biology prep

8/13/2019 Biology Prep

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I) Intro to GeneticsA) The field of Genetics concerns itself with investigating the transmission of hereditary

material(includes resemblance among relatives), the ways in which hereditary material affects

the phenotype, and an understanding of the genetic variation within and among populations.

B) For all life that we know about, the hereditary material is nucleic acid(usually DNA)1) DNA directs the production of specific proteins via RNA intermediates2) Proteins create or affect the phenotype.

II) DNA StructureA) DNA was confirmed as the hereditary material through experiments of Avery et al.(inspired by

early work of Griffiths); augmented by studies of Hershey and Chase

B) Structure of DNA determined by Watson and CrickC) DNA is a polymer of nucleotidesD) Nucleotides consist of

1) Pentagonal Sugar(deoxyribose)2) An attached base, a ring structure containing one or more Nitrogens

(a) Purines: Adenine and Guanine(i) Two interlocking rings

(b) Pyrimidines: Thymine and Cytosine(Uracil in RNA)(i) One hexagonal ring

(c) Sugar+base=nucleoside(d) Nucleoside+phosphate=nucleotide

3) One or more phosphates attached to the 5 carbon4) Hydroxyl on the 3 carbon

E) DNA is a double stranded, helical molecule1) Each Strand consists of deoxyribose sugars linked through phosphate

groups(Phosphodiester bonds)

2) Hydrogen bonding between bases hold two strands together(a) Complimentary base pairing

(i) A=T, C=G(b) Strands can be denatured by heat(melting)m but readily re-anneal at appropriatetemperatures

3) Strands are antiparallel4) Strands have a major groove and a minor groove5) Alternative forms include A-DNA, B-DNA, and Z-DNA

F) RNA is similar, but1) Uses ribose instead of deoxyribose2) Has Uracil instead of Thymine3) Usually occurs as a single strand rather than double

G) Some DNA research techniques which depend on structure1) Gel electrophoresis separates DNA molecules by size and/or structure2) Denaturation and annealing

(a) DNA hybridizationFISH is an example3) Nucleases cleave DNA, usually in a specific manner

(a) Restriction endonucleases are sequence specific(b) Single Strand nucleases are structure specific(c) Exonucleases cleave nucleotides from the ends of chains(d) General nucleases(DNase) are not specific

III) The GenomeThe total of all genetic material of an organism


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A) Consists of1) Prokaryotes

(a) One or several circular chromosomes(b) Optional plasmids(c) Typically supercoiled and bound to proteins to form nucleoid region

2) Eukaryotes(a) Linear Chromosomes

(i) Variable in size and placement of centromere(ii) Plasmids known but rare

(b) Mitochondrial and chloroplast DNAcircular found in these organellesB) How much DNA?

1) VariableSEE TEXT2) Chromosome Number varies widely among eukaryotes

C) In prokaryotes, the chromosome may be supercoiled and complexed with some proteins toform the nucleoid region.

D) DNA Packing(Eukaryotes)1) HistonesForm octamers(2 H2A, 2 H2B, 2 H3, 2 H4) around which DNA is wound, these

nucleosome beads are linked by H1 linker histones2) Nucleosomes wound into a 30nm solenoid, which is then coiled and supercoiled.3) Packing for cell division includes looping of DNA from a central scaffold, plus supercoiling

E) Chromatin is a synonym for DNA+Associated Proteins(Proteins=histone+non-histone)1) Heterochromatin is tightly packed and densely staining, but has little metabolic activity2) Euchromatin is more loosely packed and stains less; contains active genes

IV) Inheritance of GenesDNA ReplicationA) Replication is semiconservativeB) Begins at origins of replicationOriC in E.Coli

1) DnaA proteins being by binding to DNaA boxesC) Know how leading and lagging strands are constructed

1)

Know importance of helicases, single-strand binding proteins, primase, ligase,topoisomerases, etc.

D) DNA polymerases1) Link triphosphate nucleotides, work only in 5 to 3 direction; only extend existing polymers2) Prokaryotic-

(a) DNA polymerase III(Poly III)major enzyme at replication fork(b) Pol I-Primer removal and gap filling in lagging strand(c) Poly II, IV, Vrepair

3) Eukaryotic DNA Polymerases(a) , , most important

(i) Alpha has built in primase; polymerase switching hands growing strand off to:(ii) Delta for lagging strand; epsilon for leading strand

4) Eukaryotic Chromosomes are long, and contain many origins of replication.5) Eukaryotes also must deal with chromosome endsthe telomeres, which are produced by

telomerase

6) In eukaryotes, nucleosomes are displaced by the replisome, but reform almost immediatelypost-replication; old histone subunits are recycled, more are produced prior to replication

V) Two important lab techniques depend on DNA Polymerases, and have features similar to replicationA) PCR(text pg. 263-264)B) DNA sequencing


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1) Sanger Di-deoxy(185-187)2) Pyrosequencing(187-189)

VI) Gene FunctionA) The phenotype of cells(and thus organisms) depends on the proteins produced by the cellsB) Beadle and Tatum used nutritional mutants in Neurospora to dissect biochemical pathways,

testing the earlier suggeestions of Garrod that specific phenotypes were cause by enzyme

deficiencies.

