BIMM 100 Exam 1

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covers LE 1-6, week 2-3 DI notes & slides, exam 1 review notes & slides (incl. practice problems), problem sets 1-3 (3rd is the optional problem set, but likely include flashcards on it)

Last updated 4:28 AM on 4/21/26
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111 Terms

1
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def. gene

  • 2 conditions for it to be the molecular basis of a gene

a unit of hereditary trait that shapes a specific trait of a cell or organism

  • (hereditary traits - traits passed through generations)

_

DNA

  • should be transmitted from cell-to-cell (or parent-to-child)

  • should generate a trait

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what did Griffith’s experiment discover? (1)

_

describe rough vs. smooth bacteria (← 2 each)

(& what happened to mice infected with each?)

_

what happens to:

  • mice w/ smooth cell bacteria introduced to heat (←explain w/ 2)

  • mice w/ smooth cell bacteria introduced to heat + rough cells (← explain w/ ~3, where 1 of them def. diff. b/w genes and a specific term)

_

are dead bacteria virulent or non-virulent?

_

genes shape __, while chromosomes do what? (1)

discovered that a trait, like bacterial virulence (cause death/disease), can be transferred from cell-to-cell

_

rough bacteria is non-virulent

  • do NOT have polysaccharide “cloak”

  • are destroyed by the mouse immune system

    • SO mice infected with R cells survive

smooth bacteria is virulent

  • have polysaccharide “cloak”

  • sneak past the mouse immune system

    • SO mice infected with S cells die

__

heat-killed smooth cells → mice survive

  • b/c heat-killed (dead) smooth cells are non-virulent

    • heat kills bacteria

    • dead bacteria are non-virulent

dead smooth cells + live rough cells → mice die

  • b/c live smooth cells are recovered from the dead mouse

    • dead cells can transfer their traits to live cells

    • ^ virulence can be transferred from cell-to-cell

  • SO something else is transmitted from S cells → R cells in order to change the virulence of R cells, which are CHROMOSOMES

    • genes shape traits, while chromosomes carry genetic info & have both nucleic acids AND proteins

<p>discovered that <strong><u>a trait, like bacterial virulence (cause death/disease), can be transferred from cell-to-cell</u></strong></p><p>_</p><p><u>rough </u>bacteria is <u>non-virulent</u></p><ul><li><p>do NOT have polysaccharide “cloak”</p></li><li><p>are destroyed by the mouse immune system</p><ul><li><p>SO mice infected with R cells survive</p></li></ul></li></ul><p><u>smooth </u>bacteria is <u>virulent</u></p><ul><li><p>have polysaccharide “cloak”</p></li><li><p>sneak past the mouse immune system</p><ul><li><p>SO mice infected with S cells die</p></li></ul></li></ul><p>__</p><p>heat-killed smooth cells → mice <u>survive</u></p><ul><li><p>b/c heat-killed (dead) smooth cells are non-virulent</p><ul><li><p><strong><mark data-color="blue" style="background-color: blue; color: inherit;">heat kills bacteria</mark></strong></p></li><li><p><strong><mark data-color="blue" style="background-color: blue; color: inherit;">dead bacteria are non-virulent</mark></strong></p></li></ul></li></ul><p>dead smooth cells + live rough cells → mice <u>die</u></p><ul><li><p>b/c live smooth cells are recovered from the dead mouse</p><ul><li><p><strong><mark data-color="blue" style="background-color: blue; color: inherit;">dead cells can transfer their traits to live cells</mark></strong></p></li><li><p><strong><mark data-color="blue" style="background-color: blue; color: inherit;">^ virulence can be transferred from cell-to-cell</mark></strong></p></li></ul></li><li><p><u>SO something else is transmitted from S cells → R cells in order to change the virulence of R cells, which are </u><strong><u>CHROMOSOMES</u></strong></p><ul><li><p><strong><u><mark data-color="red" style="background-color: red; color: inherit;">genes shape traits, while chromosomes carry genetic info &amp; have both nucleic acids AND proteins</mark></u></strong></p></li></ul></li></ul><p></p>
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<p>def. chromosomes</p><ul><li><p>describe (2)</p></li></ul><p>___</p><p>label which is nucleic acid vs. amino acid </p><p>_</p><p>identify 5 parts of an amino acid</p><p>#of common amino acids</p><p>_</p><p>identify 3 parts of nucleic acid</p><p>name the 5 nucleotides (which 2 belong to what specifically)</p>

def. chromosomes

  • describe (2)

