warrior final review

7/30/2019 Warrior Final Review

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Cell Division

Dewey Nguyen


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Mitosis somatic/body or boring cell

division!

Two identical daughter cells, just like parent

cell

Diploid, 2n, two copies of each chromosome


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Interphase- Prep phase for mitosis

G1 make organelles and producemacromolecules

SDNA is decondensed form of

chromatin,DNA replication

G2 additional growth


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The 5 Mitotic Phases - Prophase

Formation of mitotic spindles

Centrosomes are moving to opposite ends

Sister chromatids are condensed


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Prometaphase

Nuclear envelope is broken down

Kinetochore microtubules are attached tocentromeres

Non-attaching kinetochore microtubuleselongate the cell


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Question

The ____ end of the kinetochore microtubules

requires _______ to pull the chromosomes

toward the plasma membrane.

a) Positive, dynein

b) Postive, kinesin

c) Negative, dyneind) Negative, kinesin

e) Actin, myosin


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The final steps

Metaphase plate formed in metaphase

Anaphase sister chromatids are separated

Telophase and Cytokinesis cleavage using

actin filaments and myosin, nuclear envelopeformation


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Question

Sister chromatids would first be seen at this

stage:

a) G1

b) S

c) Prophase

d) Metaphasee) B and C


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Question

A drug is found to disrupt the function of kinesin,

how would this drug affect mitosis?

a) The cell cannot elongate for cell division

b) Sister chromatids cannot be separated

c) Cytokinesis cannot take place

d) A and B

e) None of the above


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Oh Meiosis - the sex cell division

Four different daughter cells

Haploid, n, one copy of each chromosome

All stages are very similar to Mitosis


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Oh Meiosis

Metaphase I homologous chromosomesformed tetrads and crossing over occurs here!

Independent assortment- Metaphase I and II

increases genetics variation


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Sister homologous what!?!?


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Oh I get it!

Sister chromatids are the exact same thing!

Homologous chromosomes carried thedifferent variations of the same gene!

Non-homologous chromosomes are differentchromosomes!


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Crossing-Over is what?!?!

Yes, crossing-over is chromosomal sex, where

genetic materials are exchanged! So sister

chromatids cannot do it together! Only do it

with the same chromosome or homologouschromosome!

Must hold hand first, so tetrad are formed,

and then Chiasmata! Lets take a closerlook!


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So this is how they do it!


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Mendel

Law of segregation

Two alleles will separate

into two different

gametes

Law of independent

assortment

When two or more

characters are inherited,each alleles separated into

gametes independently

Wait what is the point of these two law?

- To create more variations!


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Test Cross

To test if a phenotype is homozygous

dominant or heterozygous dominant.

Cross with a homozygous recessive


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Non-Mendelian Inheritances

Incomplete dominance Co-dominance

Multiple Alleles

Pleiotropy

Epistasis

Polygenic inheritance Environmental factors/impacts


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Non-Mendelian Inheritances

Lets say R = red, and r = white. So Rr would

yield:

Incomplete dominance

Co-dominance


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Recombination

Map units dont add up!

Offspring show different phenotype fromparents!


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Abnormalities

Karyotopic Abnormalities having an

abnormal number of chromosomes

Chromosomal aberrations a part of a

chromosome is messed up!

Only translocation can involve two chromosomes


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Question

Your father is color blind, and your sister isalso color blind, but you are not. What is theprobability that you will have a diseased

brother?a) 0

b) 1/2

c) 1/4

d) 1/8



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Question

Your father has achondroplasia, and yoursister also has achondroplasia, but you arenot. What is the probability that you will have

a diseased brother?a) 0

b) 1/2

c) 1/4

d) 1/8



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Question

A female duck has a mosaic feathers, what

could be one explanation for this?

a) Barr body

b) Incomplete dominance

c) Co-dominance

d) Epistasise) More than one possible answer


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Bio Sci Peer TutoringFINAL Review: Bio93 Sec A+B

Arun Manmadhan

UNIT VI: Molecular Biology


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Frederick Griffith


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So what did this show/mean??

S (Smooth) cell = DEADLY smooth slimy coat gives the bacteria protection

R (Rough) cell = Harmless

Heat killed S cell = Harmless The cell is killed but not the DNA

Heat killed S cell + R cell = DEADLY

DNA from heat killed S cell transforms the R cell

into an S cell KEY POINT: Showed that there exists a

transforming principle but not necessarilyDNA, RNA or Protein


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Showed that DNA was the transforming agent

and thus the unit of hereditary.

