TEST 3 STUDY GUIDE

Bateson and punnett refers to what?

CH 6 GENETIC LINKAGE AND MAPPING IN EUKARYOTES 

Linkage and Crossing Over

- Linkage: 2 meanings

o Two or more genes can be located on the same chromosome (syntenic) o Genes that are close together tend to be transmitted as a unit

- Chromosomes = linkage groups

o The number of linkage groups is the number of types of chromosomes of the  species

▪ Ex: humans

∙ 22 autosomal linkage groups  

∙ An X chromosome linkage group

∙ A Y chromosome linkage group

- Genes that are far apart on the same chromosome may independently assort from each  other

Who are bateson and punnett?

Crossing Over May Produce Recombinant Phenotypes

- Genetic recombination: groups of linked alleles change due to crossing over  - Linkage can be altered curing meiosis as a result of crossing over

o Prophase I of meiosis in pachytene phase during bivalent stage

o Non sister chromatids exchange DNA segments

- Bateson and Punnet Don't forget about the age old question of Why are information technology ethical issues important?

o Discovered two genes that did not assort independently Don't forget about the age old question of Who is immanuel kant?

o Crossed sweet pea for flower color and pollen shape

o Expected a 9:3:3:1 phenotypic ratio but got a much greater proportion of the two  types found in the parental generations

o This was due to the fact that flower color and pollen shape are linked on the same  chromosome

Who is thomas hunt morgan?

Evidence for Linkage of Several X-linked genes: Thomas Hunt Morgan  

- Wanted to find out where genes were and how they recombined

- Looked at several traits following an X-linked pattern of inheritance If you want to learn more check out The money multiplier refers to what?

o Body color

o Eye color

o Wing length

- Found a much higher proportion of the combinations of traits found in the parental  generation

- Explanation

o All three genes are located on the X chromosome and therefore, they tend to be  transmitted together as a unit

o Due to crossing over, homologous X chromosomes can exchange pieces of  chromosomes

o Likelihood of crossover depends on the distance between two genes

▪ More likely to occur on two genes that are far apart from one another

Chi Square Analysis for Linkage

- Hypothesis: the genes for eye color and body color of drosophila are X-linked and,  somehow, independently assorting

TEST 3 STUDY GUIDE

- Calculated Chi square value for this hypothesis is giant so most of the time this will not  occur by chance and is not the explanation for the observed results We also discuss several other topics like What are the three types of temperament?

- Hypothesis is rejected so it is concluded that the genes are linked

Creighton and McClintock Corn Experiments

- Made crossed involving two linked genes whose chromosomes had unusual structural  features; knob and a translocated piece

o Created parental and recombinant offspring

- Were able to correlate the occurrence of recombinant offspring with observable changes  in pieces of homologous chromosomes Don't forget about the age old question of What are the ways to prevent fatigue?

- Nonparental offspring are the product of a crossover

Genetic Mapping in Plant and Animals

- Purpose: determine the linear order of linked genes along the same chromosome and  distance of separation among the linked genes along the same chromosome - Uses

o Understanding complexity and genetic organization of certain species

o Improve our understanding of evolutionary relationships between different  species

o Diagnose and one day treat human diseases

o Help predict the likelihood that a couple will produce children with certain  inherited diseases If you want to learn more check out Interpret data from epidemiology curve as, what?

o Provide information for improving agriculturally important strains through  selective breeding

- Also allow us to estimate the distances between linked genes, based on the likelihood that  crossover will occur between the two

- The farther apart the genes are, the more likely they will recombine

- Map distance = (# of recombinant offspring/ total # of offspring) x 100 - Units are map units (mu) or centiMorgans (cM)

- One mu = 1% recombination frequency

- Mapping typically carried out through a test cross and the data is used to calculate the  distance using the formula