___

label which is nucleic acid vs. amino acid

_

identify 5 parts of an amino acid

#of common amino acids

_

identify 3 parts of nucleic acid

name the 5 nucleotides (which 2 belong to what specifically)

are carriers of heritable traits/genes & are made of nucleic acids AND proteins

  • when a cell divides, chromosomes are transmitted to both daughter cells

  • inheritance of specific traits correlates to inheritance of chromosomes

__

amino acid has central carbon (alpha carbon), attached to:

  • amino group (NH2)

  • side chain (R)

  • carboxylic acid group (COOH)

  • H

__

nucleic acid has multiple:

  • phosphate group

  • sugar ring (is deoxyribose sugar for DNA)

  • Nitrogenous base

nucleotides (have one of each of the above)

  • adenine, cytosine, guanine, thymine (DNA), uracil (RNA)

<p>are <u>carriers of heritable traits/genes</u> &amp; are made of <u>nucleic acids AND proteins</u></p><ul><li><p>when a cell divides, chromosomes are transmitted to both daughter cells</p></li><li><p>inheritance of specific traits correlates to inheritance of chromosomes</p></li></ul><p>__</p><p>amino acid has central carbon (alpha carbon), attached to:</p><ul><li><p>amino group (NH<sub>2</sub>)</p></li><li><p>side chain (R)</p></li><li><p>carboxylic acid group (COOH)</p></li><li><p>H</p></li></ul><p>__</p><p>nucleic acid has multiple:</p><ul><li><p>phosphate group</p></li><li><p>sugar ring (is deoxyribose sugar for DNA)</p></li><li><p><u>N</u>itrogenous base</p></li></ul><p>nucleotides (have one of each of the above)</p><ul><li><p>adenine, cytosine, guanine, thymine (DNA), uracil (RNA)</p></li></ul><p></p>
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<p>review: what molecule is needed to transmit S cells → R cells in order to change the virulence of R cells</p><p>conclusion of Avery-Macleod-McCarty experiment (1)</p><p>_</p><p>what happens to R cells if you:</p><ul><li><p>break open/lyse S cells, eliminate protein w/ proteinase, add R cells to S-cell lysate that lacks protein</p></li><li><p>lyse S cells, eliminate DNA w/ DNase, add R cells to S-cell lysate that lacks DNA</p></li></ul><p></p>

review: what molecule is needed to transmit S cells → R cells in order to change the virulence of R cells

conclusion of Avery-Macleod-McCarty experiment (1)

_

what happens to R cells if you:

  • break open/lyse S cells, eliminate protein w/ proteinase, add R cells to S-cell lysate that lacks protein

  • lyse S cells, eliminate DNA w/ DNase, add R cells to S-cell lysate that lacks DNA

chromosome, that contain nucleic acids (RNA, DNA) & proteins

  • which led to hypothesis that nucleic acid or protein was the genetic molecule

_

conclusion:

  • absence of DNA (the genetic material) will block rough/non-virulent cells from transforming into smooth/virulent cells (whereas, smooth cell DNA transforms rough cells into smooth cells)

  • (aka smooth cell traits are transmitted to rough cells by DNA)

__

w/ proteinase

  • S-cell lysate can still cause R/non-virulent cells to transform into S/virulent cells B/C the lysate doesn’t have protein BUT does have nucleic acid

  • protein is not the genetic material

w/ DNase

  • R/non-virulent cells do NOT transform into S/virulent cells B/C the lysate does not contain DNA, even though it does contain RNA and proteins