C) The central dogma DNARNAproteins1) Info in DNA is transcribed into RNA2) Info in RNA(specifically mRNA is translated into polypeptides

D) Transcriptionprokaryotes1) RNA is a nucleotide polymer similar to DNA, but it uses different bases(U for T), is built on

nucleotides with a slightly different sugar(ribose) and is usually single-stranded

2) RNA messages are made by base pairing with the template strand of DNA( other strand isthe sense or coding strand)

3) DNA is read 3 to 5 and the RNA is made 5 to 3 by RNA polymerase4) RNA polymerase

(a) One RNA polymerase in prokaryotes(i) 4 subunits, including the sigma factor

5) 3 steps(a) InitiationRNA polymerase finds promoter region of DNA which signals where

transcription begins(review promoter structure in your test. subunit of RNA

polymerase needed for binding but released after enzyme binding and the DNA helix is

opened

(b) Elongationtriphosphate bases added to growing chain, bases added according DNAsequence

(c) Termination(i) Rho-inddependentRNA forms a hairpin loop(due to particular sequence of bases)

RNA poly stalls and falls off DNA(ii) Rho-dependentprotein factor(e.g. rho) needed, protein binds to RNA, pulls it off

the DNA

E) TranscriptionEukaryoes1) Initiation

(a) Promoter similar, but longer, differs in detail(b) Basal eukaryotic transcription complex several initiation factors(including the TATA box

binding protein) plus RNA polymerase

(c) Enhancers(DNA sequences) may also play a role2) Elongationsimilar3) Termination

(a) No clear termination sequences(b) A cleaveage site downstream from protein coding regions signifies a cleavage or end

to the molecule.

4) mRNA processingin eukaryotes onlyprimary transcript is modified(a) end modification

(i) 5:7-methyl-G phosphate cap added(ii) 3 end is clipped downstream of AAUAAA sequence, the poly-A tail is added by poly

A polymerase


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(b) Splicingintrons are spliced out and discarded, remaining exons are joined together,splice junctions are determined by sequence.

(i) Group 1 introns: self-splicing known from ciliates; the RNA acts as a ribozyme tocleave and ligate itself.

(ii) Group 2 introns.removed with help of spliceosomes Contain proteins and snRNA in both eukaryotes and prokaryotes

(iii)Intron ends have conserved sequence; splicing proceeds by creating a lariat of theintron

(c) RNA editing including substitution or insertion/deletion5) rRNA-3 of 4 done by RNA polymerase I

(a) genese for these are transcribed as a single unit, then pieces cleaved and processed toform 18S, 5.8S, and 28S rRNA.

(i) Note NTS, ETS, ITS(b) This unit is repeated many times in the genome

6) tRNAtranscribed by RNA Polymerase II(a) The promoter is internal to the transcribed sequence, not upstream, called internal

control region

(b) Some eukaryotic tRNA genes contain introns(c) Base modification(chemical modification of transcribed bases) is also typical

F) Protein structurereview proteins structure: primary, secondary, tertiary, quaternary1) Know amino acid structure and the major groups( dont memorize all the structures)2) Amino Acids linked via the peptide bond3) Levels of structure

(a) Primaryorder of amino acids(b) Secondaryalpha helix or beta pleated sheet(c) Tertiarythree dimensional organization, held together by hydrophobic/ philic

interaction, ionic bonds, salt bridges, etc.

(d) Quaternary3 dimensional organization of two or more polypeptidesEXAM 2

I. Translation

A. Protein structurereview protein structure: primary - secondary - tertiaryquaternary

B. the genetic code was broken quickly using a variety of clever experimental

techniques, and has many general properties (e.g. nonoverlapping, unambiguous,

degenerate, nearly universal, etc. etc.). Know how to use the standard table of the genetic code.

C. Translation - the production of polypeptides

1. mRNA provides instructions for the primary structure of proteins

2. The machinery of translation includes tRNA and ribosomes

3. tRNA

a. Produced via transcription of tRNA genes by an RNA polymerase


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(RNA Poly III in euks)

b. 3 main stem-loops (hairpins)

c. contain unusual bases like inosine, pseudouridine, produced by posttranscriptional modification of

initial transcripts

d. anticodon loop allows specific binding to codons

(1) a single anticodon may match more than one codons due to

relaxed pairing in the last codon position; this is known as wobble

e. amino acids added to the 3' end by a family of enzymes known as

aminoacyl tRNA synthetases

D. Ribosomes

1. large + small subunits

a. each composed of numerous proteins plus rRNA molecules

(1) rRNA molecules produced via transcription of rRNA genes by

RNA Polymerase (RNA poly I in euks); review gene organization

in text

(2) proteins produced via translation (like all proteins!)

b. contains E, P, and A sites for tRNAs

E. the process

1. Initiation

a. small subunit + mRNA + initiation factors

b. add initiator tRNA: fmet tRNAi

c. add large subunit

d. initiator tRNA starts in P site, A site vacant

e. in prokaryotes, the Shine-Dalgarno sequence (in 5'UTR) interacts with

a complementary sequence on one of the rRNAs to provide correctalignment

f. in eukaryotes, a cap-binding protein helps the small subunit bind to the

5' end, and the mRNA is scanned for the first AUG

2. Elongation


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a. next tRNA enters A site assisted by EF-Tu

b. amino acid (or chain) on P site tRNA passed to amino acid on A site

tRNA; this peptidyl transferase activity resides in one of the rRNA

molecules

c. translocation: mRNA, with bound tRNA from A site shifts causing

this tRNA to end up in the P site; tRNA from P site enters E site and is

then released

(1) requires EF-G + GTP

3. Termination

a. when a stop codon appears in A site, it is bound by a release

(termination) factor; cleave peptide chain from P-site tRNA; causes

release of mRNA and disassociation of large and small subunits

F. simultaneous transcription and translation is typical in prokaryotes

G. polyribosomes are typical in both prokaryotes and eukaryotes

H. in eukaryotes, all transcription occurs in nucleus, the mRNA (and tRNA, rRNA) is

exported to the cytoplasm, where translation occurs.