Phage = a virus

SULFUR is FOUND in Protein but NOT inDNA

Phosphorous is FOUND in DNAbut NOT inProtein

Alfred Hershey and Martha Chase

Experiment


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Rosalind Franklin, Watson & Crick

Rosalind Franklin used X-ray crystallography toget a real picture of DNA

Watson and Crick then analyzed the picture to

come up with a structure that FIT the data Basic Postulates:

1. DNA is LINEAR

2. DNA base pairing rules

A=T, C G

3. HELICAL structure


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Properties of DNA

Purines = Guanine (G) and Adenine (A)

Pyrimadines = Thymine (T) and Cytosine (C)

Phophate group = 5 end

Sugar group = 3 end DNA is anti-parallel one strand = 53

the other = 35

DNA replication is SEMI-CONSERVATIVE

DNA is replicated with DNA POLYMERASE


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One parental isPreserved, the second strand

(red) is new

Semiconservativemodel:Type of DNA

replication in which thereplicated double helixconsists ofone oldstrand, derived fromthe old molecule, andone newly made strand

Definition G-33


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Question

In the Hershey and Chase experiment with thebacteriophages, the most important clue tothe chemical nature of the hereditarymaterial was the:

a)incorporation of radioactive sulfur in the bacterialgenome

b)incorporation of radioactive phosphorus in the bacterialgenome

c)transformation of R cells to S cells via heat killed S cells

d) accumulation of phosphorus on the surface of thebacteria


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Question

If one strand of a DNA molecule has the basesequence 5-ATTGCAT-3, its complementarystrand will have the sequence:

A) 5-ATTGCAT-3

B) 5-ATGCAAT-3

C) 5-TAACGTA-3D) 5-CGGTACG-3

E) None of the above


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Solution Sequence given:

5-ATTGCAT-3

3-TAACGTA-5

5-ATGCAAT-3

This is the complementary strandBut since all options are written53 flip it!

Same as choice B


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The Central Dogma

DNA DNA

RNA=Transcription

mRNA 5 cap Poly A Tail RNA Splicing

Protein RNA

Protein =translation

Lecture 22


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Prokaryotes vs Eukaryotes

Prokaryotes Eukaryotes

MORECOMPEX!


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Transcription

Initiation: Promoterregion + transcriptionfactors

Promoter region =DNA sequences whichprovide a binding sitefor RNA polymerase

TranscriptionFactors= recruit RNApol to the promoterregion


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Transcription continued

Elongation = RNApolymerase adds RNAnucleotides until it hits atermination spot

Termination = RNApolymerase disassociatesfrom the template DNA

strand and primary transcriptis released to be processed.


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QuestionIn order for a particular drug to directly inhibit

RNA polymerase activity in animals it must

have:A) Non-polar functional groups

B) Nuclear import signal

C) Compatible transcription factorsD) Complementary sequences to the promoter

E) None of the above


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QuestionPromoters for eukaryotic mRNA synthesis:

A) are more complex than prokaryotic promoters

B) can require binding of multiple transcriptionfactors to form a transcription complex

C) have specific DNA sequences such as the"TATA" box that are recognized by proteins

D) are the stretches of DNA to which RNApolymerase binds to initiate transcription

E) All of the above


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Everythingoccurs

in thecytoplasm

Transcription

+ RNAprocessing intheNUCLEUS

Translationin

CYTOPLASM

Bacteria!

YOU &ME!!


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RNA processing 5 cap added - Capping ensures the messenger

RNA's stability while it undergoes translation

Mostly Guanine (G) repeats

Poly A Tail - important for the nuclear export,translation and stability of mRNA. The tail isshortened over time and when it is short enough,the mRNA is enzymatically degraded.

Adenosine (A) repeats


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RNA splicing

exon intron Exon Intron Exon Intron

Exon Exon Exon

Introns REMOVED

INTrons = INTerfere

5 cap Poly ATail

Primary Transcript

Mature Transcript

+ 5cap + poly A Tail


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Genetic Code

Memorize!!!: Start Codon: AUG Stop Codons: UAA, UGA, UAG

The genetic code is read in triplets (codons)

It is non-overlapping The genetic code is unambiguous

Each codon specifies a particular amino acid, andonly one amino acid.