  • DNA is the genetic material

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<p>include C1-5</p><p>_</p><p>DNA is 2 strands of __ in a helix</p>

include C1-5

_

DNA is 2 strands of __ in a helix

DNA is 2 strands of nucleotides in a helix

(picture should be deoxyribose sugar ring)


(remember that a nucleotide is made up of 1 phosphate group, 1 sugar ring, 1 nitrogenous base)

(a nucleic acid is made of multiple nucleotides)

<p>DNA is 2 strands of <u>nucleotides</u> in a helix</p><p>(picture should be deoxyribose sugar ring)</p><div data-type="horizontalRule"><hr></div><p>(remember that a nucleotide is made up of 1 phosphate group, 1 sugar ring, 1 nitrogenous base)</p><p>(a nucleic acid is made of multiple nucleotides)</p>
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diff. b/w results & conclusion

results = data/ / observations

conclusion = interpretation

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state what is attached at positions C1-3 and C5 in a pentose sugar ring

  • for 1 of them, state its importance

C1’ — where nitrogenous base attaches (purine or pyrimidine base)

C2’ — determines if molecule is RNA or DNA (w/ hydroxyl group)

  • 2’ OH means RNA (think RUH “ROH”)

  • not 2’ OH, but 2’ H, means DNA

C3’ — where OH/hydroxyl group attaches, to form a polynucleotide strand (3’ OH)

C5’ — where phosphate group attaches

<p>C1’ — where nitrogenous base attaches (purine or pyrimidine base)</p><p>C2’ — determines if molecule is RNA or DNA (w/ hydroxyl group)</p><ul><li><p>2’ OH means RNA <mark data-color="#eaa3df" style="background-color: rgb(234, 163, 223); color: inherit;">(think RUH “ROH”)</mark></p></li><li><p>not 2’ OH, but 2’ H, means DNA</p></li></ul><p>C3’ — where OH/hydroxyl group attaches, to form a polynucleotide strand (3’ OH)</p><p>C5’ — where phosphate group attaches</p>
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diff b/w purine vs. pyrimidine

which of the 5 bases are a part of each?

___

draw all the 5 bases (might help to think about the basic structure of purines & pyrimidines)

  • & highlight which parts are hydrogen bond donors & acceptors

  • ^ which 2 bases have 3 total donors/acceptors vs. 2

(MB know #s w/in a ring)

^^ can also refer to drawings in midterm notebook for better visualization

purine - 2 rings

  • A, G

pyrimidine - 1 ring

  • C, T, U

__

C-G have 3 H bonds

A-T have 3 H bonds

<p>purine - 2 rings</p><ul><li><p>A, G</p></li></ul><p>pyrimidine - 1 ring</p><ul><li><p>C, T, U</p></li></ul><p>__</p><p>C-G have 3 H bonds</p><p>A-T have 3 H bonds</p>
9
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def. glycosidic bond

  • enzyme that cleaves it

bond b/w C1’ of sugar ring & N-base

  • cleaved by glycosylase

<p>bond b/w C1’ of sugar ring &amp; N-base</p><ul><li><p>cleaved by glycosylase</p></li></ul><p></p>
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nucleoside vs. nucleotide (1 each)

  • name 4 nucleosides

t/f: different #s of phosphates can compose a nucleotide (meaning?)

nucleoside has attached OH (hydroxyl) group

  • adenosine, thymidine, guanidine, cytidine

nucleotide has attached phosphate group

(^ both have sugar ring & N-base)

__

true

  • dAMP (deoxy””-5’-monophosphate), dADP (bi-), dATP (tri-)

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name 4 things that make RNA diff. from DNA

RNA

  • ribose sugar, where 2’ OH (think RUH “ROH”)

  • has uracil, not thymine

  • usually single stranded

  • can function as an enzyme: “ribozyme”

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def. phosphodiester bond

  • what enzymes forms it (2) ← & name b/w what 2 things the bond forms between

  • what enzymes cleave it (2)

this bond forms (1)

bond b/w 3’ OH & 5’ of 2 nucleotides (specifically b/w 2 sugar rings)

