath cells through plasmodesmata

3. in bundle sheath cells 4 carbon compounds release CO2 -­> reassinated by rubisco /calvin cycle (regenerates pyruvate);; ATP used to convert pyruvate to PEP;; cell carries out PSI (cyclic electron flow)-­ only photosynthetic method of making ATP;; mesophyll cells of C4 plant keep bundle sheath with high CO2 concentration so rubisco cannot bind to CO2;; **CO2 concentrating pump powered by ATP;; ** C4 photosynthesis minimizes photorespiration and enhances sugar production (advantage in dry/hot climates)

X. Explain how CAM plants respond to hot, dry weather and include the following in your explanation: carbon fixation, mesophyll cells, temporal separation, Calvin cycle

CAM (Crassulacean Acid Metabolism) -­ adaptation to arid conditions. Plant opens stoma during night and close them during the day [temporal separation], which helps desert plants conserve water. Plants take in CO2 at night and incorporate it into organic acids [carbon fixation] and store them in the vacuole till morning. During the day, light reactions supply ATP/NADPH for calvin cycle which gets CO2 released from organic acids. This way the CAM plants always have a

supply of CO2, even with closed stoma. This all happens in the mesophyll. XI. State functions of the three basic plant organs, identify adaptations /features associated with each organ, and relate organs to role in photosynthesis: roots, stems, and leaves Roots: multicell organ that anchors vascular plant to soil and absorbs minerals and water and stores carbs

root hairs = the absorption of water and minerals are in tips of roots that have increased surface area -­ short lived -­ thin, tube extension of root epidermal cell -­ main fxn = absorb

modified roots = make more support and anchorage or store water and nutrients or absorb oxygen

examples: monocots (ex. grasses) and seedless plants have small, adventitious

(plant organ grows in weird place) roots example: fibrous root system = mat of generally thin roots spreading out below

soil with no main root;; only in shallow soils example: taproot system = taproot (main root that comes from embryo root) gives

rise to lateral roots (branch roots) and it stores sugars and starches that plant will eat during flowering prod.

Stems = organ with alt system of nodes (points at which leaves are attached) and internodes (stem segments between nodes) -­ upper angle formed by each leaf and stem is axillary bud (structure can form lateral shoot, called branch) -­ most axillary buds of young shoots = dormant so the growth of a shoot is usually at the apical bud with growing leaves and nodes/internodes

some stems = food storage and asex reprod are sometimes mistaken for roots Leaves = main photosynth organ = flattened blade and petiole (stalk) which joins leaf to node

grasses and monocots don’t have petioles -­ base of leaf forms a sheath envelops stem

veins = vascular tissue of leaves variations in leaf morphology (leaf shape, branching patterns of veins,

arrangement) lead to differences in use -­ ex. big leaves = withstand wind all leaves can photosynth but some can also support, store, protect, sex

XII. Explain the adhesion-­cohesion-­transpiration mechanism for transport of water through plant from roots, through stems, to leaves in xylem based on water potential gradients adhesion: the bond between water molecules and xylem wall cohesion: hydrogen bond between water molecules so it continuously pulls water up transpiration: water evaporating out of plant pulls water up through plant roots exert a pushing pressure up;; only pushes water up a little up the stem, most water moves up due to transpiration leaves exert an upwards negative pull as water molecules are pulled up due to transpiration, some water is evaporated via transpiration and some water goes to the leaves to be used in photosynthesis XIII. Identify and describe several environmental factors that alter rate of transpiration, and affect the shape of guard cells to cause stomata open or close

Feature How this affects transpiration

Number of leaves More leaves (or spines, or other photosynthesizing organ) will have more stomata on their surface for gaseous exchange. This will result in a greater amount of water loss and an increased surface area for evaporation.

Number of stomata More stomata will provide more pores for transpiration.

Size of the leaf Leaves with bigger surface will transpire faster and leaves with smaller surface will transpire slower.

Presence of plant cuticle A waxy cuticle is relatively impermeable to water and water vapour and reduces evaporation from the plant surface. A reflective cuticle will reduce solar heating and temperature rise of the leaf, helping to reduce the rate of evaporation. Tiny hair-­like structures called trichomes on the surface of leaves also can inhibit water loss by creating a high humidity environment at the surface of leaves. These are some examples of the adaptations of plants for conservation of water that may be found on many xerophytes.

Light supply Stomata are directly related to the rate of transpiration, and these small pores open especially for photosynthesis. While there are exceptions to this (such as night or "CAM photosynthesis"), in general a light supply will encourage open stomata.

Temperature Temperature affects the rate in three ways: 1) An increased rate of evaporation due to a temperature rise will hasten the loss of water. 2) Decreased relative humidity outside the leaf will increase the water potential gradient. 3) Increased kinetic energy of water vapour particles aids diffusion out of the leaf.

Relative humidity A drier external surrounding will make a steeper water potential gradient, and so increase the rates of transpiration.

Wind Water lost from transpiration is often left in a residual layer just beneath the leaf. If left alone, this can reduce the amount of water loss as the water potential gradient from inside to outside the leaf is slightly less, due to the accumulation of water vapour there. If there is wind, this is blown away and the gradient remains higher.

Water supply Less water available means there is less to lose. The lack of supply can also prompt other changes that reduce the rates of transpiration.

XIV. Calculate and explain how solutions affect the water potential, pressure potential, and solute

potential causing plant cells to plasmolyze, be turgid, or be flaccid water potential=solute potential + pressure potential pressure potential is dependent on the pressure relative to the atmospheric pressure

(can be positive or negative) solute potential is dependent on the solute concentration (higher solute concentration →

more negative solute potential)

Unit 7 Cellular Respiration Targets I. Describe the location, function, reactants, products, enzymatic actions, and ATP production for Glycolysis.