II. Mutations

A. definition: any change in sequence or organization of genome

B. Classifying mutations

1. extent: point vs chromosomal

2. causation: spontaneous (from natural chemical activity in cells) vs induced

3. bodily location: somatic vs gametic

4. effect on protein activity: loss of function, gain of function

5. effect in heterozygotes: recessive, dominant

6. effect on fitness: positive vs deleterious vs neutral

7. effect of phenotype: morphological, nutrtional, biochemical, behavioral

C. thinking about mutations...

1. What is the DNA level change?


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2. What is the change in protein structure?

3. How is the function of the protein affected?

D. Single gene mutations - DNA level

1. base substitution

a. transitions

b. transversions

2. insertions/deletions - indels

E. Single gene mutations - protein structure and function

1. single base changes

a. silent - codon different, but still codes for same amino acid

b. missense - one amino acid switched for another. Physiological

consequences range from "none" (aka neutral genetic variation) to"severe" depending on which a.a. is

changed

c. nonsense - changes an amino acid codon to a stop codon. Consequences

usually severe.

2. frameshift (additions or deletions of 1-several bp, not in multiples of 3)

a. drastically changes amino acid sequence following mutation site

b. often introduces premature stop codon, and so produce shortened

protein products.

F. Consequences of mutations outside of protein coding region

1. might occur in transcription regulatory regions (promoter, operators,

enhancers, silencers), intron splice sites; translation regulatory regions (ShineDalgarno, in mRNA UTR

regions)

2. may have effects on gene regulation

G. Causes

1. spontaneous

a. tautomeric shifts cause mispairing during replication (result in

transitions)


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b. slip-strand mispairing during replication results in insertions/deletions

c. base deamination (may also be induced), depurination

2. induced - cause by base alteration, replacement, or damage

a. radiation

(1) UV radiation causes thymine dimers

(2) ionizing radiation (e.g. X rays) can create free radicals and

reactive ions which damage DNA; can cause breaks in sugarphosphate backbone

b. Chemical mutagens

(1) base analogs (e.g. 5-bromouridine)

(2) base alterations - e.g alkalating agents, changes base pairing

(3) intercalating agents - typically cause single base indels

(4) the Ames test can be used to test for mutagenicity

H. Mutation Rates - vary widely among organisms (e.g. prokaryotes versus eukaryotes);

can be measured per base pair, per gene, per genome over number of cell divisions,

generations, year

1. prokaryotes

a. single base errors in range of 10 - 10 per site per cell division, e.g. -9 -10

0.36 x 10-9

b. 10 - 10 per gene per cell division/generation -5 -7

2. eukaryotes

a. 10 - 10 per gene per generation -4 -6

b. 0.19 x 10 in yeast -9

c. 2 - 9 x 10 in invertebrates and plants -9

d. 22 x 10 in humans -9

I. Limiting mutational impact - most mutations that effect the phenotype are deleterious,

so cells that can reduce the number of mutations are favored by natural selection

1. reduce rate through avoidance, e.g. superoxide dismutase, skin pigmentation

etc2. Proofreading during replication


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3. directly repair damage, e.g. split T dimers, remove methyl groups, etc.

4. Excision Repair pathways

a. Base excision repair pathway - can fix incorrect or damaged bases

(1) uses a glycosylase to remove damaged or altered base

(2) AP endonuclease cuts the affected DNA strand

(3) DNA polymerase replaces several nucleotides using 3' end

produced by endonuclease as priming site

(4) ligase seals any nicks

b. Nucleotide Excision repair - used to remove and replace damaged

DNA, e.g. lesions caused by UV, such as thymine dimers

(1) a lesion or mismatched pair is recognized, and one of the

strands is cut and removed, the remaining strand serves as a

template for DNA polymerase

(2) see E. coli example (Uva, Uvab, etc)

c. Mismatch Excision Repair -

(1) for mismatched bases, methylation of old strand may enable

cell to retain original information (in prokaryotes), typified by the

MutS system

(2) occurs in eukaryotes, but mechanisms of identifying old

strand unclear

III. Transposable elements - included with mutation since they are known to cause mutations

when they are active. Many of the replicative ones depend on an RNA intermediate -

retrotransposons.

A. prokaryotes

1. Insertion sequences

2. Transpososons: composite or non-composite

3. may be replicative or conservative

B. Eukaryotes


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1. Corn - McClintock studied Ac and Ds elements

2. Yeast - Ty elements

3. Humans - review LINEs and SINEs

a. over half our genomes are derived from transposable elements!

IV. Inheritance of genes - cell division and mendelian inheritance

A. Prokaryotes - binary fission

B. Eukaryotes

1. know general outline of cell cycle (G16S6G26mitosis6G1 etc.)

2. Mitosis

a. purpose - create new cells which are genetically identical to the original

parent

b. know overview of stages, what goes on in each

V. Inheritance of genes - cell division and mendelian inheritanceA. Eukaryotes

1. Meiosis

a. purpose - create haploid cells involved in sexual reproduction

b. one diploid cell (MeiosisI) 2 haploid cells (MeiosisII) 4

haploid cells

c. know overview of steps (prophase I, metaphase I,... anaphase II,

telophase II)

d. be able to relate to Mendelian inheritance

VI. Mendelian Inheritance

A. Some of Mendel's main conclusions

1. Each individual has two copies of each gene

2. The pair is divided during production of gametes. This is the rule of

segregation, and corresponds to the separation of homologues during anaphase

3. Each pair of alleles is divided independently of other pairs - the rule of

independent assortment

B. Monohybrid crosses (e.g. Aa x Aa) yield 1:2:1 genotypic ratios, and 3:1 phenotypic


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ratios

C. Dihybrid crosses (AaBb x AaBb) yield 9:3:3:1 phenotypic ratios

D. Some useful tools for these crosses

1. Punnett squares

2. path diagrams

3. probability - product and sum rules

VII. Chromosomal Inheritance - genes reside on a limited number of long chromosomes

A. Thomas Hunt Morgan studied Drosophila

1. He examined the famous red eye / white eye example, which was consistent

with Mendel, but included the added complication of sex chromosomes and sexlinkage

VIII. Sex linkage and sex determination

A. Sex determination can be environmental or genetic

1. Genetic systems include single locus, haplo-diploidy, and sex chromosomes,

(XX and XY or ZZ and ZW)

a. note that among XX/XY systems, the details may vary widely; e.g.

