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Biol102 Midterm II Review

Keira Lucas

May 13th, 2013

Determine dominance (multiple alleles)

Biochemical pathways

Complementation Test

Modified Mendelian ratios

DNA structure and replication

RNA

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Determining Dominance

Interactions between allelesMost mutant alleles we have discussed for Midterm I have been dominant orrecessive - the heterozygote has a phenotype like that of one homozygote.

Explanation for recessive mutations:

Haplosufficientcy: One copy of the mutant gene and one of the normal alleleconfers a normal phenotype because the heterozygote has 50% of normalenzyme activity which (for this trait) is enough for a normal phenotype

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Interactions between allelesExplanations for dominant

mutations:

Haploinsufficiency: one copyof the normal gene does notspecify enough enzyme for anormal phenotype

Dominant Negatives: alteredgene product that actsantagonistically to the wild-type allele. Presence ofmutant allele inactivatesentire complex.

Incomplete Dominance Heterozygote has an

intermediate phenotype

Mechanism: heterozygote hasan intermediate level ofenzyme activity and an

intermediate amount ofproduct is synthesized

F2 has 1:2:1 genotypic ANDphenotypic ratio

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Codominance

Both alleles are detected at the

same time in the heterozygote

Codominance usually involves a

system in which the 2 alleles of

a single gene have slightly

different products, both of

which appear in the expression

of the phenotype

Heterozygote also has anintermediate phenotype

Best example is ABO blood

type

Lethal Alleles

Expression of the lethal allele results in death ofan individual, which can affect the genotypic andphenotypic ratios!

Dominant lethal: aa normal, AA and Aa die

Recessive lethal: AA and Aa normal, aa die

Aa x AA (all progeny survive)

Aa x Aa ( progeny die, genetic ratio is 1:2)

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Thalassemia could be caused by partially recessive lethal gene in which TT is normal, Tt

is thalassemia, and tt is lethal to the embryo or child so these individuals are not

observed in offspring of Tt x Tt matings. The underlying genotypic and phenotypic ratio

is 1:2:1, but one class expected at of the total is missing because the gene is lethal

when homozygous.

1 TT; 2 Tt; 1tt

Normal thalassemia Die

Ae = elongate

Ab = bulbous

At = tapered

As = stubby

Cross #1: Gives stubby but neither parent is stubby.As must be recessive to Ae and Ab, Ae must be dominant over Ab. The cross must be

Ae/As x Ab/AsCross #2: Gives bulbous but both parents are tapered.

At must be dominant to Ab. The cross must be At/Ab x At/Ab

Cross #3: Gives elongate but neither parent is elongate.At/Ae x Ab/Ab if Ae is recessive to At but dominant to Ab.

Cross #4: Gives half of each parental type.At/Ab x Ab/Ab, where Ab is recessive to At.

Cross #5: Gives bulbous but neither parent is bulbous.As must be recessive to both Ae and Ab. The cross must be Ae/Ab x As/As.

Cross #6: Give stubby but neither parent is stubby.As must be recessive to both Ae. The cross must be Ae/As x Ae/As.

At>Ae>Ab>As

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Biochemical Pathways

Understanding biochemical pathwaysusing Neurospora crassa

Auxotrophica strain that grows only if an

additional nutrient is in the medium

Prototrophica strain (like wild type) that

grows on minimal medium

Synthesis of a particular nutrient by a

prototrophic strain occurs through a

biochemical pathway with discrete steps

controlled by genes

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Understanding biochemical pathways

using Neurospora crassa

A block near the end of the pathway (argenine) cannot be "fixed"by adding a substance that occurs earlier in the pathway

A block near the beginning of the pathway can be fixed by addinganything that occurs later in the pathway

Ch 6 #15

A B C D E G

1 - - - + - +

2 - + - + - +

3 - - - - - +

4 - + + + - +

5 + + + + - +

Compound

Mutant

Sever mutants are isolated, all of which require

compound G for growth. The compounds (A to E) in

the biosynthetic pathway to G are known, but their

order in the pathway is not known. Each compound

is tested for its ability to support growth of each

mutant (1 to 5). In the following table, a plus sign

indicates growth and a minus sign indicated no

growth.

a. What is the order of compounds A to E in the

pathway?b. At which point in the pathway is each mutant

blocked?

EACBDG

The more mutants the compound supports, the later it is in the pathway.

5 4 2 1 3

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Complementation Test

How to identify genes that contributeto a particular trait?

1. Obtain many single-gene mutations bytreating cells with mutagensidentifymutants with mutant phenotype for yourtrait

2. Test the mutants for allelism (are two

mutants from the same locus?) use theCOMPLEMENTATION TEST

3. Combine mutants to form double mutants totest for gene interaction

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The Complementation Test

Intercross individuals that display the samerecessive mutant phenotype (homozygousrecessive individuals)

Do progeny have a wild-type phenotype?