Glycolysis is the breakdown of sugar Located inside the cytosol/cytoplasm Glucose, a 6 carbon sugar, is split into

two three carbon sugars and 2 h20 molecules These smaller sugars are oxidized and to

form pyruvate (the ionized form of pyruvic acid) The net energy from glycolysis per

glucose is 2 ATP plus 2 NADH Phosphofructokinase is used to transfer

phosphate groups from ATP to the sugar Does not require oxygen ((Look at the diagram for exact net

inputs/outputs)) II. Describe the location, function, reactants, products, enzymatic actions, and ATP production for Intermediate Step (Bridge Reaction).

if oxygen is present, the pyruvate will move into the mitochondria

the movement from glycolysis to the citric acid cycle by three steps:

pyruvate’s carboxyl group (oxidized) has little energy, so it removed as CO2

remaining 2 carbon fragment is oxidized, forming acetate and an enzyme transfers electrons extracted to NAD+, storing energy as NADH

coenzyme a (CoA), is attached to acetate that makes the acetyl group very reactive (so the acetyl CoA

has high PE) III. Describe the location, function, reactants, products, enzymatic actions, and ATP production for Krebs (Citric Acid) Cycle.

note: while it is not needed to know every

single step, it is very important to understand what happens in the chemical reactions between each step

Acetyl CoA combines with oxaloacetate to form citrate

The unstable citrate molecule has a high potential energy;; it degenerates into lower energy molecules, forming 3 NADH, 1 ATP, 1 FADH2 for every acetyl CoA

Final molecule is oxaloacetate;; regenerated by acetyl CoA to start cycle over again

Process occurs in the mitochondrion

IV. Describe the location, function, reactants, products, membrane proteins, & enzymatic actions for Electron Transport Chain. Explain how the chemiosmosis model generates ATP.

Electron carriers (NADH, FADH2) attracted “downhill” to oxygen, causing protons to be pumped into intermembrane space and create an H+ gradient

Chemiosmosis: H+ gradient causes protons to flow back, powering ATP synthase to create

ATP ~10 NADH, 2FADH2 produces 32-­34 ATP This process requires oxygen

V. Identify mathematical relationships describing conversions from one molecule to another. Examples: 1 glucose yields 2 pyruvate, 1 NADH makes 3 ATP in ETC, 1 FADH2 makes 2 ATP in ETC, each pyruvate makes 3 NADH and 1 FADH2 in Kreb’s, etc…

Glycolysis 1 glucose = 2 pyruvate, 2 H20, 2 ATP, 2 NADH, 2H+

Bridge 1 pyruvate = 1 NADH, 1 Acetyl CoA

Krebs Cycle 1 Acetyl CoA = 3 NADH, 1 ATP, 1 FADH2

ETC 1 FADH2 = 2 ATP 1 NADH = 3 ATP

VI. Describe the connection between glycolysis and the fermentation reactions (alcoholic and lactic acid) in anaerobic respiration. Describe the location, function, reactants, products, and enzymatic actions for each step. Be able to summarize inputs and outputs for the entire anaerobic process including ATP production.

Anaerobic respiration is the breakdown of food without oxygen Comes after glycolysis creates pyruvate Two types:

Alcoholic: 2 pyruvate + 2 NADH = 2 Ethanol + 2NAD+ Lactic: 2 pyruvate + 2 NADH = 2 Lactate + 2 NAD+

NAD+ can be regenerated in glycolysis to produce 2 ATP and pyruvate again VII. Compare and contrast the processes of aerobic cellular respiration in the mitochondria and photosynthesis in the chloroplast.

cellular respiration both photosynthesis

NADH, FADH2 as electron carriers

Breaks down glucose Occurs in

mitochondria and cytoplasm

takes in O2 and gives off CO2

Uses ETC to synthesize ATP

Electron transport molecules

Chemiosmosis in ETC

Oxidation (loss of electron) and

NADP+ as electron carriers

Creates glucose Occurs in

chloroplasts takes in CO2 and

gives off O2 Light reaction

Kreb Cycle Catabolic oxyidatmephosrylation

reduction (gain of electrons) [REDOX]

seen in plants/autotrophs

Anabolic Photophosphorylation

VIII. Identify and explain the significance of redox reactions, oxidative phosphorylation, substrate-­level phosphorylation, decarboxylation, and hydrolysis in the biochemical pathways of aerobic cellular respiration. Redox reactions are significant in many steps of cellular respiration. They occur in glycolysis to make NADH, in the bridge reaction to generate NADH, in the Krebs cycle to make NADH and FADH2, and finally in chemiosmosis to release electrons from the electron carriers. Redox reactions involve the gaining of an electron from a molecule (reduction) and the simultaneous losing of an electron from a molecule (oxidation). Oxidative phosphorylation only occurs in chemiosmosis in the electron transport chain and involves the production of ATP using energy derived from the redox reactions of an electron transport chain. Since the energy comes from the oxidation of the electron carriers, and the process involves an addition of a phosphate group, the name "oxidative phosphorylation" is logical in context. Substrate-­level phosphorylation similarly involves the addition of a phosphate group to ADP to create ATP, however the phosphate group comes from a substrate, hence the name "substrate-­level phosphorylation". This occurs in both glycolysis and the Krebs Cycle. Decarboxylation is the process of releasing a CO2 molecule. This occurs in the Bridge Reaction and the Krebs Cycle. Finally, hydrolysis is the process of breaking a polymer into a monomer with a water molecule. It is a catabolic process and occurs in glycolysis and the Krebs Cycle. All of these processes are significant because they cause a necessary transfer of energy to either make molecules more stable, less stable, or a certain shape to fit an enzyme during all steps of cellular respiration. IX. Apply the first law of thermodynamics to the relationship between exergonic reactions, endergonic reactions, catabolism, anabolism and energy transfer (ATP & heat).

Catabolism, exergonic releases energy, heat Anabolism, endergonic absorb energy, heat

X. Describe the structure of ATP. Explain how the hydrolysis and regeneration of ATP relates to metabolism.

ATP is made up of adenine, and three phosphate groups;; highly reactive Can break down into lower energy forms such as ADP or AMP through hydrolysis

(splitting by water) Hydrolysis: Releases energy, so catabolism Dehydration synthesis/Regeneration: Stores energy, so anabolism

PROCESS-­BASED UNIT 7 TARGETS

A. Conduct an experiment and analyze the data to determine the effect of various factors on the rate of cellular respiration using colorimeters and Logger Pro computer software.

Unit 8 Targets Protein Synthesis I Compare and contrast structures/functions of mRNA, tRNA, and rRNA. mRNA (messenger RNA):

single stranded RNA with a phosphate group, ribose sugar, and RNA nucleotide complementary to the DNA on the template strand carries the genetic code from the nucleus into the cytoplasm where protein

synthesis occurs makes the “blueprint” for protein sythesis gives instructions to create the protein has nucleotide base pairs on the strand (3 nucleotide bases= a codon) structure: linear

rRNA (ribosomal RNA): makes up the ribosome ribosome consists of a large and small subunit reads cd from tRNA and mRNA to make the protein

tRNA (transport RNA): anti-­codon at bottom of tRNA structure;; amino acid on top of tRNA structure anti-­codon of tRNA attaches to codon on mRNA;; amino acids link to create the

polypeptide structure: clover-­leaf shaped

II Understand the significance of the 3 base pair genetic code (including the redundancy) and be able to use it to determine the amino acid sequence of a protein.