The genetic code is degenerate. each amino acid can be specified by more than one

codon.

The code is nearlyuniversal.


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QuestionWhich of the following is NOT a feature of eukaryotic

gene expression?

A) Random repeat segments at mRNA polesB) many genes are interrupted by non-gene coding

DNA sequences

C) RNA synthesis and protein synthesis are coupled

D) mRNA is often extensively modified beforetranslation

E) Addition of a nuclear export tag to mature mRNA


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Translation important points tRNA molcules are the translators

Purpose??? MAKE PROTEIN, PROTEIN, ANDMORE PROTEIN!!!!!

PossesANTI-CODONS to the CODONS found in

mRNA Thus if mRNA has codon: 5-AUG-3, tRNA has

the complementary anti-codon: 3-UAC-5

Ribosomes read mRNA in 53 direction

E site: Exit site, tRNAs leave P site: Peptidyl site, amino acid chain lives/

grows here

A site: Acceptor, temporary home for tRNAs


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What does that spell?! E-P-A

Environmental ProtectionAgency


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Mutations Oh Nooooo! Base pair substitution

Types: Silent: mutation has no effect on amino acid

added b/c genetic code is DEGENERATE

Missense: mutation changes one amino acidbecause of mutation

Nonsense: mutation results in a STOP codon,prematurely cuts the amino acid chain

Normal Wildtype Mutant

5-AUGAAGUUUGGCUAA-3 5-AUGAAGUUUGGUUAA-3


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Mutations continued

If mutations add or remove nucleotides inanything BUT multiples of three, the mutation isa FRAMESHIFT

Wildtype:

THE|FAT|CAT|ATE|THE|WEE|RAT

Mutant (frameshift mutation): THE|FAT|CAT|ATE|THZ|ZWE|ERA|T

Insert ONE or TWO Nucleotides


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QuestionA nonsense mutation generally results in:

A) Translation of incorrect amino acids

B) a shorter polypeptide

C) no change in expressed protein

D) a frameshift


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QuestionWhich of the following is NOT correct?

A) There are sixty-four different codons.

B) All codons specify a specific amino acid.

C) Some codons are used for initiation or

terminationD) There are more codons than amino acids

E) All of the above are correct


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DNA Cloning In Bacteria

First Step: IsolateDNA/Gene of interest

Second Step: Insert

DNA into a PLASMID

Third Step: AllowBacteria to incorporateDNA and multiply

Fourth Step: Observe/Extract protein of interestsynthesized


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Cutting DNA

You need palindrome sequences IE racecar Read from both directions = same word

Restriction enzymes cut DNA at thesepalindrome sequences

Create Sticky Ends Once DNA has been inserted into a plasmid,

covalent bonds are reformed using DNA Ligase


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PCR Reaction Start with one piece of

DNA

Can amplify and duplicatemillions and billions ofcopies in a few hours.


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Gel electrophoresis

Purpose: Separate DNA sequencesBased on size.

Allows for identification of certainSequences of DNA.

DNA is negatively charged, so movesTo positive end. Smaller DNAMolecules feel less resistance in the

Gel and thus move a greater distance.

What makes DNANEGATIVELYCharged????


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Continued (Copied from your lecture

notes) KNOW THIS!

1. A SMALL fraction of the genes in the genomeare expressed in any given cell

2. Gene expression can also be regulated at manyother steps such as

1. RNAprocessing,

2. RNAtransport,

3. RNAstability/degradation,4. RNAtranslation,

5. Proteinmodification

6. Proteinstability/degradation.


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Continued

Transcription Factors: Tell the cell/polymerasesWHEN andWHERE transcriptioncan occur

Enhancers: Work with transcription factors andthe promoter region to stimulate transcription.

Micro RNA (miRNA) attach to MATUREmRNA and either DEGRADE or PREVENTTRANSLATION


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Gene Expression

So MANYcheck points that can control how the gene is expressed


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Question

Which of the following tools of recombinant DNAtechnology is INCORRECTLY paired with one of itsuses?

A) restriction enzyme - production of DNA fragmentsfor gene cloning

B) DNA ligase - enzyme that cuts DNA, creating stickyends.