  • (where 3’ OH becomes 3’ O that’s part of phosphate, which binds to the 5’ of another ring)

  • 3” OH can be either with :

    • 5’ phosphate w/in the strand — DNA ligase

    • 5’ phosphate of incoming dNTP — DNA polymerase

__

made from DNA ligase & DNA polymerase

cleaved from exonuclease & endonuclease

__

phosphodiester bond forms the sugar-phosphate backbone w/in the SAME strand

<p>bond b/w <strong><u>3’ OH &amp; 5’ </u></strong>of 2 nucleotides (specifically b/w 2 sugar rings)</p><ul><li><p><span style="color: rgb(176, 174, 174);">(where 3’ OH becomes 3’ O that’s part of phosphate, which binds to the 5’ of another ring)</span></p></li><li><p>3” OH can be either with :</p><ul><li><p>5’ phosphate w/in the strand — DNA ligase</p></li><li><p>5’ phosphate of incoming dNTP — DNA polymerase</p></li></ul></li></ul><p>__</p><p>made from DNA <u>ligase</u> &amp; <u>DNA polymerase</u></p><p>cleaved from <u>exonuclease &amp; endonuclease</u></p><p>__</p><p>phosphodiester bond forms the <u>sugar-phosphate </u><strong><u>backbone</u></strong><u> w/in the SAME strand</u></p>
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describe 4 protein structures

  • bond of secondary? structures (can only name 1)?

___

name if single vs. double-stranded nucleic acid chains

primary - linear polypeptide chain of amino acids

secondary - folding of backbone w/ H bonds

  • alpha helix, etc.

tertiary - 3D folding of chain

quaternary - multiple chains into a protein

__

single - RNA

double - DNA

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for discovering structure of DNA

brief description of Florence Bell vs. Linus Pauling model for DNA ← both were incorrect models of DNA

  • what’s incorrect about Pauling’s model (2)

3 parts of correct DNA structure? ← by Franklin-Wilkins-Watson-Crick

direction of DNA movement

Bell — stacked nucleotides

Pauling — triple helix

  • wrong about:

    • 3 strands of DNA in the helix (triple helix) — should be double-helix

    • bases being outside the helix — should have bases from opp. strands pair w/ e/o in the center of the helix

__

  • 2 strands of DNA (double-helix)

  • base pairs, where nucleotide bases on opposite strands form hydrogen bonds

  • the 2 strands are anti-parallel to e/o (5’→3’, 3’→5’ which refer to the C in the sugar ring)

DNA moves from 5’ → 3’ end

<p>Bell — stacked nucleotides</p><p>Pauling — triple helix</p><ul><li><p>wrong about:</p><ul><li><p>3 strands of DNA in the helix (triple helix) — should be double-helix</p></li><li><p>bases being outside the helix — should have bases from opp. strands pair w/ e/o in the center of the helix</p></li></ul></li></ul><p>__</p><ul><li><p><u>2 strands</u> of DNA (<u>double-helix</u>)</p></li><li><p><u>base pairs</u>, where nucleotide bases on opposite strands form <u>hydrogen bonds</u></p></li><li><p>the 2 <u>strands are anti-parallel</u> to e/o <span style="color: rgb(158, 154, 154);">(5’→3’, 3’→5’ which refer to the C in the sugar ring)</span></p></li></ul><p>DNA moves from 5’ → 3’ end</p>
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<p></p>

<p></p>
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can RNA be antiparallel? explain

yes

  • even though it’s single-stranded, it can become a complex structure, where the single strand folds in on itself & the complementary bases can pair together ← anti-parallel

<p>yes</p><ul><li><p>even though it’s single-stranded, it can become a complex structure, where the single strand folds in on itself &amp; the complementary bases can pair together ← anti-parallel</p></li></ul><p></p>
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name the 2 types of intermolecular interactions present in DNA (& what are the interactions)

what about another one w/ what specific types of bonds?