Recessive mutations MUST be on different genes!

Mutations have COMPLEMENTED each other

Do they have a mutant phenotype?

Recessive mutations MUST be on the same gene!

3, mutant 2 and 4 are on the same gene

Line 1 : a/a ; B/B ; C/C

Line 2 : A/A ; b1/b1 ; C/C

Line 3 : A/A ; B/B ; c/c

Line 4 : A/A ; b2/b2 ; C/C

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Modified Mendelian Ratios

Gene Interactions

9:3:3:1 ratioNo gene interaction

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9:7 ratiocomplementary gene interaction

Two gene loci complement each other to produce aparticular phenotype

One dominant allele at each locus for expression of thephenotype

P: white flower x white flower

F1: purple flowers

F2: 9 purple flowers : 7 white flowers

9 A-;B- Purple3 A-;bb White

3 aa;B- White

1 aa;bb White

9:3:4 ratioRecessive Epistasis Double mutant shows phenotype of one mutation but not the

other

P: white flower x magenta flower

F1: blue flowers

F2: 9 blue flowers; 3 magenta flowers; 4 white flowers

9 w+-;m+- blue

3 w+-;mm magenta

3 ww;m+ white

1 ww;mm white

Colorlessmagentablue

Magenta is blocked by a mutant allele for gene w, blue is blocked bya mutant allele for gene m. White (w) is epistatic.

Gene w+ Gene m+

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12:3:1 ratioDominant Epistasis

The dominant allele for a gene (the epistatic gene) may mask the effect of

another allele for another gene

P: dark red flower x white spotted flower

F1: all white flowers

F2: 12 white spotted flowers; 3 dark red flowers; 1 light red flowers

9 W-;D- White spotted

3 W-; dd White spotted

3 ww; D- dark red

1 ww; dd light red

White (W) is epistatic and must prevent to synthesis of the red pigment!

PrecursorAnthocyanin (red pigment)Restricts red pigment to spots

D

d

Lots of red

Less red

W

13:3 ratioSuppressor activity One gene masks the expression of a specific allele of another

gene.

Mutant allele reverses the effect of a mutation of anothergene

P: Red flower x purple flower

F1: All red flowers

F2: 13 red flowers: 3 purple flowers

9 A-;B- red

3 A-;bb red

3 aa; B- purple

1 aa;bb red

Mutant b reverses the affect of the a mutation!

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Penetrance and Expressivity

Penetrance - the frequency with which agenotype actually expresses aphenotype. A trait has low penetrance iffew individuals with the gene express it.

Expressivity - the degree to which a traitis expressed in individuals having thegene.

Reasons:

1. Environment can modify expression

2. Other interacting genes(suppressors etc.) obscurephenotype

3. Mutant phenotype can be difficultto measure accurately

The F2 ratio is about 9:7, which indicates

complementary gene interaction

BBrr

bbRR

BbRr

B_R_

B_rr bbR_ bbrr

This is complementary gene action in which only individuals with at least one

dominant allele at each loci can complete the pathway and produce the runner

phenotype; all other genotypes are bunch.

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a. How many gene loci appear to be involved in the production of this flower color

phenotpye in snapdragon?

b. This is an example of what type of genetic phenomenon?

c. Using your own clearly defined genetic symbols, give the genotypes of the parental, F1

and F2 plants.

d. Provide a possible biosynthetic pathway that could explain these results.

2 gene loci

12:3:1 ratioDominant epistasis

Dominant epistasis

9 R_W_ White

3 rr W_ White

3 R_ww Dark red

1 rrww Light red

W is epistatic to R and R. W inhibits the expression of the red pigment

PrecursorAnthocyanin (red pigment)White

D/d W

Recessive epistasis!!

B b

brown Yellow y

B_Y_

bbY_

bbyy B_yy

Double mutant shows the expression

of single y mutant!!!

Y B

Yellowbrownblack

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DNA

You should probably know and understand all

experiments talked about in class regarding DNA

(All in lecture 9):

Griffiths transformation experiment

Avery's Experiments

Hershey-Chase Experiment

Chargaff

Franklin and Wilkins

Watson and Crick

Meselson and Stahl

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The Nucleotide

DNA is a polymer (chain)composed of monomers callednucleotides. A nucleotide iscomposed of 3 parts:

1. a pentose (5-carbon) sugar(deoxyribose in DNA)

2. a nitrogenous basepurines - with 2 rings: adenine andguanine

pyrimidineswith 1 ring: cytosine,thymine, uracil

3. A phosphate group

You can remember C and T and pyrimidines because they have a

Y in the name!!