64 codons code for only 20 different amino acids, so there is a lot of redundancy

Some mutations are silent mutations due to redundancy III Outline the steps and procedures in transcription. Include discussion of RNA polymerase, promoter region, TATA box, 3’, 5’, terminator region, and transcription factors.

1. Transcription factors, groups of enzymes that regulate the RNA Polymerase transcription rate, bind to the promoter region on the DNA sequence in a region called the TATA box. These transcription factors allow the binding of RNA polymerase, an enzyme that pries the two strands of DNA apart and joins the RNA nucleotides as the base-­pair along the DNA template

a. promoter region: DNA sequence where RNA polymerase attaches and

initiates transcription b. TATA box: made up of a nucleotide sequence (TATA) and allows several

transcription factors to be bound to it before RNA Polymerase II 2. The RNA polymerase transcribes the DNA in the 3’ to 5’ direction, and synthesizes

RNA in the 5’ to 3’ direction 3. Transcription ends at the terminator region, the sequence that signals the end of

transcription. Here the RNA transcript is released and the RNA polymerase detaches from the DNA

These steps can also be described as Initiation, Elongation, and Termination IV Identify methods of RNA modification: processing and editing. Relate to these terms: 5’ cap, poly A tail, RNA splicing, spliceosome, introns, exons, alternative RNA splicing. *RNA processing and editing is only in eukaryotes in order to make mRNA usable for protein synthesis

5’ cap is added to 5’ end and poly(A) tail is added to 3’ end RNA splicing: spliceosome cuts out introns and puts remains (exons) together

to form one mRNA strand Alternative RNA splicing: genes can be spliced in multiple different ways, so

various sets of exons can be expressed, leading to synthesis of different proteins V Describe the steps and procedures in translation: initiation, elongation and termination. Include the following terms in your explanation: tRNA, rRNA, codon, anticodon, start codon, stop codon, wobble, ribosome (large and small subunits), aminoacyl-­tRNA synthetases, E site, P site, A site, mRNA, 5’, 3’, amino acids (20 total).

Initiation: the small ribosomal subunit attaches to

the mRNA at the start codon AUG tRNA enters through the P-­site and

attaches the anti-­codon complementary to the mRNA start codon;; tRNA also brings in an amino acid

the large ribosomal subunit attaches to the small subunit to create the initiation complex Elongation:

a new tRNA enters through the A-­site the original tRNA moves from the P-­site to the E-­site where it is discharged from the

ribosome the new tRNA that was in the A-­site moves the the P-­site and attaches its anti-­codon

to the complementary codon on the mRNA;; the tRNA also brings in an amino acid that attaches onto the growing polypeptide

the process continues as a tRNA goes from A-­site to P-­site to E-­site Termination:

a release factor attaches to the mRNA at the stop codon (UAA,UGA, or UAG) the synthesis of proteins end ribosomes separate from the mRNA and the pieces float off in the cytoplasm

VI Describe ways that proteins can be altered following translation.

modifications are necessary for the protein to be able to correctly reach its can function properly.

sugars can be added to help the protein reach its final destination. lipids and phosphate groups might also be added to the polypeptide chain

ex. of post-­translational modifications: glycosylation: tagging of sugar molecules to proteins, which helps enhance the

specificity and function of the protein phosphorylation: tagging of phosphates to proteins to help with tagging.

other modifications include cleaving and creating disulfide bridges ( for insulin). Enzymes may remove one or more amino acids. Or two polypeptide chains can come together.

VII Relate the following terms: point mutation, insertion, deletion, frameshift, nonsense, missense, mutagen, chromosomal mutation. For single-­base mutations, the first type of mutation is a missense mutation. This involves the incorrect

transcription of a single nucleotide that in turn, leads to the incorrect codon translation. Another type of

mutation is a nonsense mutation. Like the missense mutation, only one nucleotide is incorrectly copied, but

it has a more damaging effect on the gene function because the incorrect nucleotide leads to the copying of

a premature stop codon. Less detrimental than these, a silent mutation involves the incorrect transcription of

one nucleotide but it does not affect the protein sequence because both the original codon and the codon

affected by the mutation code for the same amino acid. Insertions and deletions are mutations that can

involve 1 nucleotide (or more, if there are 3 an entire codon is lost/added and results in no frameshift

mutation). One nucleotide is either added or deleted and the "reading frame" of the codons is altered which

most likely will lead to completely different amino acids and thus a dysfunctional protein.

Chromosome mutations occur during meiosis and include translocations, inversions, deletions, duplications,

and nondisjunctions. The names are self-­explanatory for what happens and are similar to mutations in the

transcription of DNA to mRNA, however instead of nucleotides being affected, it's pieces of chromosomes.

Chromosomal mutations can lead to the loss of genes, duplication of genes, and various genetic

syndromes. VIII Explain what the universality of the genetic code implies about evolutionary history.

Genetic code is shared by organisms from the simplest bacteria to the most complex plants and animals-‐-‐> genes can thus be transcribed and translated after being transplanted from one species to another

RNA codon CCG, is translated as the amino acid proline in all organisms whose genetic code has been examined

Bacteria can be programmed (by insertion of human genes) to synthesize certain human proteins

EX: Insulin Exceptions to the universality are translation systems in which a few codons differ

from the standard ones; slight variations also exist in unicellular eukaryotes and in organelle genes of some species

genetic code near universality means that a common language between all living things must have been operating very early in the history of life-‐-‐early enough to be present in the common ancestor of all present-‐day organisms

Unit 9 Meiosis and Inheritance

I Describe the process of meiosis using the terms: meiosis I, meiosis II, crossing over, synapsis, chiasma, tetrad, sister chromatids, haploid, diploid, and homologous pair. General overview of meiosis: a diploid cell, or a cell that contains a doubled set of each chromosome type, is divided into four haploid cells, or cells that contain only a single set of each chromosome type

first begins in the Meiosis I In Prophase I, a process called synapsis occurs, which means that each pair of

homologous chromosomes attach to each other at an X-­shaped spot called a “chiasma.”

The structure formed by these two homologous chromosomes is called a tetrad, as each homologous chromosome contains two identical sister chromatids, which adds up to a total of four chromatids. This structure facilitates a process called ‘crossing over,’ where specific portions of the chromosome switch places with its chromosome pair, which gives the chromosomes their genetic diversity.

After Prophase I, the homologous chromosomes line up side by side along the metaphase plate in Metaphase I, and in Anaphase I, each chromosome is split up from its homologous pair, and become separated at opposite ends of the cell.