- Look at Strurtevant’s test cross

- When finding the distance between two genes that are very far apart, it is much more  accurate to add up the distances between other genes between them

o This is due to the likelihood of multiple crossovers

- Trihybrid crosses

o Also used to find distance

o Cross two true breeding strains that differ in three alleles

o Perform testcross  

o Analysis of F2 generation allows you to map the three genes

o Collect data for F2 generation

o Combination of traits in the double crossover tells which gene is in the middle o Calculate map distance between pairs of alleles

▪ Regroup the data according to pairs of genes

o Construct the map

- Interference

o First crossover prevents another from occurring

TEST 3 STUDY GUIDE

o Product rule allows us to predict the likelihood of a double crossover from the  individual probabilities of each crossover

o Positive interference: The first crossover decreases the probability the at a second  will occur nearby  

Crossing over in Mitosis

- Rare

- Does not involve the homologous pairing of chromosomes to form bivalents - May produce a pair of recombinant chromosomes with a new combination of alleles (mitotic recombination)

- If it occurs at an early stage of embryonic development, it may result in a patch of tissue  with characteristics that are different from the rest of the cells in the organism - Curt Stern and patches on Drosophila

o X-linked genes for body color and bristle morphology

o “Twin spots” due to a single mitotic recombination in one cell during embryonic  development

CH 11 DNA REPLICATION 

Structural Overview of DNA Replication

- DNA replication relies on the Chargaff Rule (AT/CG)  

o AT has 2 H bonds and CG has 3 H bonds

- Strands of DNA are antiparallel

o One in 5’ to 3’ direction, the other in 3’ to 5’direction

- Summary of process

o 2 DNA strands come apart

o Each strand is a template for the synthesis of new strands

o The two new strands are daughter strands

o The two original strands are parental strands

- Mechanisms of DNA Replication

o Conservative

o Semiconservative

o Dispersive

- Matthew Meelson and Franklin Stahl’s Experiment with E. Coli

o Label DNA with heavy 15N  

o Put E. Coli in medium with light 14N  

o Collect cells after allowing to reproduce

o Analyze Density of DNA  

o Conclusion: 50% of DNA was labeled with heavy 15N and light 14N; 50% was  labeled with only light 14N. DNA replicates following the semiconservative  mechanism.

Bacterial DNA Replication

- Origin of Replication: 1 in Bacteria

- Synthesis is Bidirectional

- Replication forks meet at the opposite side of the bacterial chromosome - Catanase: ligation flipped between paternal and daughter strands

TEST 3 STUDY GUIDE

- Replication Process

o Origin of replication in E. Coli is called the oriC

o 3 sequences in oriC are significant

▪ AT rich region (13 nucleotides)

∙ Site of the first break

▪ 5 The DnaA boxes (9 nucleotides)

∙ Strand Unwinding

▪ GATC methylation sites (11 nucleotides)

∙ Regulation of replication

- Initiation:  

o Adenines are methylated in both strands by DNA Adenine Methyltransferase o 20 DnaA protein – ATP complexes bind to DnaA box sequences

o AT rich regions unwind (in a lefthand direction with positive supercoiling which  is relaxed by topoisomerase II / DNA Gyrase

o DnaA protein – ATP complex with the help of DnaC recruits 20 more proteins  and Helicase (also known as DnaB; to break H bonds) to bind

▪ Helicase made up of 6 regions powered by ATP  

o Single Strand binding proteins keep single strands from re-binding - Synthesis:

o DNA Primase synthesizes RNA primers (10-12 nucleotides)

▪ Needed to allow DNA polymerase to function

o DNA polymerase I (1 subunit) removes the primer from 5’ to 3’ and replaces it  with DNA from 5’ to 3’

o DNA polymerase III (10 diff. subunits) continues synthesis towards replication  fork

▪ α subunit synthesizes DNA (catalyzing ester bonds)

▪ β subunit (ring like) is the clamp that helps in constant synthesis process  (continual addition of enzymes)

▪ γ subunit is the clamp loader (other subunits can carry out this function) ▪ ε subunit is exonuclease (proof reading)