XXY are male in humans, female in drosophila

b. Non-genetic sex determination is also well known

B. genes on sex chromosomes are sex-linked

1. for mammals (and many other species), there are relatively few genes on the Y

chromosome (and none needed by females), so sex-linkage is essentially

X-linkage

C. Review examples of sex linkage in your text

1. phenotype may be restricted to one sex, depending on cross performed

2. Reciprocal crosses typically give different results

3. sex-linkage may be deduced from pedigreesD. other details

1. typically, the X and Y (or Z and W) chromosomes have a short region of

homology, the pseudoautosomal region, which allows pairing in meiosis

2. not to be confused with sex-influenced or sex-limited traits


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3. dosage compensation - how is this achieved? e.g. Barr bodies in mammals,

transcription rates in drosophila This outline is a conceptual overview/framework of the material since

the last exam; use it to

structure your thinking/studying. This should help you know what to study, but does not contain

all of the relevant information! You should certainly know the processes, terms, etc. listed

below. Warning: There may be terms/minor concepts not included in the outline which may

appear on the exam.

A major skill in genetics is problem solving. Particularly for Mendelian genetics, I urge you to

review appropriate problems at the end of each chapter.

I. Mendelian Inheritance

A. Some of Mendel's main conclusions (included on last exam)

1. Each individual has two copies of each gene

2. The pair is divided during production of gametes. This is the rule of

segregation, and corresponds to the separation of homologues during anaphase

3. Each pair of alleles is divided independently of other pairs - the rule of

independent assortment

B. Monohybrid crosses (e.g. Aa x Aa) yield 1:2:1 genotypic ratios, and 3:1 phenotypic

ratios (included on last exam)

C. Dihybrid crosses (AaBb x AaBb) yield 9:3:3:1 phenotypic ratios

D. Some useful tools for these crosses

1. Punnett squares

2. path diagrams

3. probability - product and sum rules

E. Pedigrees can be analyzed using Mendelian principles

II. Extending Mendels simple predictions

A. Single gene cases

1. dominance variations: partial or incomplete dominance, codominance

2. multiple alleles


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3. lethal or deleterious alleles

B. multiple gene examples

1. complementation - you should consider this as a test to determine if two

phenotypes which look identical are actually due to the same genotype, that is,

share the same mutation

a. complementation occurs if alleles at 2 different genes can produce the

same phenotype.

b. F1's usu. not like parents, F2's show a 9:7 ratio

2. epistasis

3. novel phenotypes

4. genetic background can affect expression of a phenotypic trait

a. Penetrance - percentage of individuals with a given genotype which

show the typical phenotype. Due to environmental and genetic

differences among individualsb. Expressivity - among individuals who show a trait, there is variation in

the degree of expression.

III. Recombination of genes and gene mapping

A. genetic recombination: any process in a meiocyte that generates a haploid product of

meiosis with a genotype differing from both the haploid genotypes that originally

combined to form the diploid meiocyte

B. two different mechanisms

1. independent assortment - Mendels second rule: from random, independent

behavior of chromosomes during meiosis (interchromosomal)

a. in Mendels standard dihybrid cross, produces 9:3:3:1 ratio

2. crossing-over - allows recombination of genes on same chromosome

(intrachromosomal)

C. the recombination frequency (RF value) quantifies recombination

1. expected to be 50% for unlinked genes

2. expected to be less than 50% for genes linked on same chromosome


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D. Standard 2-gene cross: AaBb x aabb (test cross: a double heterozygote crossed

with a homozygous recessive)

1. estimate % of recombinant gametes produced by mother

2. the recombination frequency (RF) gives a measure of linkage between genes

3. know difference of linkage in cis (coupling) and trans (repulsion)

IV. Chromosomal mutations

A. Mutations in Chromosome Structure

1. Morphology of chromosomes may be identified cytogenetically, through size,

position of centromere (metacentric, telocentric, or acrocentric), chromosome

banding patterns

2. Types

a. deletions

b. duplications - tandem, inverse tandem, terminal

c. inversions

(1) pericentric - spans centromere

(2) paracentric - does not span centromere

(3) in inversion heterozygotes, recombination in region of

inversion is "suppressed" - no recombinant gametes are produced.

This is due to extensive duplication/deletion with pericentric and

deletions with paracentric

d. translocations - intrachromosomal vs interchromosomal, nonreciprocal,

reciprocal

3. The above are formed through

a. chromosome breakage and rejoining

b. unequal crossing over

B. changes in number

1. aneuploidy - differs from wildtype (e.g. diploid) by less than a full set; usually

result from nondisjunction during meiosis (or mitosis)a. trisomy: one extra chromosome, 2n+1


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b. monosomy: missing one chromosome, 2n-1

c. nullisomy: missing both of one type of chromosome, 2n-2

2. Euploidy - changes in numbers of whole sets of chromosomes present

a. monoploid, triploid, tetraploid, etc.

b. autopolyploid vs. allopolyploid

c. may cause problems in meiosis, e.g. triploids are usually sterile due to

dosage problems

V. Bacterial and Viral genetics

A. Bacteria are haploid, clone-forming organisms with circular chromosomes.

B. We are interested in

1. organization of the genome

2. transfer of genetic info from one individual to another

C. Recombination

1. Conjugation - an orderly and cell mediated process of genetic exchange

between cells

a. controlled by the F (fertility factor), an episome containing -100 genes,

replicated during cell division, passed to other cells during conjugation

b. F+ cells have this genetic element, F- do not, F- are converted to F+ by

conjugation

c. occasionally, the F plasmid inserts into the chromosome, producing an

Hfr cell.