Telophase I then begins the process of separating the cell into two distinct cells. In Meiosis II, the second stage, these two haploid cells are further separated into four

haploid cells. Meiosis II is identical to Meiosis I except that in Anaphase II, instead of separating

each chromosome from its homologous pair, it is the sister chromatids that get separated from each other and placed into opposite ends of the cell.

II. Compare and contrast sexual and asexual reproduction. Describe examples of organisms that perform each type and the impact on each on genetic diversity within a population.

asexual reproduction: single individual is the sole parent and passes copies of genes to offspring

in plants, bacteria, and animals sexual reproduction: fusion of haploid gametes forms a diploid cell (a zygote)

in sperm and eggs in animals and in pollen and ovules in plants

Asexual Reproduction Sexual Reproduction

Number of parents 1 (either male of female) 2 (male and female)

Makeup of offspring Genetically identical (to parent and other offspring)

Genetically different

Cell division process Normal cell division following nuclear division (by mitosis)

Special cell division following nuclear division (by meiosis) producing sex cells (gametes): after fertilization subsequent divisions: normal

Advantages Quick Produces variation-­ the basis of evolution

Disadvantages Disease may affect all slower

III. Describe how segregation, crossing over, and random fertilization produce genetic variation in a population.

Genetic variation via sexual reproductions is produced during independent assortment and crossing

over in meiosis 1 and through random fertilization in meiosis 2.

Independent assortment means that maternal and paternal chromosomes can assort independently to form 2^n different kinds of gametes.

Crossing over is when non-­sister chromosomes of a tetrad from opposite chromosomes “cross over” at a chiasma to exchange genes, so the chromatids become genetically

different. This will produce unique haploid gametes.

Random fertilization means that any of a man’s 8.4 million types of sperm can combine with any of a woman’s eggs, meaning there are trillions of different types of gamete

combinations formed by a couple. IV. Relate karyotypes to nondisjunction of homologous chromosomes, gender, and their diagnostic capabilities. Karyotypes of chromosomes will decided if a child is a boy or girl based on the x and lack or there of Y chromosomes. Sometimes humans have extra chromosomes. For example an extra chromosome at location 21 leads to Down syndrome. Down syndrome is an example of non-­disjunction in the chromosomes. This occurs when during during either meiosis I or meiosis II when metaphase or anaphase are occurring the chromosomes may split up incorrectly and an extra chromosome or one less chromosome will form in the gamete cell created. V. Compare and contrast meiosis and mitosis.

Mitosis Both Meiosis

·∙ Divides somatic cells ·∙ Produces 2 new cells ·∙ Goes through the cell cycle once (prophase, metaphase, anaphase, telophase) ·∙ Asexual reproduction

·∙ Go through the cell cycle ·∙ Use spindle fibers ·∙ Go through cytokinesis during telophase ·∙ Produce new cells ·∙ Require DNA replication prior to each process

·∙ Divides reproductive cells ·∙ Produces 4 new cells ·∙ Goes through PMAT twice ·∙ Makes gametes ·∙ Involves a tetrad, synapsis, and crossing over ·∙ Has the phase Interkenesis (rest stage between the two PMATs that occur) ·∙ Sexual reproduction

VI. Define, describe and relate to meiosis each of Mendel’s three laws: Law of Dominance, Law of Independent Assortment, and Law of Segregation. The following only occur during Mendel's conditions and are known as Mendelian laws.

The law of segregation states that during the creation of gametes that a trait will randomly separate.

achieved by the end of meiosis II;; ex.: if the trait alleles are Aa, there is a 50% chance for each one will be in the

gamete formed. Law of Dominance occurs when two alleles for a trait are different. One allele will be

expressed and the the other allele has no phenotypic effect. only one dominant allele is needed for a gamete to express the phenotype that corresponds to that dominant trait.

The Law of Independent Assortment states that alleles sort independently because the gene of each is located on a specific chromosome. Each homologous chromosome with an associated allele is segregated into a separate gamete.

If two or more traits are expressed in a genotype, then the law of independent assortment assumes that each allele is on a separate chromosome.

PROCESS-­BASED TARGETS A. Create a website (wiki) to educate a family about inheritance and reproductive options to a particular genetic disorder. B. Use bead chromosomes to model the process of meiosis including crossing over. C. Perform, interpret results, determine probability, genotypic ratios, phenotypic ratios, and recognize patterns for a variety of monohybrid and dihybrid crosses that follow Mendelian patterns of inheritance. D. Perform, interpret results, determine probability, genotypic ratios, phenotypic ratios, and recognize patterns for a variety of monohybrid and dihybrid crosses that do NOT follow Mendelian patterns of inheritance: Sex linkage, mitochondrial inheritance, multiple alleles, co-­dominance, incomplete dominance, sex limited. E. Draw, interpret, and determine inheritance patterns for traits presented in a pedigree.

Unit 10 Regulation of Gene Expression

I Explain the process of in vivo DNA cloning and its application for copying eukaryotic genes in prokaryotic organisms. Use the following terms: plasmid, rDNA, eukaryotic gene of interest, bacterium, eukaryotic cell, ampicillin resistance (ampR) gene, ampicillin, mitosis, bacterial clones, pure culture, agar plate, mRNA, reverse transcriptase, transformation

To start, cDNA for the eukaryotic gene of interest is isolated from the eukaryotic cell by taking the mRNA and using reverse transcriptase, the cDNA does not contain the introns because prokaryotes cannot duplicate those. Then, using the process of transformation, the bacteria is cloned. This process entails the following steps:

1) the plasmid and the cDNA are both cut by the same restriction enzymes, thus allowing for the cDNA to enter the plasmid [the sticky ends of the DNA and plasmid will attach and ligase will then join them?]

2) a different restriction enzyme then cuts the plasmid again, allowing for the ampR to enter the plasmid

3) then place the plasmids into the bacterium, i.e. e coli. 4) to test which bacterium successfully received the gene of interest ( through the

plasmid), place them on the agar plates and apply ampicillin. the cells that are able to perform mitosis and survive the ampicillin, successfully received the gene of interest because if the ampicillin resistance genes were transferred, then the gene of interest had to have been transferred as well. **the bacteria cells are in a pure culture, meaning that the culture is created to have predetermined environmental controls and only one specie.** II Describe how DNA microarray assays are used to study gene expression. Use the following terms: single-­stranded DNA fragments, glass slide, cDNA, mRNA from cells of interest, fluorescent dye, reverse transcriptase, hybridize, fluorescence detector, intensity of fluorescence, level of expression, target DNA Steps to determine if study expressions of large groups of genes:

1. isolate mRNA from cell of interest, and use the mRNA as templates to perform reverse transcriptase to form the cDNA