DNA Structure Double helix

Sugar-phosphatebackbone (sugars linked tophosphates though the 3and 5 carbon atoms of thesugar)

Base pairs between CGand A=T

Antiparallel chains(opposite polarity) one strand is 5' to 3' and

the other 3' to 5'

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each strand of the helix is composed of deoxyribose sugars linked to phosphates thoughthe 3 and 5 carbon atoms of the sugar.

The bases are stacked in the center of the helix and form base pairs; CG and A=T

The bases of different strands form H bonds which hold the two strands together, 2 bonds for

A=T pairs and three for CG pairs. All bonds within a single strand of DNA are covalent bonds.

determined by the orientation of the deoxyribose sugars of adjacent nucleotides, a

phosphate group on the 5 carbon and an OH group on the 3 carbon. This

means that the 5 end of each DNA chain will have a phosphate group, and the 3 end an OH group. The two strand of DNA have opposite polarity.

DNA Replication Semiconservative replication:

each new molecule contains oneold strand and one new

Conservative replication: the twooriginal strands remain togetherafter replication - one all "old"molecular, one all "new

Dispersive replication: newmolecules are composed onsome new and some oldnucleotides

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Meselson - Stahl Experiment

Labeled DNA with heavy isotope (15N)

Allowed a few round of replication andtransferred to media with (14N)

Allowed more replication and then separatedDNA according to density

The density of the DNA reflects the distribution ofthe original labeled DNA into the new strands ofDNA

1

2

2

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Prokaryotic DNA Replication Replication is initiated at a specific

site on the chromosometheorigin of replication ORI is AT-rich and so more easily

denatured

Origin of Replication Complex(ORC) binds ORI

DNA helicase denatures DNA

Primase complexes with helicaseand creates and RNA primer

Replisome is recruited and DNA polIII extends the RNA primer

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Replication Fork

DNA pol III: primary enzyme responsible for DNA replication. Adds

nucleotides to the 3 tip of the new strand (synthesizes DNA 5-3)

Beta clamp protein: Encircles the DNA and keeps pol III attached to the DNA

Primase: In order for synthesis to begin, a primer must be made. Primasesynthesizes a short RNA primer complementary to the region beingreplicated. Leading stand: 1 primer. Lagging strand: 1 primer/okazakifragment.

DNA pol I: Removed RNA primer with a 5-3 exonuclease activity and fills inthe gaps using a 5-3 polymerase activity

Ligase: Joins the 3 end of the gap filling DNA to the 5 end of the downstreamokazaki fragment

Helicase: disrupts the H bonds holding the double helix together

Topoisomerase: Removes twists and supercoils in the DNA

SSBP: stabilizes ssDNA

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Eukaryotic DNA Replication

Replisome contains more subunits thanprokaryotic replisome because nucleosomes butbe removed and reassembled

Chromatin assembly factor 1 (CAF-1) brings newhistones to the DNA

Replication of DNA ends Replication of ends is

complex because the 5'

end of a strand will be

an RNA primer which is

removed

It cannot be replaced

because there is no 5'

sequence at which new

synthesis can initiate

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Telomeres and Telomerase

Telomerase addsnucleotides to the endof the overhanging DNA,lengthening thetelomere sequence

Synthesizes DNA usingan RNA template(reverse transcriptase)

Gap is filled in bynormal DNA synthesis

DNA polymerase III is the enzyme responsible for synthesis of new strands of DNA (adding

nucleotides onto the 3 end of the new strand extending the RNA primer) using an existing

strand as template (reading the template strand 3-5). The two strands of DNA in a double

helix are replicated simultaneously, one by each molecule of DNA polymerase III.

DNA polymerase III can only read the template strands 3 5, adding nucleotides only to

the 3 end of a nucleotide chain. The two strands of DNA in a double helix are of oppositepolarity, so only one can be replicated as a single molecule, the other strand (lagging strand)

has the wrong orientation relative to movement of the replication fork.

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RNA

Classes of RNAs Messenger RNA (mRNA): RNAs that specify the amino acid sequence of a polypeptide

Transfer RNA (tRNA): Function as intermediates in translation, bringing amino acids to thepolypeptide

Ribosomal RNA (rRNA): a major component of the ribosomes that are involved in translation ofmRNA to polypeptides

Small nuclear RNA (snRNA): in eukaryotes only, component of the spliceosome that removesintrons from the pre-mRNA

MicroRNAs (miRNA): small RNAs that regulate gene expression through mRNA degradation and/ortranslational inhibition

Small interfering RNAs (siRNA): small RNAs that function in protecting the integrity of genomes.Functioning in anti-viral RNAi and transposon silencing.