__

(2) are important for stability of DNA

hydrogen bonds — base pairing (aka b/w bases of opposite strands)

Van der Waals forces — base stacking

  • (does NOT include any of the 3 bonds talked about earlier)

_

covalent bonds

  • glycosidic bonds

  • phosphodiester bonds

__

base pairing & base stacking for stability

<p>hydrogen bonds — base pairing (aka b/w bases of opposite strands)</p><p>Van der Waals forces — base stacking</p><ul><li><p>(does NOT include any of the 3 bonds talked about earlier)</p></li></ul><p>_</p><p>covalent bonds</p><ul><li><p>glycosidic bonds</p></li><li><p>phosphodiester bonds</p></li></ul><p>__</p><p>base pairing &amp; base stacking for stability</p>
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what 2 types of hydrogen bonds are for DNA (aka the molecules that base pair for DNA) vs. 1 type of H bond for proteins

DNA

  • O — H(N)H

  • N — H(N)

proteins

  • O — H(N) ← for alpha helix, beta sheet

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<p>relation b/w H bonds &amp; methyl group (CH3)</p>

relation b/w H bonds & methyl group (CH3)

C

thymidine (“-side”)

__

hydrogen bonds aren’t usually formed w/ methyl groups (CH3)

<p>C</p><p>thymidine (“-side”)</p><p>__</p><p><strong><u>hydrogen bonds aren’t usually formed w/ methyl groups (CH<sub>3</sub>)</u></strong></p>
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review(ish)

locate hydrogen bond acceptors vs. donors for 5 bases (probably draw it out)

__

t/f: molecules have additional H-bond donors and acceptors that aren’t highlighted by arrows

t/f: base pairs form due to alignment of hydrogen bond donors and acceptors

true

(i.e. the other H in NH2)

_

true

<p>true</p><p>(i.e. the other H in NH<sub>2</sub>)</p><p>_</p><p>true</p>
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the secondary structure of DNA and RNA, specifically (1), is sensitive to (2 ← specify a bit)

__

def. Tm

what happens as temp increases

  • relate this to Tm (melting temperature) (2) ← & explain each

what happens as temp decreases

H bonds of secondary structure is sensitive to temperature & pH

  • high temp & extreme pH (very high or very low pH), disrupt the secondary structure

_

Tm (melting temp)

  • temperature where the helix is half double-stranded & half single-stranded (aka temp when 50% denatured)

_

as temp increase, the strands of DNA helix will denature/separate

  • stable DNA helix = high Tm

    • ^ stable helix needs more heat to denature, so high Tm

    • (denatures 50% at higher temp, so more stable)

  • unstable DNA helix = low Tm

    • ^ unstable helix needs less heat to denature, so low Tm

    • (denature 50% at lower temp, so more unstable)

as temp decreases, DNA strands w/ complementary sequences will anneal/zip up

<p><u>H bonds</u> of secondary structure is sensitive to <u>temperature &amp; pH</u></p><ul><li><p><strong><u>high temp &amp; extreme pH </u></strong>(very high or very low pH), disrupt the secondary structure</p></li></ul><p>_</p><p>T<sub>m</sub> (melting temp)</p><ul><li><p>temperature where the helix is <u>half double-stranded &amp; half single-stranded</u> (aka<strong><u> temp when 50% denatured)</u></strong></p></li></ul><p>_</p><p>as temp increase, the strands of DNA helix will <strong>denature/separate</strong></p><ul><li><p><strong><u>stable </u></strong>DNA helix = <strong><u>high Tm</u></strong></p><ul><li><p>^ stable helix needs more heat to denature, so high Tm</p></li><li><p>(denatures 50% at higher temp, so more stable)</p></li></ul></li><li><p><strong><u>unstable </u></strong>DNA helix = <strong><u>low Tm</u></strong></p><ul><li><p>^ unstable helix needs less heat to denature, so low Tm</p></li><li><p>(denature 50% at lower temp, so more unstable)</p></li></ul></li></ul><p>as temp decreases, DNA strands w/ complementary sequences will <strong>anneal/zip up</strong></p>
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t/f: DNA sequence also affects DNA stability & Tm of DNA (in addition to temp and pH)