2. place the single stranded DNA fragments on a glass slide on a tightly spaced array or grid (look at fig 20.15 for rest)

III Identify components of an operon model used to describe regulation of gene expression in prokaryotes. Use the following terms: promoter, regulatory gene mRNA, protein, RNA polymerase, operator, genes of operon Microarrays can be used to look for mutations in a gene, to measure how active a gene is in a cell or tissue type, to measure how gene activity changes, and to compare the genotypes of different samples. Single-­stranded DNA fragment sequences that can account for thousands of different genes, each acting as a probe for a particular gene, are individually placed in small wells on a glass slide, These glass slide, or DNA

chips, are used to analyze the expression of certain genes to a particular gene of interest. A particular cell of interest to be analyzed in the study, (ex. sample of a plant infected with a virus) has its mRNA extracted. Using reverse transcriptase, the mRNA is made into more stable cDNA. This sample is then labeled with a fluorescent dye. The sample of cDNA is applied to the DNA microarray chip. The fluorescently labeled cDNA will bind the the complementary single stranded DNA probes that are in each well if the nucleotides match up as complementary base pairs (hybridize). Those that do not hybridize are washed off. The DNA micoarray chip is then analyzed using a fluorescence detector. The hybridization of fluorescently labeled cDNA and DNA in the wells will show up on the detector. The intensity of the fluorescence determines the level of gene expression. This allows for an understanding of what particular genes are expressed or affected when given an virus or particular sample. IV Distinguish between inducible and repressible operon function in negative gene regulation. Provide examples of each type of operon. Use terms: corepressor, active repressor, inactive repressor, no mRNA produced, mRNA produced, tryptophan, allolactose, lac operon, trp operon, inducer, “turn on”, “turn off”

In inducible operons, the normal state of the operon is “off,” which means that a repressor is actively blocking RNA Polymerase from passing the operator. This prevents mRNA from being transcribed;; thus, the protein and trait that the gene codes for is not expressed. When an inducer binds with the repressor, the repressor is inactivated, which allows RNA Polymerase to transcribe mRNA and the gene to be expressed. An example of an inducible operon is the lac operon. Normally, a repressor prevents the gene for lactose to be transcribed into mRNA, therefore preventing lactose from being produced;; however, when lactose is present in the environment, the inducer allolactose binds with and inactivates the repressor, opening up the operator and allowing mRNA coding for lactose-­utilizing enzymes to be produced.

In repressible operons, a similar process of activation occurs. The normal state of a repressible operon is “on,” which means that mRNA is transcribed. When a corepressor binds with a repressor, the repressor is activated and latches onto the operator. This action prevents RNA Polymerase from passing the operator and transcribing mRNA, which effectively means that the operon is now “off.” An example of a repressible operon is the tryptophan operon.

Normally, tryptophan is produced;; however, when tryptophan is present in the environment, it acts as a corepressor, binding and activating a repressor, which closes the operator, causing mRNA coding for tryptophan to stop being produced. V Describe how the presence of cAMP promotes positive gene regulation in the lac operon. Use terms: CAP-­binding site, inactive CAP, active CAP, activator, cAMP, stimulate transcription, and other relevant terms for operation of lac operon

E. Coli will only use lactose when it is present and glucose levels are low (see fig 18.5)

cAMP accumulates with glucose is scarce CAP is an activator that binds to DNA and starts

transcription of DNA CAP when activated goes to the lac promoter site to

help RNA polymerase attach to the promoter easier

CAP directly stimulates gene expression (positive regulation)

if amount of glucose in cell increases cAMP concentrations fall and w/o cAMP CAP won’t attach at the operon, so lac translation would be very slow [even with lactose in the cell environment]

so lac operon is under negative control as well [by the lac repressor]

lac repressor determines if transcription happens or no (on-­off switch)

CAP controls the RATE of transcription (volume switch) VI Explain how histone acetylation affects gene regulation in eukaryotic cells. Use terms: histones, acetylate, stimulate transcription Histones (proteins around which the DNA is wrapped in

nucleosomes) play direct role in regulation of gene transcription. As seen in fig 18.7a, histone tails are accessible to enzymes which means they catalyze the addition or removal of specific chemical groups.

During histone acetylation, acetyl groups: (-­COCH3) attached to lysines in histone tails, deacetylation = removal of acetyl groups.

Acetylated lysines go from a positive charge to neutral so histone tails no longer bind to neighboring nucleosomes, which promotes compact folding of chromatin. (fig 18.7b)

Result: transcription proteins easily access genes in acetylated region histone acetylation enzymes may promote initiation of transcription not

only by remodeling chromatin structure by ALSO binding to & thus recruiting components of transcription machinery

VII Explain how DNA methylation affects gene regulation in eukaryotic cells. Use terms: histones, methyl groups, repress transcription

Set of enzymes can methylate certain bases in DNA itself. Most inactive DNA has more methylation than DNA that is actively

transcribed. Genes more heavily methylated in cells in which they are not expressed, removing the

extra methyl group can turn some of these genes back on. Proteins that bind to methylated DNA, can be repress transcription by sending histone

deacetylation enzymes to the site. essential for some cells that DNA methylation occurs or else it leads to abnormal

embryonic development in organisms VIII Describe how the organization of a typical eukaryotic gene lends itself to regulation. Use terms: enhancer, proximal control elements, distal control elements, promoter, exon, intron, poly-­A signal sequence, termination region, activators, transcription factors, mediator proteins, RNA polymerase II, transcription initiation complex

Cluster of proteins (RNA poly 2, helicase, etc) , transcription initiation complex, assemble on the promoter sequence.

RNA poly 2 proceeds to transcribe the gene, synthesizing the pre-­mRNA. Splicing, addition of poly-­A tail and 5’ cap which make mature mRNA.