(1) Hfr cells donate genetic material other than the F factor; in

fact, the whole chromosome is transferred before the complete F

factor.

(2) Once a piece of a chromosome is inserted, it can cross-over

with the recipient chromosome, creating a recombinant

chromosome, extra-chromosomal pieces are lost

(3) Mating Hfr strains with F- strains can give linkage maps via


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interrupted mating technique

d. F cells efficiently move specific sequences (normally, genes flanking

regions adjacent to the insertion site of the plasmid in the host

chromosome

2. Transformation is the second way genetic information is transferred between

cells

a. competent cells take up relatively small pieces of DNA from their

surroundings

b. Incorporated in host chromosome by single strand displacement,

followed by recombination and lost of single stranded, linear fragment

c. requires a cell division to produce one original and one

recombinant type cell

3. Transduction

a. depends on viral intermediate

(1) review lytic and lysogenic cyclesb. generalized: pieces of bacterial DNA mistakenly incorporated into

viral capsules, infection by the virus particle injects this piece into a cell,

where it can be incorporated through crossing over

c. specialized: when prophage (viruses inserted in bacterial chromosome)

loop out, occasionally a small piece of the bacterial chromosome is carried

along

VI. Gene regulation in prokaryotes (What kinds of proteins are made? How much of each?)

A. Cells / organisms have lots of genes, and many of these are not needed all the time

(needed for specific environmental conditions, tissue-specific differences, etc.).

B. major control points

1. transcription (most important)

2. translation

C. transcriptional control

1. review lac operon


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a. three genes necessary for lactose metabolism are grouped together in an

operon, and transcription is controlled through a single promoter

2. lac operon, an inducible operon

a. lactose enters the cell via lactose permase, and is typically cleaved into

simple sugars by $-galactosidase; a small amount of allolactose is also

made

b. three genes necessary for lactose metabolism are grouped together in an

operon, and transcription is controlled through a single promoter

c. transcriptional control through I gene product

(1) in the absence of lactose, repressor binds to operator,

preventing transcription

(2) allolactose is the inducer, prevents I protein from binding to

operator, allowing transcription

d. negative control through I gene product (repressor)

e. positive control through CAP-cAMP binding

f. both cis (operator, promoter, CAP site) control elements and trans

(repressor) control elements

g. know mutations and their effects

3. Review trp operon

a. a repressor interacts with co-repressor (tryptophan) to regulate initiation

of transcription

b. early termination of transcription (before structural genes), called

attenuation, is controlled by hairpin loops of the nascent RNA and

simultaneous translation of the leader peptide

VII. Control of gene expression in Eukaryotes.

A. Eukaryotes contain many thousands of different genes; cells may express only a

subset of these genes at any given time.

B. This section seeks to answer the questions "What proteins are made?" and "How


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much of each?"C. Some major differences between proks and euks

1. DNA is physically segregated from the cytoplasm (nucleus)

2. DNA is packed in histone proteins

3. the genome has a much different organization

a. many chromosomes

b. much noncoding DNA

c. genes contain numerous exons and large introns (RNA requires

processing)

4. mRNA is more stable - it hangs around the cell longer

D. gene expression could be controlled at many points, e.g. number of mRNA transcripts

which are made, how many times each RNA is translated, etc.

1. Chromatin remodeling - sequence specific directed changes in the way DNA is

packed

a. HAT - histone acetyl transferase - provides looser more open

chromatin conformation, works with HDACs (histone deacetylases)

2. Transcription initiation

a. transcriptional control - the major level of control; control includes both

cis (DNA sequences like enhancers and silencers) and trans components

(like transcriptional activators and repressors)

b. transcription initiation is complicated, and requires many accessory

factors in addition to the RNA poly II and general (basal) transcription

factors

c. DNA regulatory sequences (cis control)

(1) promoter - RNA polymerase II binding region upstream of

gene

(2) enhancers

(a) binding region for transcription factors

(b) associated with positive control


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(c) structure (placement, sequence) quite variable

(upstream, downstream, long distance...)

(3) silencers

(a) provide negative control

(4) insulators

(a) protect regions from influence from nearby enhancers

or silencers

d. Regulatory proteins (trans control)

(1) activators bind DNA regions and some part of the basal

transcription complex (RNA poly + TFs)

(a) coactivators bind to activators and some part of the

basal transcription complex

(2) These regulatory proteins typically have several domains

(a) one of which interacts directly with the DNA (inc. zincfingers, leucine zippers, helix-turn-helix, helix-

loop-helix)

(b) other domains may interact with other mediator or

affector molecules (e.g. hormones, transcription factors,other regulatory proteins)

(c) Review: glucocorticoid hormone regulation.

(3) Often work in a combinatorial fashion

Bio 220 Exam 4 Study Outline (Winter 2013) Cabe

This outline is a conceptual overview/framework of the material since the last exam; use it to

structure your thinking/studying. This should help you know what to study, but does not contain

all of the relevant information! You should certainly know the processes, terms, etc. listed

below. Warning: There may be terms/minor concepts not included in the outline which may

appear on the exam.