  • explain

true

  • A/T and C/G have/form different #s of H bonds (2 vs. 3) and base-stacking interactions

  • where C/G pair have a higher Tm

    • meaning C/G is more stable at higher temp than A/T which is more unstable at higher temp / more stable at low temp)

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term image

low pH is unstable

unstable helix is low Tm (needs less heat to denature)

buffer 1 has the lower Tm

<p>low pH is unstable</p><p>unstable helix is low Tm (needs less heat to denature)</p><p>buffer 1 has the lower Tm</p>
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3 conformations/forms of DNA in tertiary structure

which one allows nucleic acid bases & proteins to interact

__

function of quaternary structure

A-, B-, Z-DNA (have diff. configurations)

_

B-DNA conformation allows proteins to bind to specific DNA sequences

  • this is essential for gene expression

____

quaternary structure of DNA helps in packing genomic DNA in cells

<p>A-, B-, Z-DNA (have diff. configurations)</p><p>_</p><p>B-DNA conformation allows proteins to bind to specific DNA sequences </p><ul><li><p>this is <u>essential for gene expression</u></p></li></ul><p>____</p><p>quaternary structure of DNA helps in <u>packing genomic DNA in cells</u></p>
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DNA is stored as __ in EUkaryotic cells

  • describe

what about how DNA is packaged in PROkaryotic cells?

DNA is stored as chromatin in eu- cells

  • DNA helix

  • DNA helix + histones (are proteins) = nucleosomes

  • nucleosomes are compacted into chromatin fiber

__

pro- genomic DNA is supercoiled

<p>DNA is stored as <u>chromatin </u>in eu- cells </p><ul><li><p>DNA helix</p></li><li><p>DNA helix + histones (are proteins) = nucleosomes</p></li><li><p>nucleosomes are compacted into chromatin fiber</p></li></ul><p>__</p><p>pro- genomic DNA is <u>supercoiled</u></p>
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review:

state central dogma w/ processes (which process allows genetic info to be transmitted from parent cell to daughter cells via cell division)

genes are encoded in __

DNA → RNA → proteins

replication (of DNA), transcription (DNA ←→ RNA), translation (into protein)

  • DNA replication allows genetic info to be transmitted from parent cell to daughter cells

__

genes are encoded in DNA

<p>DNA → RNA → proteins</p><p>replication (of DNA), transcription (DNA ←→ RNA), translation (into protein)</p><ul><li><p>DNA replication allows genetic info to be transmitted from parent cell to daughter cells</p></li></ul><p>__</p><p>genes are encoded in <u>DNA</u></p>
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discovery from Meselson-Stahl experiment (1)

what serves as template to make daughter strand?

DNA replication is semi-conservative

  • after replication, EACH helix has 1 strand of template parent DNA & 1 strand of newly synthesized daughter DNA

parent strand is the template for daughter strand

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state key steps & enzymes involved in DNA replication (8 steps)

  1. helicase denatures/unwinds DNA

  2. topoisomerase relieves helix strain that occurs when unwinding DNA

  3. RNA primase (aka primase) creates a short RNA primer to start DNA synthesis

  4. DNA polymerase (DNA pol III) adds nucleotides via phosphoryl-transfer rxn

  5. DNA pol “proofreads” via exonuclease activity

  6. DNA pol (DNA pol I) removes RNA primer & replaces it w/ DNA

  7. DNA ligase (aka ligase) joins together the DNA fragments by forming a phosphodiester bond

  8. telomerase acts at the end of chromosomes to maintain their/the chromosome’s length