Control elements= segments of noncoding DNA;; help regulate transcription by binding to certain proteins, and the proteins they bind are critical to making sure gene regulations are precise.

transcription factors: help initiate transcription (RNA poly II requires this to start transcription)

enhancer: segments of eukaryotic DNA containing multiple control factors usually located far from the gene whose transcription regulates

distal control elements grouped together

poly-­a signal sequence: signals where transcription is cleaved & poly-­a tail added (its in last exon of gene)

mediator proteins: bend DNA & add mediator proteins which help assemble the initiation complex on the promoter

activators: attach to enhancers & together help transcription go faster euk version of CAMP & CAP only in eukaryotic cells

promoter: region in all eukaryotic/prokaryotic DNA that just tells RNA poly II to start transcribing there

required in all DNA to start transcription termination region: RNA poly II stops coning at this region of the mRNA code

IX Explain how the process of RNA interference (RNAi) is used to silence expression for certain genes. Use terms: miRNAs, Dicer, degrade mRNA, block translation, “turn off” gene expression RNAi interference is the process which involves RNA molecules to inhibit gene expression, which is usually done by the destruction of other specific mRNA molecules. RNAi requires two types of RNA molecules which are miRNA and siRNA (microRNA and small interfering RNA). RNA interference can defend cells against parasitic nucleotide sequences and also aid in development of gene expression as well. In eukaryotes, RNAi is initiated by the Dicer enzyme which cuts long double stranded RNA molecules into fragments of 20 nucleotides, whwhichcih is

called block translation. These are called siRNA. The siRNA is degraded, which allows for post transcriptional gene silencing, or “turns off” gene expression. A. Transform E. coli bacteria with a recombinant DNA plasmid containing the GFP gene to alter the phenotype of bacteria cells under UV light B. Analyze the results of simple DNA microarray to compare gene expression between two different environmental conditions, cell types, or organisms

Unit 11 Targets Immune System

I Explain how barrier defenses prevent pathogens from entering body: skin, mucus membranes, ciliated cells, & secretions (lysozymes of tears & saliva, ear wax, stomach acid, oil, & sweat) External nonspecific defenses skin: prevents microbes from obtaining needed water and nutrients;; prevents microbes from entering the body sweat/tears/saliva etc.: secrete acids and natural antibiotics that inhibit the growth of bacteria;; also traps bacteria mucus membranes: have anti-­bacterial enzymes that desitroy the cells walls of bacteria and trap bacteria cilia (ciliated cells): are on the mucus membranes;; they sweep up mucus that has trapped microbes and the mucus is coughed up, sneezed out, or swallowed II Explain how phagocytic leukocytes attack pathogens that enter the body: neutrophils, natural killer (NK) cells, macrophages, and eosinophils. Internal nonspecific defenses neutrophils: phagocytic white blood cells that travel thorugh capillaries to wounds;; (most numerous) macrophages: also phagocytic;; engulf and digest bacteria;; BIG eaters natural killer cells: type of white blood cell;; (compared to killer T cells, natural killer cells are NONSPECIFIC);; do not engulf invading microbes, but they identify other body cells that have been invaded by viruses;; they bind to the compromised cell’s membranes and release proteins that pierce holes in the membrane;; infected cells lyses;; in the end natural killer cells destroy virus-­infected cells and cancer cells eosinophils: white blood cells easily stained by eonsin;; important thing is that they kill parasites

III Describe the events of an inflammatory response including the roles of: histamine, mast cells, interferons, cytokines, and pyrogens.

a foreign object penetrates the skin and passes the external nonspecific defense bacteria is released and mast cells and other cells are damaged damaged mast cells release histamine

histamine causes increased blood flow (redness) and causes capillaries to dilate and become more permeable (leaky)

lymphocytes and macrophages in the capillaries can leak out chemokines (chemical signal) is also released;; there is a high concentration at the

area of injury;; provides a location or “road map” for the macrophages and lymphocytes to follow to attack the bacteria

phagocytic cells finally digest pathogens Fever may occurs Fever: certain white blood cells, in response to an infection, release hormones called pyrogens to increase body temperature

-­pyrogens increase activity of white blood cells -­fever increases production of interferon: travels to others cells and increases

resistance to viral infection IV Describe the events of a humoral (antibody-­mediated) response pathway. Explain the roles of B-­cell receptors called antibodies (a.k.a. immunoglobulins-­Ig), B-­Cells, plasma cells, Macrophages, Lymphokines (Cytokines), and memory B-­Cells. 1) First exposure to antigen 2) eaten by macrophage and displayed on the body of the macrophage 3) helper T cell binds to the antigen on the macrophage

4) B cells attach the the helper T cells and this causes the B cells to replicate 5) some become memory B cells that will remember to produce the antibody in case of a second infection and some become plasma cells that produce specific antibodies that target the antigen-­-­the antibodies bind to the antigen and neutralizes it and marks it for consumption of the macrophage *lymphokines(cytokines) are produced by t-­cells that attract macrophages and other lymphocytes V Describe the events of a cell-­mediated response pathway. Explain the roles of Helper T-­Cells, Killer (Cytotoxic T) Cells, T-­Cells, Macrophages, Antigens, Lymphokines (Cytokines), and memory T-­Cells. 1) First exposure to antigen 2) eaten by macrophage and displayed on the body of the macrophage 3) helper T cells binds to the antigen on the macrophage 4)helper t cells then activate cytotoxic T cells (killer cells) 5) these bind to infected cells and secrete the chemical perforin that eats away at the cell walls of the infected cell 6) memory T cells also are formed which will later be activated in case of a second infection and will be able to kill the infected cells VI Compare and contrast humoral (antibody-­mediated) immunity and cell-­mediated immunity. **Text with background of pink means the same in both humoral and cell-­mediated

Humoral response Similarities Cell-­mediated response

·∙ Produces antibodies to attach to antigens ·∙ Attacks pathogens directly in blood or humors mostly involves B

cells B cells differentiate

into plasma cells and memory B cells

·∙ activated by helper T cells using cytokines

·∙ helper T cells can be activated by antigen-­presenting cells

·∙ both use CD/MHC molecules to bind to other cells

·∙ in the cell-­mediated response, the infected cell has an mhc 1 molecule which displays

·∙ cell mediated will destroy the infected cells

·∙ cell-­mediated attacks already infected cells mostly involves

cytotoxic killer T cells when normal cells

become infected, it presents the antigen on the cell surface for recognition of the

an antigen then, the cytotoxic t cell sees the antigen displayed by the mhc molecule so it binds to the infected cell with the help of cd8 and destroys it

·∙ in the humoral response the b cell can display an antigen with an mhc 2 molecule and then the helper t cell can bind with the b cell with a cd4 and this binding activates the b cell to replicate and turn into the plasma and memory cells

cytotoxic killer T cells cytotoxic killer T cells

bind onto the antigen perforin is released

to cause cells to lyse by creating pores in the cell

memory t cells can also be created

VII Describe how specific immunity can be passive or active, and natural or artificial. Address vaccines and immunity passed from mother to baby.