A major skill in genetics is problem solving. Particularly for Mendelian genetics, I urge you to

review appropriate problems at the end of each chapter.

I. Control of gene expression in Eukaryotes.


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A. Eukaryotes contain many thousands of different genes; cells may express only a

subset of these genes at any given time.

B. This section seeks to answer the questions "What proteins are made?" and "How

much of each?"

C. Some major differences between proks and euks

1. DNA is physically segregated from the cytoplasm (nucleus)

2. DNA is packed in histone proteins

3. the genome has a much different organization

a. many chromosomes

b. much noncoding DNA

c. genes contain numerous exons and large introns (RNA requires

processing)

4. mRNA is more stable - it hangs around the cell longer

D. gene expression could be controlled at many points, e.g. number of mRNA transcripts

which are made, how many times each RNA is translated, etc.

1. Chromatin remodeling - sequence specific directed changes in the way DNA is

packed

a. HAT - histone acetyl transferase - provides looser more open

chromatin conformation, works with HDACs (histone deacetylases)

b. nucleosome remodeling - activators help nucleosome remodeling

complexes target specific areas

(1) complexes slide, transfer, or restructure nucleosomes in

some way.

2. Transcription initiation

a. transcriptional control - the major level of control; control includes both

cis (DNA sequences like enhancers and silencers) and trans components

(like transcriptional activators and repressors)

b. transcription initiation is complicated, and requires many accessory


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factors in addition to the RNA poly II and general (basal) transcription

factors

c. DNA regulatory sequences (cis control)

(1) promoter - RNA polymerase II binding region upstream of

gene(2) enhancers

(a) binding region for transcription factors

(b) associated with positive control

(c) structure (placement, sequence) quite variable

(upstream, downstream, long distance...)

d. Regulatory proteins (trans control)

(1) activators bind DNA regions and some part of the basal

transcription complex (RNA poly + TFs)

(a) coactivators bind to activators and some part of the

basal transcription complex

(2) These regulatory proteins typically have several domains

(a) one of which interacts directly with the DNA (inc. zincfingers, leucine zippers, helix-turn-helix, helix-

loop-helix)

(b) other domains may interact with other mediator or

affector molecules (e.g. hormones, transcription factors,

other regulatory proteins)

e. Gene silencing - DNA methylation: highly methylated CpG islands act

to inhibit transcription

(1) genomic imprinting is a special case

3. post-transcriptional controls

a. Alternative splicing - can yield different proteins from the same

transcript

b. mRNA must be transported to cytoplasm - another regulatory step;

known to be important in some examples


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c. mRNA degradation

(1) may be controlled by elements (sequences) within them

(2) half-life may be altered by presence of regulatory chemical

signals (e.g. see Table 20.2)

d. Example - transferrin and ferritin mRNA

(1) both have Iron Response Elements (IRE)

(2) Iron Response Binding Protein (IRP) binds to IRE in the

absence of iron, and blocks degradation of transferrin mRNA and

blocks translation of ferritin mRNA; when iron is present, it binds

to IRP, preventing binding to RNA, allowing degradation of

transferring mRNA and translation of ferritin mRNA

E. Tissue specific regulation, regardless of which process is controlled, requires that

different tissues contain different regulatory substances (usually proteins), since the

genetic information is largely identical in each cell.

F. RNA interference (RNAi) is in a special category of gene regulation, and can

influence mRNA degradation and translation

1. probably arose (and still functions in) response to viral infection and

transposable element activity

2. also has a major role in regulating genes, which requires the transcription of

endogenous microRNAs

3. Fire and Mellow awarded Nobel prize for first clear description of the system4. Requires the

production/presence of double-stranded RNA to initiate this

process, from microRNA, short interfering RNA, artificially produced antisense

RNA

5. review process in your text

II. Applying the lessons of gene regulation to larger questions - Drosophila development

A. Sex determination in Drosophila - review figures in text, or animation!

1. The autosomal:X chromosome ratio is key in producing active and inactive


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dimer proteins which are transcriptional activators

a. females have more of the functional (sisterless) gene product, and thus

have active transcription of the sxl early promoter

2. The sex lethal (sxl) gene is under different transcriptional control in the two

sexes

3. The sxl protein regulates splicing of the transformer (tra) gene

a. in males, a default splice site yields an early stop codon, and thus a

short, nonfunctional protein

b. in females, an alternate splice site (regulated by sxl) produces a fulllength, functional protein

4. the tra protein, found only in females, is a splicing regulator of the doublesex

(dsx) gene

a. in males, a default splice pattern yields a male specific dsx protein,

which is a transcription regulator (inhibits female-specific genes)

b. in females, an alternate splice pattern yields a female specific dsx

protein, which is a transcription regulator. (inhibits male-specific genes)

B. Body plan

1. Drosophila are characterized by segmentation; developing embryos need to

specify at least three kinds of information

a. where each cell is located (positional information)

b. the number of segments which will be produced

c. the identity of each segement

2. Position information (usually in terms of protein concentration gradients)

important in development. Anterior/Posterior axis very important and best

studied

a. three proteins which are important: bicoid (BCD), caudal, and

hunchback maternal (HB-M)

b. these are transcription factors

c. egg is created with bicoid mRNA localized in anterior and nanos (nos)


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mRNA localized in posterior; these mRNAs were transcribed from the

mothers genes

(1) these mRNA's are anchored using the 3' untranslated region

d. proteins from these RNA's create a gradient through egg cell and

developing embryo

e. the nanos protein acts as a translational inhibitor of HB-M (mRNA for

this is spread evenly throughout the egg cell): the protein is produced most

in anterior region, and less in posterior regionf. caudal protein concentration is regulated by the bicoid

protein, which is

a translational inhibitor of caudal mRNA (which is distributed uniformly)

3. Establishing the number of segments - a hierarchy of genetic controls works

through a cascade effect.

a. A/P gradient genes (bicoid/hunchback-M, caudal) regulate

expression of the ...