<ol><li><p><strong><u>helicase </u></strong>denatures/unwinds DNA</p></li><li><p><strong><u>topoisomerase </u></strong>relieves helix strain that occurs when unwinding DNA</p></li><li><p><strong><u>RNA primase</u></strong><u> (aka primase) </u>creates a short RNA primer to start DNA synthesis</p></li><li><p><strong><u>DNA polymerase</u></strong><u> <mark data-color="red" style="background-color: red; color: inherit;">(DNA pol III)</mark> </u>adds nucleotides via phosphoryl-transfer rxn</p></li><li><p><u>DNA pol</u> “proofreads” via <u><mark data-color="blue" style="background-color: blue; color: inherit;">exonuclease </mark></u>activity</p></li><li><p><u>DNA pol <mark data-color="red" style="background-color: red; color: inherit;">(DNA pol I)</mark></u> removes RNA primer &amp; replaces it w/ DNA</p></li><li><p><strong><u>DNA ligase</u></strong><u> (aka ligase)</u> joins together the DNA fragments by forming a phosphodiester bond</p></li><li><p><strong><u>telomerase </u></strong>acts at the end of chromosomes to maintain their/the chromosome’s length</p></li></ol><p></p>
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each human cell has ~__ feet of DNA, which takes ~__ hours for the human cell to replicate its genome b/c during mitosis, the cell copies the entire length of DNA before it divides

where does genome replication start (1)

  • describe (1)

  • in pro- vs. eukaryotes

~6 ft of DNA, ~8 hours

__

DNA replication starts at an origin of replication (“ori”)

  • at an ori, 2 strands of DNA helix are denatured

pro- have single ori

eu- have multiple ori

<p>~6 ft of DNA, ~8 hours</p><p>__</p><p>DNA replication starts at an origin of replication (“ori”)</p><ul><li><p>at an ori, 2 strands of DNA helix are denatured</p></li></ul><p>pro- have single ori</p><p>eu- have multiple ori</p><p></p>
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why do eu- have multiple ori?

multiple ori allows for FAST genome replication

  • (would take ~1 year for human cell to divide if human genome only had 1 origin of replication)


(OVERALL, not only regarding multiple ori, you want DNA replication to have a balance b/w being both fast & accurate)

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function of helicase (~2), which causes __ that looks like __

  • direction of movement

  • forms (1 w/ 1)

helicases bind the ori & denature/unwind DNA by breaking the H-bonds, which causes tension that looks like supercoils

(locally denature DNA at the ori)

  • move in opposite directions to unwind the helix & separate the strands

  • forms the “replication bubble”, where there are 2 “replication forks” that move in opposite directions away from the ori

<p>helicases bind the ori &amp; denature/unwind DNA by breaking the H-bonds, which causes <u>tension</u> that looks like <u>supercoils</u></p><p><span style="color: rgb(157, 153, 153);">(locally denature DNA at the ori)</span></p><ul><li><p>move in opposite directions to unwind the helix &amp; separate the strands</p></li><li><p>forms the “<u>replication bubble</u>”, where there are <u>2 “replication forks</u>” that move in opposite directions away from the ori</p></li></ul><p></p>
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term image
knowt flashcard image
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function of topoisomerase (1)

  • function of SSB (is NOT an enzyme, but a protein)

as helicase unwinds DNA, topoisomerase relieves torsional stress that was caused by strand separation (prevent supercoils from forming)

__

SSB (single-strand binding protein ← not an enzyme)

  • stabilizes single-stranded DNA (works together w/ topoisomerase in DNA replication)

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<p>function of primase (← which is a  __ __)</p><p>def. (RNA) primers</p><ul><li><p>why is it needed?</p></li></ul><p></p>

function of primase (← which is a __ __)

def. (RNA) primers

  • why is it needed?

RNA primase (an RNA polymerase)

  • makes RNA primers based on DNA template

primers - short piece of DNA or RNA (DNA/RNA primer) that DNA polymerase needs to make DNA

  • BUT we use RNA primers in genome replication

__

is needed b/c DNA polymerase (next step) CANNOT bind DNA and initiate DNA synthesis w/o a primer

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replication forks move in what direction?

move in same direction as helicase moves aka same direction as DNA separating

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state function of DNA polymerase, where it adds nucleotides (1 ← long)

daughter strands are made __

__

DNA pol. can only add nucleotides to the (1), so if it is the very beginning of genome replication w/o some synthesized part, what is needed?