Active Passive

Natural Exposure to a disease leading to antibody creation

Immunity acquired by a newborn baby from a mother via breast-­feeding

Artificial Vaccine injection with antigen, immune system produces antibodies

Injection of already-­made antibodies to defeat a pathogen

VIII Apply the concepts of immunity to the following: blood type, organ transplantation, allergies, autoimmune disorders, and HIV infection.

autoimmune disorder: when your immune system can’t distinguish between self and non-­self cells, so self cells are destroyed (targeted) in addition to non-­self cells

ex. lupus, rheumatoid arthritism diabetes, multiple sclerosis allergies: when your immune system overreacts to harmless non-­self cells as being

dangerous, esp. environmental antigens stimulates release of histamine

matching compatible blood groups (types) is critical for blood transfusions because a person produces antibodies against foreign blood antigens

Type O = universal donor, Type AB = universal acceptor HIV: AKA Human Immunodeficiency Virus

virus infects helper T cells -­-­> helper T cells don’t activate rest of immune system (T cells and B cells)

organ transplantation: crucial for MHCs AND blood type to match so new cells aren’t targeted by immune system

IX Explain how bacteria use quorum sensing to establish an infection. Be sure to include: cytokines, inducer, receptor, concentration gradient, positive feedback, operon, gene regulation

Bacteria use quorum sensing to communicate and make sure there is high enough density of bacteria to produce genetic products

quorum sensing allows bacteria to know how many bacteria cells are present process of quorum sensing:

bacteria produce small signal molecules or cytokines once a concentration gradient of the cell realizes that a high density of the

bacteria are present this will lead to the cytokines attaching to the receptors, activating the

operons, and the gene regulation will begin. so it can lead to positive feedback of more of the cytokines or pathogens being created

X Describe the lifecycle of a T-­4 bacteriophage virus. Discriminate between the lysogenic cycle and the lytic cycle. Explain how one cycle can shift to the other. http://biology.kenyon.edu/HHMI/Biol113/2virus.htm

Cell can go from lysogenic cycle to lytic cycle if the prophage separates from the host genome

Lysogenic Cycle Lytic Cycle

“dormant” stage virus incorporates DNA into the

genome of the host cell new DNA replicates

phage rapidly replicates host cell eventually bursts (lyses) viruses are released to infect other

cells

Cell Signaling (Nervous System & Hormones) Unit 12 Targets

I Name & state parts and functions of a neuron (include the following: dendrites, axon, cell body, axon hillock, synaptic terminals, nucleus). Distinguish the difference in functions between a myelinated and unmyelinated neuron.

dendrites: branching structure of a neuron that receives messages

axon hillock: cone shaped region where the axon joins the cell body

synaptic terminals: bulb at the end of an axon where neurotransmitters are stored and released, this is where synapse

occurs axon: long extension of a neuron that carries nerve impulses away from the cell body to

target cells cell body: part of neuron that hosts nucleus and other organelles nucleus: contains the genetic info regarding the neuron/cell schwann’s cells: cells that produce the myelin and are located within the myelin sheath myelin sheath: found surrounding the axon, electrically insulates cells, allows for saltatory

conduction node of ranvier: gap in myelin sheath, where action potential occurs unmyelinated neurons are found when the neuron doesn’t have to travel a very long

distance, ex: gray matter of brain II Explain how a chemical signal originates from a neuron, crosses a synapse, and becomes an action potential along the next neuron. Include the following terms: synaptic terminals, synapse, [Ca2+], dendrites, vesicles, exocytosis, “All or nothing”, neurotransmitter, synapse, threshold, stimulus, impulse transmission, action potential, resting potential, depolarization, sodium-­potassium pumps, [K+], [Na+], gated ion (voltage-­gated, ligand-­gated) channels, mV units, refractory period Part of the Central Nervous System (CNS), neurons go through synapses to signal each other and other cells in the body. These are the steps of synapse:

1. The process begins with a wave of electrochemical excitation called an action

potential traveling along the membrane of the presynaptic cell, until it reaches the

synapse.

2. The electrical depolarization of the membrane at the synapse causes channels to

open that are permeable to calcium ions.

3. Calcium ions flow through the presynaptic membrane, rapidly increasing the calcium

concentration in the interior.

4. The high calcium concentration activates a set of calcium-­sensitive proteins attached

to vesicles that contain a neurotransmitter chemical.

5. These proteins change shape, causing the membranes of some "docked" vesicles to

fuse with the membrane of the presynaptic cell, thereby opening the vesicles and

dumping their neurotransmitter contents into the synaptic cleft, the narrow space

between the membranes of the pre-­ and postsynaptic cells.

6. The neurotransmitter diffuses within the cleft. Some of it escapes, but some of it binds

to chemical receptor molecules located on the membrane of the postsynaptic cell.

7. The binding of neurotransmitter causes the receptor molecule to be activated in some

way. Several types of activation are possible, as described in more detail below. In any

case, this is the key step by which the synaptic process affects the behavior of the

postsynaptic cell.

8. Due to thermal shaking, neurotransmitter molecules eventually break loose from the

receptors and drift away.

9. The neurotransmitter is either reabsorbed by the presynaptic cell, and then

repackaged for future release, or else it is broken down metabolically.

III Distinguish between and provide an example of an EPSP and an IPSP and relate them to the concepts of spatial and temporal summation. EPSP’s, or excitatory postsynaptic potentials, involve the depolarizaton of a postsynaptic neuron by the flow of positively charged ions into the postsynaptic cell as a result of opening of ligand-­gated ion channels. EPSP’s make the charge of the neuron more positive from the resting potential of -­70 mv to trigger an action potential. On the contrary, IPSP’s, or inhibitory postsynaptic potentials, are when the postsynaptic neuron is hyperpolarized due to the flow of negative ions into the cell or positive ions out of the cell. IPSP’s make the neuron more negative to prevent (inhibit) an action potential.Unlike the “all or nothing” of action potentials, postsynaptic potentials differ by strength. ESPS’s have varying positive values and ISPS’s are negative values.Their magnitudes are affected by the amount of neurotransmitters released at the chemical synapse. Therefore, usually more than one EPSP is needed to stimulate an action potential. The sum of the postsynaptic potentials must reach a certain threshold to cause an action potential. One way this occurs is through temporal summation, which is when two ESPS’s arrive at the postsynaptic terminal at different times, but one follows shortly after the other and they combine to depolarize the neuron to the threshold level. Another way of triggering an action potential is through spatial summation. This is when ESPS’s arrive simultaneously

through different synapses and combine to produce a more positive charge. ISPS’s can counteract the sum of the EPSP’s by bringing a more negative charge. Examples:

EPSP-­ Glutamate, an amino acid, is the main excitatory neurotransmitter in the central nervous system of vertebrates.

IPSP-­ GABA is the most common neurotransmitter associated with IPSPs in the brain.