(1) gap genes - divide developing embryo into large zones.

Examples include:

(2) hunchback (zygotic form) is expressed in the anterior end; (fyi

this gene has 5 binding sites for the bicoid protein, transcription

levels depend on number of bound bicoid molecules)

(3) Kruppel - expressed in middle section

(4) Knirps - strongly repressed by BCD: made in posterior portion

(5) The gap genes make transcription factors which control the...

b. pair-rule genes - specify the number of segments. Make transcription

factors which control the...

c. segment-polarity genes - specify an anterior and posterior end to each

segment. Produce transcription factors and other regulatory proteins.

4. Establishing the identity of each segment - what structures (wings, legs) on

each segment?


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a. Homeotic genes: two clusters of genes (the bithorax gene complex and

the Antennapedia complex) together comprise the homeotic gene

complex

(1) these genes are controlled by the gap-gene and pair-rule gene

products

(2) the order of the genes on the chromosome mirrors the order of

body segments they control

(a) gene expression is (in part) controlled via the polycomb

and trithorax gene complexes, which close and open

chromatin strucuture, respectively

(3) each contains a similar homeobox - a 180 bp sequence coding

for a 60 a.a. protein domain (homeodomain) which has DNA

binding function - the implication is that these genes produce

transcription factors which regulate other genes

(4) similar genes (Hox) with same function have been found in

vertebrates - birds, mice, humans

(5) have similar order on chromosome, appear to have tissue

specific expression as in Drosophila

C. Genetic Dissection of gene expression during development, an example of genomic

scale research

1. gene expression can be studied by analyzing mRNA pools in specific cells and

tissues.

a. mRNA can be retrieved via nucleic acid purification followed by

hybridization with a biotinylated poly-T probe

b. mRNA converted to cDNAc. cDNA can be studied via microarrays - see text

d. in situ hybridization of cellular mRNA can also elucidate tissuespecific patterns of expression

2. see link in PowerPoint collections for the published example we examined in

class.


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a. http://genomebiology.com/2007/8/7/R145

b. Genome Biol. 2007;8(7):R145. Global analysis of patterns of gene

expression during Drosophila embryogenesis. Tomancak P, Berman BP,

Beaton A, Weiszmann R, Kwan E, Hartenstein V, Celniker SE, Rubin

GM.

III. Genomics - the study of whole genomes

A. strategies

1. hierarchical shotgun sequencing, used by the HGP

a. used large-insert libraries (e.g. using BACS) and much work to

establish minimum tiling paths for each chromosome, then each of the

large inserts was fragmented and cloned into plasmids for shotgun

sequencing

b. as sequence was assembled, it was always known where on which

chromosome it came from; map first, sequence later

2. Whole genome shotgun sequencing

a. pioneered by Venter and Celera research team for use on Drosophila

and later the human genome

B. The human genome project

1. A federally funded, international team headed by Francis Collins used a

hierarchical shotgun method (map first, sequence later) to produce a human

genome sequence by 2001.

2. A privately funded initiative (Celera) headed by Craig Venter raced the

government team, and finished a genome sequence at the same time.

C. Data - the human genome

1. base pair composition: 41% G+C, but not uniformly distributed

2. transposable elements - Lessons from studying transposons

a. if you take all copies of a particular repeat, e.g. LINE1, and compare

them, you can determine the consensus sequence; if these all represent


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copies from the same ancestral sequence, you can determine the number of

mutations that separate any two copies; the number of mutational

differences is proportional to the time since the copies were made: you can

determine ages of various transposons, and reach the following

conclusions

(1) some transposon repeats are extinct fossils

(2) transposons persist in lineages (ie. mammals) for a very long

time - more than 100 million years

(3) humans have little recent transposition compared to mice

(4) some types of transposons (LTR, a type of retrotransposon) are

extinct (or nearly so) in humansb. Some chromosomal regions are nearly devoid of transposons - this

implies that they contain very important genes (or their regulatory

sequences) (HGP Fig 21)

c. comparison of transposons help us understand mutation rates, both

overall (HGP Fig 27) and within males and females (HGP Fig 29)

I. Genomics - the study of whole genomes

A. The human genome project

1. A federally funded, international team headed by Francis Collins used a

hierarchical shotgun method (map first, sequence later) to produce a human

genome sequence by 2001.

2. A privately funded initiative (Celera) headed by Craig Venter raced the

government team, and finished a genome sequence at the same time.

B. Data - the human genome

1. Protein coding genes

a. How many genes? (20,000-25,000)

b. What does a typical gene look like?

c. Where do most genes occur? (GC rich regions)

d. Gene Function - What do we know about the genes?


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(1) can compare these to other known genes in other species (or

known human genes)

(a) either the nucleotide or predicted amino acid sequence

can be compared to a variety of databases

(2) a large proportion of genes (40-50%) remain unknown in

function

e. functional categories (see HGP Fig 37)

(1) compared to other species, humans have proportionally large

numbers of genes involved in transcription/translation and

signalling, suggesting complicated gene-regulation pathways

f. comparisons with other species shows a large proportion of genes are

similar to other groups (prokaryotes + eukaryotes, just eukaryotes, just

animals, just vertebrates etc.) (see HGP Fig 38)

C. The transcriptome - evidence of gene expression

1. RNA (including mRNA, miRNA, other RNA) can be extracted from cells, and

converted to cDNA to be studied

2. Can be studied via hybridization on DNA microarrays

a. DNA chips, DNA hybridization, and Gene expression studies (see pg

216-217, also 216 Fig10.11b)

3. DNA hybridization

a. DNA chips contain many DNA sequences affixed to a solid substrate,

like a glass slide. In some chips from the company Affymetrix, there are

representative sequences from every known gene in the human genome

b. mRNA from two samples (e.g. normal and tumor tissue etc.) iscollected, copied into cDNA, and

labeled fluorescently. The cDNA is then

hybridized to the chip, and the chip is then checked for pattern of

fluorescence. Since the chips contain sequences of all known genes, you

can check global differences in gene expression. We looked at examples


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from sporulation genes in yeast and breast tumors.

