• 7/22/2019 Bio 102 Midterm Keiras Midterm II Review



    Biol102 Midterm II Review

    Keira Lucas

    May 13th, 2013

    Determine dominance (multiple alleles)

    Biochemical pathways

    Complementation Test

    Modified Mendelian ratios

    DNA structure and replication


  • 7/22/2019 Bio 102 Midterm Keiras Midterm II Review



    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

  • 7/22/2019 Bio 102 Midterm Keiras Midterm II Review



    Interactions between allelesExplanations for dominant


    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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    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


    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.


  • 7/22/2019 Bio 102 Midterm Keiras Midterm II Review



    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 + + + + - +



    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


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

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



    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


    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


    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+

  • 7/22/2019 Bio 102 Midterm Keiras Midterm II Review



    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



    Lots of red

    Less red


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


    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.


    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





    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



    bbyy B_yy

    Double mutant shows the expression

    of single y mutant!!!

    Y B


  • 7/22/2019 Bio 102 Midterm Keiras Midterm II Review




    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


    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




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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


    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


    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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    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


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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


    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


    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

  • 7/22/2019 Bio 102 Midterm Keiras Midterm II Review



    Small RNAs

    Lucas et al. 2013

    Transcription by

    RNA pol II

    Two processing

    steps, once in the

    nucleus the

    second in the


    RISC directedmRNA


    repression and/or



    The classic RNAi


    Dicer cleaves

    long dsRNA into

    short (21-25 nt)


    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.

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