Diagram:

IV Describe how the reflex arc maintains homeostasis. (include the following: sensory neuron, motor neuron, interneuron, spinal cord, brain, effector, receptors, reflex response

An explanation A of A the reflex arc: A reflex arc occurs in order to maintain homeostasis. When we try to maintain homeostasis without thinking, it is called a reflex. Nerve cells initially carry the message from stimulated receptors to the correct effectors.In this example, the sensory neuron directly connects with the motor neuron from the spinal cord, in a traditional reflex arc, the sensory neuron sends its information down to the CNS which then the sends the information to the effector via the motor neuron. V Describe how the nervous system components work together to maintain homeostasis. Define the functions of the CNS and PNS (include the following: brain, spinal cord, autonomic nervous system, somatic nervous system, sympathetic nervous system, parasympathetic nervous system, interneurons, motor neurons, sensory neurons).

Central Nervous System (CNS): includes brain and spinal cord;; neurons that carry out integration are organized here

includes: interneurons: association neurons that form synapses with sensory

and/or motor neurons and integrate sensory input and motor input-­only in the CNS (brain)

Peripheral Nervous System (PNS): comprised of nerves and ganglia (cluster of groups of neurons);; consists of the neurons that carry information into and out of the CNS

neurons in PNS include: sensory neurons: convey information from eyes and other sensors that

detect stimuli (light, sound, touch, heat, smell) to the spinal cord motor neurons: respond to signals from sensory neurons and transmit

signals to muscle cells, causing them to contract PNS composed of: autonomic nervous system: regulates the internal environment by controlling

smooth and cardiac muscles and the organs of the digestive, cardiovascular, excretory, and endocrine systems

composed of: sympathetic nervous system: activation of this corresponds to arousal

and energy generation (“fight or flight” response) parasympathetic nervous system: activation of this generally causes

opposite responses that promote calming and a return to self-­maintenance functions (“rest and digest”

somatic nervous system: (aka motor system) consists of neurons that carry signals to skeletal muscles, mainly in response to external stimuli

VI Describe how the nervous and endocrine systems are linked in vertebrates (use the terms: hypothalamus, neurosecretory cells, nerve impulses, hormones, anterior pituitary,

posterior pituitary). hypothalamus: plays a central role in integrating the endocrine and nervous systems;;

endocrine gland in brain that receives information from nerves throughout body and other parts of the brain

anterior pituitary: develops at roof (look at 45.16 45.15 and 45.14)

VII Distinguish between the mechanisms employed by protein hormones (use the terms: water-­soluble, plasma membrane, signal receptor, second messenger, enzyme cascade, signal transduction, nucleus cytoplasmic response OR gene regulation/cytoplasmic response, “short distance”) and lipid hormones (use the terms: lipid-­soluble, steroid, plasma membrane, signal receptor, nucleus, gene regulation/cytoplasmic response, “long distance”) to cause changes in a target cell. Endocrine glands make hormones, which are stored in the gland and secreted when signaled. The hormones are then transported to their target cell by the bloodstream;; the target organ has receptors for a specific hormone. If the hormone is a protein/amine/peptide hormone (water-­soluble), the hormone binds to a plasma membrane signal receptor and causes signal transduction;; the binding of hormone to receptor triggers an enzyme cascade of events involving the synthesis of cyclic AMP (cAMP) as a short-­lived second messenger. Thus, water-­soluble hormones that bind to their target cell’s cell membrane receptor can cause a nucleus cytoplasmic response or a

gene regulation/cytoplasmic response. If the hormone is a steroid hormone (lipid-­soluble), the hormone crosses the cell and nuclear membrane and goes into the nucleus of its target cell, which causes a direct change in gene expression. Thus, lipid-­soluble hormones that bind to their target cell’s cell receptor within the cell can cause a gene regulation/cytoplasmic response.

VIII Describe the antagonistic relationship between the hormones glucagon and insulin. Explain where they are produced, where they are stored, the target organs, and the mechanism for how the hormones interact with target cells. Be sure to include how negative feedback is involved and the relationship of these hormones to homeostasis in the blood Insulin is produced when blood sugar levels are higher than normal. The beta cells of the pancreas produce the insulin into the bloodstream. The insulin signals all organs of the body (except the brain) and cells to absorb the sugar and use it for energy for cellular functions. On the other hand if blood levels are low glucagon is produced through the

alpha cells of the pancreas. They signal the cells of the liver to break down glycogen (with was stored from food consumed and is a carbohydrate) into glucose and release it into the blood to increase blood sugar level. The are both hormones with negative feedback. When blood sugar goes up/down then hormone is produced to counteract that. They are also opposites of each other and are called antagonistic.

IX. Describe the antagonistic relationship between the hormones PTH and calcitonin. Explain where they are produced, where they are stored, the target organs, and the mechanism for how the hormones interact with target cells. Be sure to include how negative feedback is involved and the relationship of these hormones to homeostasis in the blood. PTH is a peptide hormone produced in the parathryroid gland (behind the thyroid), while calcitonin is a peptide hormone produced in the thyroid gland. Homeostasis is when the blood calcium (Ca2+) level is about 10 mg/100 mL. When a stimulus results in a falling blood calcium level, the parathyroid gland releases the parathyroid hormone (PTH) into the bloodstream;; PTH stimulates calcium release from the bones, and the blood calcium level rises. Then, as the blood calcium level increases (when a stimulus results in a rising blood calcium level), the thyroid gland releases the calcitonin hormone into the bloodstream;; calcitonin stimulates calcium uptake from the bones, and the blood calcium level falls. Thus, PTH and calcitonin share an antagonistic relationship because PTH and calcitonin have opposite effects on the release of the “other” substance. Negative feedback is involved in this process because if PTH is high, then blood calcium will increase, which will signal the parathyroid gland to secrete less PTH. Likewise, if calcitonin is high, then blood calcium will decrease, which will signal the thyroid gland to secrete less calcitonin. X. Describe the positive feedback hormonal relationship at work in breast feeding and child birth. Include discussion of oxytocin, estradiol, prostaglandins. Estradiol (a steroid hormone) is produced in the ovary. It signals the oxytocin receptors on the uterus-­-­-­oxytocin is simultaneously being produced from the posterior pituitary gland. This signaling of the receptors in the uterus and the production of oxytocin the posterior pituitary gland 1) makes the uterus contract 2) makes more placenta to make

prostaglandins. Prostaglandins in turn stimulate more contractions of the uterus. And these contracts cause more oxytocin to be produced. This is an example of Positive Feedback. The suckling of the baby on the mother will stimulate the hypothalamus posterior pituitary gland to secrete oxytocin. The oxytocin smoothes the muscle in the breasts (target cells) which in turn allow milk to be produced and then the production of milk will once again encourage the child to suckle more. This is another example of Positive Feedback.

PROCESS-­BASED TARGETS A. Design and perform a nervous system experime