C) DNA polymerase - copies DNA sequences in thepolymerase chain reactionD) Electrophoresis - Detect DNA lengthE) Antibiotic resistance - detect if bacteria

transformed


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Question

Which of the following is not part of the normalprocess of cloning recombinant DNA in bacteria?

A) restriction enzyme digestion of cellular andplasmid DNAs

B) production of recombinant DNA using DNA ligasesand a mixture of digested cellular and plasmidDNAs

C) separation of recombinant DNAs by

electrophoresisD) transformation of bacteria by the recombinant

DNA plasmidsE) All of the above are valid processes


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Time permitting questions

The regions of DNA in a eukaryotic gene thatencode a polypeptide product are called:

A) promoters

B) Exons

C) Introns

D) tRNA's

E) Transcription factors


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Question

Three types of RNA involved in comprising thestructural and functional core for proteinsynthesis, serving as a template for translating,and transporting amino acid, respectively are:

A)mRNA,tRNA,rRNA

B)rRNA,tRNA,mRNA

C)tRNA,mRNA,rRNAD)tRNA,rRNA,mRNA

E)rRNA,mRNA,tRNA


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Question

A messenger acid is 336 nucleotides long,including the initiator and termination codons.The number of amino acids in the proteintranslated from this mRNA is:

A) 999B) 630C) 330D) 111E) 112


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Question

A key feature for the identification of plasmidscontaining recombinant DNA is *hint white vsblue colonies):

A) weighing it

B) the DNA sequencing of recombinant plasmids

C) the production of restriction maps ofrecombinant plasmids

D) introns can be moved to new locations withinthe gene

E) the disruption of a gene on the plasmid by theinserted recombinant DNA

Recommended Animations To Watch


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Recommended Animations To Watch

(copy and paste into browser each link

separately) http://nortonbooks.com/college/biology/animations/ch12a01.htm

http://highered.mcgraw-hill.com/olc/dl/120076/bio21.swf

http://www.sumanasinc.com/webcontent/animations/content/meselson.html

http://bcs.whfreeman.com/thelifewire/content/chp12/1202001.html

http://www-class.unl.edu/biochem/gp2/m_biology/animation/gene/gene_a3.html

http://bcs.whfreeman.com/thelifewire/content/chp14/1402001.html

http://highered.mcgraw-hill.com/sites/0072556781/student_view0/chapter11/animation_quiz_4.html

http://highered.mcgraw-hill.com/sites/0072556781/student_view0/chapter14/animation_quiz_1.html


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Developmental Biology

Kevin Huang

M, F 2-3 PMSH 149

[email protected]

D l t l Th
mailto:[email protected]:[email protected]:[email protected]


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Developmental TheoryA combination of Cytoplasmic Determinants and Cell Induction

Cytoplasmic Determinants:are present in egg(maternalinheritance) before fertilization.When zygote divides, differentdeterminants yield different celllineages.

My interpretation of CD's: essentially establishthemselves during early divisions of cleavage(You likely don't have to know that bicoid is anexample of C.D.'s but it doesn't hurt for me totell you)

Cell induction: the result of

intercellular communication.Signals from one cell influencethe fate of another cell.

My interpretation of C.I.: Common examples thatyou guys see cell induction could extend tothe actions of organizers which secrete

signals (morphogens) to influencedifferentiation!

Pattern formation vs Positional


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Pattern formation vs. PositionalInformation

Pattern formation: theestablishment of axes A/P, D/V, P/D(determined by maternalcytoplasmic determinantspresentin the egg before fertilization).Howdoes this make sense? Well, you

know bicoid is a C.D. anddetermines A/P polarity!

Positional Information: directspattern formation by givingpositional values to cells.

Dumbing it down: positionalinformation --> This definition likelydoesn't make senseso think of itin terms of how morphogens work.Morphogens (like bicoid) yield aconcentration gradient. Differentlevels of bicoid yield different celltypes. HENCE, DIFFERENT


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Drosophilaand bicoid

In drosophila, cytoplasmicdeterminants are present in theegg before fertilization.Fertilization results activation ofthe egg.