▪ Bacterial DNA polymerase may vary in subunit composition

o DNA ligase (powered by NAD+ which splits into NMP and AMP to give energy)  goes through after and creates 1 covalent bond to connect the synthesized DNA  fragments (okazaki fragments: 1000-2000 bp) (happens mostly in lagging strand  due to the looping)

o Tus binds to the Ter sequence in the DNA strands to terminate replication o Leading strand (in 5’ to 3’ direction) can be synthesized normally

o Lagging strand (in 3’ to 5’ direction) needs to loop (1000 -2000 bp) so that the  DNA polymerase can read in the 3’ to 5’ direction

▪ Has multiple RNA primers

o DNA is read in 3’ to 5’ direction (towards the replication fork) but the DNA is  synthesized in the 5’ to 3’ direction.

o DNA polymerase II IV and V carry out DNA repair

o DNA Polymerase works by catalyzing a phosphodiester bond between the  innermost phosphate group on the incoming deoxynuclease triphosphate and the  3’ OH of the sugar of the previous deoxynucleotide

TEST 3 STUDY GUIDE

▪ The last two phosphates of the incoming nucleotide are released in the  form of pyrophosphate (PPi)

o DNA Polymerase III is a processive enzyme

▪ If there is a problem with the ring, the DNA polymerase keeps falling off  (will only make about 20 bp before it falls off)

o Termination of Replication

▪ When catenanes (two intertwined circular DNA molecules) form, it quickly is rectified bu topoisomerase II (DNA Gyrase)

▪ Only need 1 ter sequence (directly opposite to the oriC) to stop replication  (tus that binds will halt both strands)

▪ If one is stalled, the other will not proceed past the ter sequence

- DNA Replication Complexes

o 2 DNA pol III act together to replicate the leading and lagging strands ▪ The two DNA polymerase holoenzymes form a dimeric DNA polymerase  that moves as one unit towards the replication fork

- Proofreading  

o DNA replication has a high rate of fidelity

o Reasons for fidelity

▪ Instability of mismatched pairs

∙ Complementary base pairs have a much higher stability than  

mismatched ones making mistakes less desirable

▪ Configuration of DNA polymerase active site

∙ α subunit

∙ DNA polymerase is unlikely to catalyze bond formation between  

mismatched pairs

∙ Induced fit phenomenon

▪ Proofreading function of DNA polymerase

∙ ε subunit

∙ DNA polymerase can identify a mismatched nucleotide and take it  off the daughter strands

∙ Uses its 3’ to 5’ exonuclease activity to remove

∙ Changes direction, moving from 5’ to 3’, to continue DNA  

synthesis

- DNA replication only occurs when the cell is about to divide

o Bacterial cells regulate the replication process by controlling the initiation of  replication at the oriC

▪ 2 mechanisms

∙ DNA adenine methyltransferase only adds methyl group to adenine  of GATC after the cell has enough nutrients

∙ The amount of DNA protein can provide a way to regulate

- DNA Replication Can be Studied in Vitro

o Arthur Kornberg

o Hypothesis: deoxynucleoside triphosphates are the precursors of DNA synthesis o Experiment:  

▪ Got an Extract of proteins from E. Coli  

▪ He had template DNA and Radio Labeled nucleotides

TEST 3 STUDY GUIDE

▪ He allowed them to synthesize new strands and precipitated the DNA  strands from the solution with acid

▪ Centrifuged to separate them from the radioactive nucleotides

o Results:  

▪ Found that DNA synthesis can be studied in vitro

- Isolation of Mutants

o Mutants can help us discover various things about genes, proteins and their  functions

▪ See what does or does not change when something is mutated

o DNA replication is vital for cell division, therefore if a mutation blocks DNA  synthesis it is most likely lethal

▪ Researchers must screen for conditional mutants (loss of function  

mutations)

o Brute force genetic screen

▪ E. Coli have many vital genes that are not involved in DNA rplication

∙ Only a small amount of mutations would effect the replication  

process

∙ Therefore researchers needed to screen thousands of mutants to  

find out which mutations influenced replication

o General Strategy

▪ Use a mutagen in a sample of bacteria cells to increase the likelihood of  mutations

▪ Expose to a mutagen and put on a growth medium.