II. Population genetics

A. Hardy-Weinberg model

1. know basic assumptions

2. this model that makes two predictions

a. given allelic frequencies p=freq(A) and q=freq(a), we can predict the

genotypic frequencies: freq (AA) = p , freq (Aa)=2pq, freq(aa)=q 2 2

b. allelic frequency (p) does not change over time

3. Effect of non-random mating

a. Inbreeding - increases frequency of homozygotes

b. Outbreeding - increases freq of heterozygotes

c. Assortative mating

(1) positive- increases frequency of homozygotes

(2) negative- increases freq of heterozygotes

B. Understanding what happens within populations over time (the 2nd H-W prediction)

depends on understanding the influences of the following four basic mechanisms of

evolutionary change. Only these four processes can change allele frequencies over time:

1. selection

2. drift

3. mutation

4. migration

C. Genetic Drift

1. includes all H-W assumptions except very large (-infinite) populations

a. assumes genetic variation is neutral: no fitness differences

2. "sampling error" inherent in finite populations may change value of p every

generation

3. )p is inversely proportional to population size

4. the direction in which p moves in any one generation is random


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5. equilibrium points are p=0 or p=1 ("fixation")

a. drift eliminates variation from populations

b. if we start with a population with freq p, there is a probability of p that

it will fix at p = 1 and a prob. q = that it will fix at p = 0

6. Population Bottlenecks - a manifestation of genetic drift

a. Can affect both heterozygosity and allelic diversity

b. Starlings in North America

1 0 (1) H = H (1 - 1/2N) (change in heterozygosity due to finite pop

size)

(2) a moderate bottleneck

(3) reduced allelic diversity, but not heterozygosity

D. Autosomal selection1. A simple model (memorize this equation!)

a. AA Aa p = p w + (2pq)w / (population mean fitness)

2

b. population mean fitness = AA Aa aa p w + (2pq)w + q w

2 2

2. assumes

a. H-W: random mating, large pop's, no mutation, closed population

b. ind. of different genotypes do not contribute equally to next generation.

This is due to differences in viability and fecundity and sexual selection.

Differences in contributions are summarized by the fitness parameters

c. An alternative fitness coding scheme:

(1) AA = 1

(2) Aa = 1-hs (h = degree of dominance; A is dom if h=0)

(3) aa = 1-s (s = selection coefficient)

3. Dynamics depend on the strength of selection (ie. the difference in fitnesses)

and dominance relationships

4. modes of selection


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a. Directional - if one of the alleles is favored over the other

(1) degree of dominance may be important

(2) which allele is favored (recessive or dominant) may be

important

b. overdominance - heterozygote is most fit

c. underdominance - heterozygote is least fit

5. Realistically...

a. directional selection most important. Good examples of

overdominance (sickle-cell system) are known, but rare

b. Selection is capable of increasing alleles which confer only minor gains

in fitness in a relatively modest number of generations

E. mutation

1. the ultimate source of all new genetic variation

2. changes the freq in populations only slightly:

a. let the rate of mutation from A 6 a equal :

b. )p = -:p

F. Migration - can change allele frequencies within a population. Depends on

1. m - the migration rate parameter, the proportion of the population which are

immigrants

2. difference in allele frequencies between the immigrants and the population

which receives them

3. y x y

p = p + m(p -p ) (see text; no need to memorize this - if you need it, I will

provide it).

4. General importance

a. keeps subpopulations similar

b. can add new genetic variation to populations

G. Nonrandom mating can bias the genotypic frequencies


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1. inbreeding decreases heterozygosity; outbreeding has opposite effect

2. Assortative mating can also occur

a. positive - similar genotypes more likely to mate - probably true inhumans for genes underlying height,

IQ, skin color, etc

b. negative - different genotypes more likely to mate, may be true in

humans for red hair

H. Drift and Migration together

1. if a population is divided into a number of more-or-less isolated

subpopulations (demes), with the possibility of migration (gene flow) among

them, what do we expect?

a. Drift will tend to create allele frequency differences among populations

b. migration will tend to keep demes similar in frequency

c. these two forces work against each other

d. We can measure the genetic differences among populations using

Sewall Wrights FST statistic, which equals 0 when populations are

identical, and 1 when they are fixed for different alleles

e ST . F varies with migration and drift, and thus can be used to estimate

migration in real populations

I. Note: The simulation package I used in class is called Populus, and is available free:

http://www.cbs.umn.edu/populus/

J. Application - Are Roads Barriers to Dispersal in Woodland Salamanders? A

collaborative project with Dr D. Marsh, several technicians, and many students

1. We used a population genetics approach to estimate dispersal of red-backed

salamanders

a. along a transect in continuous habitat

b. across streams

c. across roads

2. We collected microsatellite data to e ST stimate F values (in addition to many


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other statistical approaches). We found

a. distance is a barrier to migration (isolation by distance)

b. streams are a barrier to migration

c. small roads are NOT a barrier to migration

d. large roads ARE a barrier to migration

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