The resulting activation initiatesa bicoid gradient across theembryo (highest at anterior,lowest at posterior)

High [bicoid] = head

Intermediate [bicoid] = thorax

Bicoid acts as a morphogen


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Embryonic Development - Overview

Fertilization

Cleavage

Gastrulation

Organogenesis

Fusion of haploid egg/sperm toform a diploidzygote (influx ofCa2+ results in eggactivation)

Rapid cell divisions without anincrease in overall size. Thuswith each successive cleavagedivision, each cell becomessmaller.

Cell movement, forms germlayers (ectoderm, mesoderm,endoderm)

formation of organs (formationof neural tube to become future

spinal cord/brain)

Fertilization


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e t at ohaploid sperm + haploid egg = diploid zygote

- Entry of sperm results in an influxof Ca2+ into the egg.

- This results in increased proteinsynthesis, etc.

- Egg is activatedat this point.

Acrosomal rxn: there are specificreceptors for the sperm. Thisessentially prevents interspeciesfertilization

Cortical rxn: essentially prevents

polyspermy by blocking the entry ofadditional sperm.

(acrosomal rxn occurs before corticalrxn)

Cl


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CleavageThe initial divisions of the embryo

- Rapid cell divisions that proceedafter fertilization. Involvesduplication of organelles and DNAbut no overall growth!!!

- Form of mitosis (once again: NOGROWTH)

- Morula is formed during cleavage.At the end of cleavage a blastula iscreated.

- the blastula has a cavity within itcalled the blastocoel

G t l ti


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GastrulationFormation of germ layers

- Dorsal lipinvaginatesinto theinterior of the embryo in a processcalled involution.

- The blastocoel shrinks and in itsplace another cavity called thearchenteron is present

- Blastopore lip encircles the yolk plug

- Generates the 3 germ layers:

- Ectoderm (outer)

- Mesoderm (middle)

- Endoderm (inner)

Because not a lot of background information isgiven to you guysRemember that theblastocoel is characteristic of the blastula,and the archenteron is characteristic ofgastrulation.

Know the 3 germ layer fates


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Know the 3 germ layer fates(general, not super specific!)

Ectoderm

Mesoderm

Endoderm

epithelial/outer covering like the skin (easy toremember)

Central nervous system (brain, spinal cord)How to remember this: You know that theneural plate is ectoderm, and receives signalsfrom the notochord to become FUTUREspinal cord/brain!!

Notochord (you know that the notochord ismesoderm and signals to the overlyingectodermneural plate)

Muscular, skeletal, circulatory

Gastrointestinal tract (looking at diagram onprevious slide, you see that endoderm is theprimary cell tissue type within the gastrula! Itessentially invaginates to become the gut)

Organogenesis


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Organogenesiscreation of organs

Germ layers (ectoderm, mesoderm,endoderm) develop into rudiments oforgans

Vertebrate organogenesis: neuralplate (ectoderm), notochord(mesoderm)

Mesoderm surroundingnotochord somites (whichsplit to become body cavity)

Neural

fold

Neural plate

Neural crest

Outer layer

of ectodermNeural crest

Neural tube

CNS Formation (Brain/Spinal Cord)


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CNS Formation (Brain/Spinal Cord)Organogenesis

Notochord (mesoderm)signals to overlying neuralplate (ectoderm) whichresults in the inward folding ofthe plate---forming the neuraltube (future brain/spinalcord)

Requires microtubules, actin

Somites


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SomitesKey to early embryonic segmentation

Not a lot is told to you guysabout somites

What I think you should know:

Somites = mesoderm

surrounding notochord Shows segmentation

Somites

HEAD

Neuraltube

Neural folds

Head mesoderm

Tail

Organizers


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OrganizersImportant stuff!

Organizers act sort of likecommand centers. Whereverthey are present, they enforcetheir demands

What do I mean by this?

They secrete factors (morphogens) thataffect gene expression insurrounding cells

If you graft organizers of particularregions into foreign regions, theorganizer will cause surroundingcells to change their destiny

Ex: you graft a foot organizerwhere hands normally form---youwill get feet for hands!

Organizers to know:

1) dorsal lip ofblastopore

notochord/neural tubeformation

2) apical ectodermalridge (AER)

Limb bud pattern formation

3) Zone ofPolarizing Activity(ZPA)

- limb bud pattern formation

AER/ZPA


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AER/ZPA

ZPA secretes Sonichedgehog which givespositional informationposterior to anterior.