▪ Incubate at permissive temperature (The temperature that allows the  

colony to divide and thrive)

▪ Replica plate colony into two plates; one at permissive temperature, one at  nonpermissive temperature. This allows researchers to identify mutations  that cannot grow at nonpermissive temperature.

▪ Rapid stop mutations: mutations that stop DNA synthesis quickly when  subjected to nonpermissive temperature.

∙ Ex: Helicase, DNA polymerase I and III, topoisomerase, etc.

▪ Slow-Stop mutations: mutations that are able to complete their current  round of replication before stopping when subjected to nonpermissive  

temperature

∙ Ex: DnaA, DnaC

DNA Replication in Eukaryotes

- Eukaryotic Chromosomes are much more complicated than bacterial chromosomes o Multiple Origins of Replication

▪ Origins of replication in saccharomyces cerevisiae are called ARS  

elements (Autonomously Replicating Sequence)

▪ 50 bp

▪ High percentage of A and T

▪ 3 – 4 copies of a specific sequence

▪

o Linear DNA

TEST 3 STUDY GUIDE

o DNA Associated with histones

- Proteins: DnaA, DNA Helicase, DnaC, topoisomerase, primase, polymerase I and III,  ligase (powered by ATP), Tus

- DNA replication proceeds bidirectionally

- Not as well understood as bacterial replication

- Replication starts when 14 different proteins form the prereplication complex (preRC) - Origin replication complex (ORC) – protein complex

o Part of preRC

o Six subunits

o Initiator of eukaryotic DNA replication (G1 phase)

o Uses ATP to bind to ARS elements first before the rest of preRC binds o Single stranded DNA stimulates ORC to hydrolyze ATP

- After both ORC and preRC are binded, preRC recruids others to come bind o Minisatellite chromosome maintenance helicase (MCM helicase)

- After MCM helicase bonds, S phase begins; end of DNA replication licensing - Eukaryotic DNA polymerases

o 4 subunits that have the function of replicating DNA

▪ α, δ, ε subunits: replication of nondamaged DNA in the cell nucleus  during S phase

▪ γ subunit: Replication of mitochondrial DNA

▪ α is the only polymerase associated with primase

- DNA pol α /primase complex synthesizes small DNA/RNA primer (10 mer RNA, 20-30  mer DNA)

- DNA pol α /primase complex then dissociates from the replication fork, and DNA pol α  is replaced by DNA pol ε in the leading strand and DNA pol δ in the lagging strand.  (polymerase switch)

- DNA pol β plays a role in base-excision repair

o Remove incorrect bases from damaged DNA and replace with correct one o Glycosylase removes nitrogenous base

o Ap endonuclease removes the sugar and phosphate

o DNA polymerase β comes and substitutes in the proper nitrogenous base - Lesion-replicating polymerases

o Involved in the replication of damaged DNA

o Can synthesize a complementary strand over the abnormal region

o Correction is made on the daughter strand, not the parent strand

o PCNA /lesion repair polymerase overrides the AT/GC rule to repair - Flap endonuclease removes RNA primers

o DNA pol δ elongates okazaki fragment until it reaches the primer of the next  fragment, this makes part of the primer into a flap which is then removed by flap  endonuclease. This continues until the entire primer is removed and DNA ligase  can seal the two fragments together.

o If a flap gets too long however, it is instead removed by Dna2 nuclease /helicase  which generates a short flap that is removed by flap endonuclease

o PNCA removes DNA polymerase δ backwards to start the process again on the  next fragment