Piwi-interacting RNAs (piRNAs): small RNAs that function in protecting the integrity of genomes.Preventing rampant spread of transposable elements in gonads, but have recently been identifiedin somatic cells.

Long noncoding RNAs (lncRNAs): Function of most lncRNAs is stillunknown.

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Prokaryotic Transcription

Initiation: RNA polymerase holoenzyme that

binds to the promoter is composed on 5 core

subunits plus a sigma factor ( ), which

recognizes different promoter sequences

usually at -35 to -10. Sigma is released after

transcription initiation.

Prokaryotic Transcription

Elongation: RNA polymerase moves along the

template strand, unwinding the helix and

rewinding it as it passes. Adds nucleotides

(A,U,G,C) to the 3 growing end of the

transcript.

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Prokaryotic Transcription

Termination: Two types

1. Intrinsic terminationthis depends on a specific

sequence in the RNA that forms a stem-loop

structure by internal base-pairing. The sequence

is G-C rich and ends in about 8 U's.

2. Rho dependenta "helper" protein called rho

binds to a sequence called rut. Polymerase

pauses when it reaches a sequence downstream

of rut, and rho facilitates release of the RNA

from the polymerase

Eukaryotic TranscriptionInitiation:

1. TBP, which is part of theTFIID complex, binds toTATA box ~ 30bp from TSS

2. Additional proteins bind tothis complex to form thepreinitiation complex (PIC)

3. The enzyme RNApolymerase binds to thecomplex and begins RNAsynthesis.

4. Most proteins are releasedfrom the PIC

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Eukaryotic TranscriptionElongation:

RNA polymerase reads thetemplate DNA stand reads 3' to5 adding ribonucleotides tothe 3 end of the growingtranscript

Similar to prokaryotictranscriptional elongation butthe pre-mRNA is alsoprocessed: Addition of a 7-methyl guanine

cap (modified nucleotide) tothe 5' end

Splicing to eliminate introns bysnRNPs

Eukaryotic TranscriptionTermination:

Transcription terminates

when the RNA contains

the sequence AAUAAA or

AUUAAA

Another processing step:

Then about 150-200 A's

are added to the cut end

to form the poly(A)-tail

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Why is eukaryotic transcription more

complex than prokaryotic transaction?

1. Low gene densityone gene per 1400 bp in E.coli vs one gene per 100,000 bp in humans. Howis the appropriate start point recognized? Threepolymerases specialized for different types ofgenes (rRNA, proteins-coding, small functionalRNAs)

2. Eukaryotes have a nucleus containing the DNA,but proteins are synthesized in the cytoplasm.RNA is modified (processed) in the nucleus

3. Chromatin in eukaryotes is more complex thanthe "naked" DNA in prokaryotesit can regulatetranscription

RNA polymerase promoter

template ribonucleotides

3

1) A cap composed of 7-methylguanosine is added to the 5end of the mRNA

2) Introns are spliced out of the transcript as it is being synthesized

3) A poly-A tail is added to the 3end of the transcript

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Small RNAs

Lucas et al. 2013

Transcription by

RNA pol II

Two processing

steps, once in the

nucleus the

second in the

cytoplasm

RISC directedmRNA

translational

repression and/or

mRNA

degradation

The classic RNAi

Pathway

Dicer cleaves

long dsRNA into

short (21-25 nt)

fragments

RISC directed

RNA cleavage

RNAi (siRNA) function1. Anti-viral defense mechanismreplication of some

virus may lead to double stranded RNA intermediateswhich are taken up into the classical RNAi pathway

2. Genome maintenance and stability (genome defense)siRNAs can promote specific transposable elementinactivation. TEs may integrate into important regionsof the genome, causing mutations, TE specific siRNAs

can inhibit this process.3. Now scientist can utilize the RNAi pathway for their

advantage by using a reverse genetics approach totarget specific genes by injecting or expressing doublestranded RNA. DsRNA will be picked up by the RNAimachinery and promote siRNA specific cleavage oftheir gene of interest.

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You should probably know:

Jorgensens cosuppression - attempt to overexpressa pigment synthesis enzyme, chalcone synthase, to

produce a deep purple petunia; however, instead thelarge majority of the plants harboring the transgeneunexpectedly produced completely white flowers!!Both genes (the plant's normal gene and the addedone) would sometimes be silenced.

Lin-4was the first miRNA discovered (it wasnt until 7years later that another miRNA was even discovered!!!)

Fire and Mello experiments - dsRNA initiates an RNAiresponse (gene silencing)

Baulcombe - 25nt anti-sense RNA molecules derivedfrom an RNA template. Suggested that these small RNAmolecules could be the basis of post-transcriptionalgene silencing.