AER is overlying ectoderm

ENGRAFTMENT OF NEWZPA YIELDS POSTERIORIDENTITY AT NEWLOCATION (notice in pic.That there are two pinkies,and two ring fingers :O)

Homeotic Genes vs Organizers


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Homeotic Genes vs. Organizersanatomic identity

What to know: HOMEOTIC GENES: Code

foranatomic identity (ex:tells a particular segmentyou are going to be a

hand) ORGANIZERS: secrete

signals that influence geneexpression in surroundingcells

Homeotic is at the GENElevel

Organizers at the CELLULARlevel

Tidbits you might want to be aware


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Tidbits you might want to be awareof....

Zygote (result of fertilization)

Morula (during cleavage divisions)

Blastula (the result of cleavage divisions)

Gastrula (result of gastrulation)

Neurula (result of neuralation)----this is known in our lectures as our exampleof ORGANOGENESIS....therefore know that the neural tube folding is duringorganogenesis!

Blastomeresare the individual cells that make up the morula and blastula

during cleavage...

Discussion Generation:


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The questions I present here are intended to broaden yourunderstanding and help see the bigger picture if development and how

they may apply to other topics in the class :)

Do organizers function as cytoplasmic determinants or cellinduction? What kind of signals to they secrete?

harder question: How do organizers affect homeotic geneexpression? Be as specific as possible!

Another harder question: How do organizers change theexpression of cells in the area in which an engraftment hasoccured?

Although the acrosomal rxn functions to prevent interspecies fertilization,sometimes it does not work indefinitely.


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Ex: the zorse (buahaha. . .)real example: a mule (horse + donkey)

A mule is formed by the mating of a horse and a donkey. Thismeans in this situation acrosomal rxn function failed, howeverthe species were compatible enough that an embryo wasviable enough to develop. If I told you the # of chromosomes ina horse was 64, and in a donkey 62, and in a mule 63---why do

you think mule are sterile (cannot mate)? Random and irrelevant fact:when I looked up these values it turned out to be these exact numbers...creepy.

Hint: this does not relate towards development, think cell cycle.

What could be the best way to define the sort of signaling that


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What could be the best way to define the sort of signaling thatcytoplasmic determinants and morphogens convene?

What if you could inhibit all mesoderm signaling to the ectoderm? Based


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off of what you know, what kind of long term effects can this have on theorganism?


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THE NERVOUS SYSTEM

Wendys fun-filled introduction toNeurons

Brains are cool! Girls love brains.

L b l th N


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Label the Neuron

Indicate where the concentration of the

following ions are highest in the neuron:

Na+, K+, Cl-, and anions-

L b l th A ti P t ti l


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Label the Action Potential

A:

B:

1:

2:

3:

4:


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Will the ion move in or out of the cell using thesechannels:

1. Voltage-gated Na+ Channels ___________

2. Voltage-gated K+ Channels ____________

3. Non-gated K+ Channels ______________

Label the Synapse


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Label the Synapse

A:

B:

C:

D:

E:

F:

G:

H:


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What part of the neuron relays signals from

one neuron to another neuron?

a. Dendrite

b. Axon hillockc. Synaptic Terminal

d. Axon

e. Node of Ranvier


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The part of the neuron that carries an impulsetowards the cell body is called ________.

a. A nerve

b. A neurotransmitter

c. An axon

d. A dendrite

e. White matter


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The membrane potential refers to:

a. The voltage inside a cell during an action potential.

b. The difference in voltage between the inside of the

cell and the outside of a cell

c. Ion gradients in excitable cells

d. Graded changes in polarization

e. The fact that neuronal cells must generate more

membrane when they grow


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An EPSP (excitatory postsynaptic potential)

facilitates depolarization of the postsynapticmembrane by:

a. increasing the permeability of the membrane toNa+

b. increasing the permeability of the membrane toK+

c. insulating the hillock region of the axon

d. allowing Cl to enter the celle. stimulating the sodium-potassium pump


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For each statement, determine if the neuron

membrane will be hyperpolarized, depolarizedor no change.

Binding of serotonin to ligand gated K+ channels__________________

After summation of inhibitory postsynaptic

potentials _______________


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What is the approximate threshold? _________

Which letter(s) represent inhibitory post-synaptic potential? ________Which letter(s) represent hyperpolarization? _______

The cell is depolarizing at letter: _________

Letter A is an example of? _________

Good Luck on All Your Finals!


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