- Histones must also be replicated to accommodate the increase of DNA

TEST 3 STUDY GUIDE

o follow a semiconservative model of replication

o occurs during S phase

o assemble into octamer structures that associate with the newly made DNA very  near the replication fork

- Telomeres and DNA replication

o Telomeric sequences consist of moderately repetitive tandem arrays and 3’  overhang that is 12-16 nucleotides long

CH 12 GENE TRANSCRIPTION AND RNA MODIFICATION 

2 Main Concepts

1) DNA sequences provide the underlying info  

a. Signals for start and end of transcription

2) Proteins recognize these sequences and carry out the process

a. Other proteins modify RNA transcript to make it functionally active

Base Sequences are Required for Gene Expression

- Gene = transcriptional unit

- The strand that is transcribed is the the template strand or anti-sense strand - The opposite stand is the coding strand or sense strands

o Base sequence is identical to RNA transcript except that uracil is used instead of  thymine

- Central Dogma

o Gene (transcription) ???? mRNA (translation) ???? polypeptide

o Polypeptide can become a functional protein or part of a functional protein - Sequences  

o DNA  

▪ Promoter: promote transcription

▪ Terminator: signal end of transcription

▪ Regulatory: site for binding of regulatory proteins

o RNA  

▪ Ribosomal binding site

▪ Start codons (AUG)

▪ Codons

▪ Bacterial mRNA

Different Functions of RNA Transcripts

- A structural gene encodes a polypeptide

o When structural genes are transcribed, the product is mRNA

- RNA transcripts from nonstructural genes are not translated

o Have various cellular functions

o In some cases, RNA transcript becomes part of a complex that contains protein  subunits

▪ Ribosomes

▪ Spliceosomes

▪ Signal Recognition particles

Promoter Sequences

- Promote gene expression by directing the exact location for the initiation of transcription - Located upstream of the site where transcription of a gene actually begins

TEST 3 STUDY GUIDE

o Bases in promoter sequences are numbered in relation to the transcription start  site (in a negative direction)

o Transcription start site = +1  

o -10 is the pribnow box (TATAAT)

o -35 is the first contact sequence (TTGACA)

o Sequence elements = short sequences critical to promoter recognition o Weak promoters: transcribe every 10 minutes and have substitutions in the -35  and -10 sequences

o Strong promoters: Transcribe one every 2 seconds. Have unaltered -35 and -10  sequences.

o Negatively labeled sequences will not be transcribed

o Modification in -10 and -35 sequences greatly decrease the rate of transcription Stages of Transcription

- Initiation

o RNA polymerase catalyzes synthesis of RNA

o In E.Coli, RNA polymerase holoenzyme is composed of a core enzyme (5  subunits) and a sigma factor (1 subunit: s)

▪ Without sigma factor, polymerase cannot bind  

o Step one: Sigma factor binds lightly to RNA and runs along it until it detects -35 and -10. It then binds tightly, with the help of a helix-turn-helix structure, forming a closed complex

o Step 2: Sigma factor binding causes tension in pribnow box, unwinding it and  forming an open complex (17 bases long).

o Step 3: A small strand of RNA is synthesized to check if RNA polymerase can  synthesize there. Sigma is released which indicated the end of initiation.

o

- Elongation

o Core enzyme slides down DNA to synthesize an RNA strand

o Either strand of DNA can be a template for transcription but the direction will be  different for either one

o Behind the open complex, the DNA rewinds back into a double helix

- Termination

o Occurs when the short RNA/DNA hybrid of the open complex is forced to  separate.

o E. Coli has 2 diff. mechanisms

▪ Rho-independent (intrinsic termination): 50% of structural genes

∙ Does not need (ρ) rho protein (same function as helicase)

∙ Stem loop forms

∙ URNA – ADNA hydrogen bonds are very weak and break apart easily

∙ NusA stabilizes RNA pol pausing

∙ No protein required to physically remove RNA from DNA

▪ Rho-dependent: 50% of structural genes

∙ Needs (ρ) rho protein

∙ Rho binds at rut site in RNA and moves towards 3’ end

∙ RNA polymerase reaches terminator sequence, a stem loop forms

attaches to RNA pol, and RNA pol stalls.

TEST 3 STUDY GUIDE

∙ Rho catches up to open complex and separates the DNA/RNA  

complex

Transcription in Eukaryotes

- Much more complicated

- Eukaryotic RNA polymerases

o RNA pol I: synthesizing all rRNA except 5S rRNA

▪ Found in nucleolus region

o RNA pol II

▪ Transcribes all structural genes  

▪ Synthesizes all mRNAs  

▪ Found in nucleus

▪ Transcribes some snRNA genes

o RNA pol III

▪ Transcribes all tRNA genes and 5S rRNA

o Structure

▪ Right angled turn allows nucleotides to easily polymerase (guided by  bridge)

▪ Rudder helps dissociation of RNA from DNA at RNA 3’ end

▪ Clamp controls rate of movement  

▪ 12-14 nucleotides in open structure

- Structural Genes

o TATA box: -25  

▪ Core promoter sequence

▪ Core promoter by itself produces a low level of transcription called basal  transcription

o Regulatory elements

▪ Vary in location but are often found in the 50-100 region

▪ Cis- acting elements

∙ DNA sequences that exert their effect only on nearby genes

∙ If there is a problem, nothing can be done with the gene

∙ Ex: TATA box, enhancers, silencers

▪ Trans-acting elements

∙ Regulatory proteins that bind to such sequences

∙ Usually found w/in stem cells, unless talking about hormones  

(estrogen)

▪ Enhancers

∙ Stimulate transcription

▪ Silencers  

∙ Inhibit transcription

- Transcription Factors in Polymerase II

o Three categories of proteins are required for basal transcription to occur at the  promoter

▪ RNA polymerase II

▪ 5 different general transcription factors (GTFs)

∙ TFIID

o TATA box binding factors

TEST 3 STUDY GUIDE

o Recognize TATA box

o Bind First  

o Recruits TFIIB

∙ TFIIB

o Acts as a bridge to recruit RNA pol II/TFIIF

∙ TFIIF

∙ TFIIE

o Comes after TFIIF

∙ TFIIH

o Last transcription factor to come in

o 10 polypeptides

▪ Some act as ATPase (hydrolyze ATP)

▪ Some act as protein kinase: transfer phosphate from  

hydrolyzed ATP to carboxyl terminal domain  

(CTD) in RNA pol II

▪ Some act as helicases (breaking H bonds to make  

open complex)

∙ TFIIB/TFIIH are released after formation of open complex

o This signals the end of initiation

▪ Mediator

∙ Binds transcription factors and RNA pol II to the template

∙ Mediates effect of transcription factors onto RNA pol II

∙ Can also phosphorylate CTD domain (to form an open complex)  

by itself (regulate TFIIH)

∙ If silencer is attached to mediate, other transcription factors cannot  dissociate (no transcription)

∙ Enhancers increase function of CTD

- RNA Modification

o The RNA that is made has to be modified to become mRNA  

o Pre mRNA is made before this

o Eukaryotes: polyA signal sequence at 3’ end of mRNA

▪ This is after the termination sequence

▪ Once RNA polymerase moves past poly A sequence, there are 2 methods  of termination

∙ Dislocation of elongation factors

∙ Termination factors added on  

o Creates instability between DNA/RNA complex

o Gene expression is collinear in prokaryotes due to the small amount of introns ▪ The RNA is the same length as the DNA at the end of the process

o Eukaryotes have too many introns for gene expression to be collinear ▪ Splicing: removes introns from RNA and splices the exons back together ▪ Trimming: Large strands of RNA are transcribed and then cut apart (not  splicing exons back together)

∙ tRNA and rRNA

▪ Riboendonuclease digests the bonds between the nucleotides in the introns