  • explain

  • is the primer part of the immediate product (not talking abt final product)?

adds the complementary nucleotide to the template base

by forming phosphodiester bond b/w 3’ OH & 5’-phosphate / alpha-phosphate of incoming dNTP (polymerization, via phosphoryl transfer rxn)

^^^ (daughter strands are made simultaneously)

^^^ dNTP as in dATP, dGTP, dCTP, dTTP

__

DNA pol. can ONLY add nucleotides to the 3’ OH at the 3’ end of a primer (at the primer’s 3’ end)

  • if starting new DNA strand (5’ → 3’), then need primase to form a primer, so that primase can add the first nucleotide to initiate a strand

primer is part of immediate product, located at the 5’ end of the synthesized DNA strand

<p><strong><u>adds the complementary nucleotide</u></strong> to the template base</p><p>by <u>forming phosphodiester bond</u> b/w <mark data-color="blue" style="background-color: blue; color: inherit;">3’ OH </mark>&amp;<mark data-color="blue" style="background-color: blue; color: inherit;"> 5’-phosphate / alpha-phosphate of incoming dNTP</mark> (<strong><u>polymerization</u></strong>, via <u>phosphoryl transfer rxn</u>)</p><p>^^^ (daughter strands are made <strong><u>simultaneously</u></strong>)</p><p>^^^ dNTP as in dATP, dGTP, dCTP, dTTP</p><p>__</p><p>DNA pol. can ONLY add nucleotides to the <u>3’ OH at the 3’ end of a primer (at the primer’s 3’ end)</u></p><ul><li><p>if starting new DNA strand (5’ → 3’), then need primase to form a primer, so that primase can add the first nucleotide to initiate a strand</p></li></ul><p>primer is part of immediate product, located at the 5’ end of the synthesized DNA strand</p><p></p>
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t/f: DNA polymerase can’t initiate synthesis, but (RNA) primase can

true

  • DNA polymerase can’t initiate synthesis, so it needs RNA primer

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def. nucleoside analogs (2)

  • function ← how?

give example & describe for (viruses):

  • HIV infections

  • Covid-19

describe for cancer

compounds that are structurally similar to nucleosides & that are used to combat DNA & RNA viruses and cancer

  • block replication of viral genomes

    • nucleoside analog is incorporated into the strand during DNA (or RNA) synthesis, but the analog stops further synthesis which prevents viral replication

__

ex:

  • AZT treats/combats HIV infections, where AZT is an analog of the nucleoside thymidine

  • b/c thymidine has 3’ OH for adding nucleotides during DNA synthesis, but AZT does NOT have 3’ OH so no nucleotides added → DNA synthesis stops

    • AZT inhibits viral replication

ex:

  • Remdesivir combats Covid-19, where Remdesivir is an adenosine analog

__

slow mitosis & proliferation of cancerous cells

<p>compounds that are structurally similar to nucleosides &amp; that are used to combat DNA &amp; RNA viruses and cancer</p><ul><li><p>block replication of viral genomes</p><ul><li><p>nucleoside analog is incorporated into the strand during DNA (or RNA) synthesis, but the analog stops further synthesis which prevents viral replication</p></li></ul></li></ul><p>__</p><p>ex:</p><ul><li><p><u>AZT </u>treats/combats HIV infections, where AZT is an <u>analog of the nucleoside thymidine</u></p></li><li><p>b/c thymidine has 3’ OH for adding nucleotides during DNA synthesis, but AZT does NOT have 3’ OH so no nucleotides added → DNA synthesis stops</p><ul><li><p>AZT inhibits viral replication</p></li></ul></li></ul><p>ex:</p><ul><li><p><u>Remdesivir </u>combats Covid-19, where Remdesivir is an <u>adenosine analog</u></p></li></ul><p>__</p><p>slow mitosis &amp; proliferation of cancerous cells</p>
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how are RNA primers removed & replaced? (& by what?)

DNA polymerase 1 (DNA pol I) removes RNA primers & fills in the gap w/ DNA

  • (aka simultaneously removes the primer from the 5’ end of newly synthesized strand & replaces it w/ DNA)


(DNA pol binds at the 3’ end of the RNA primer & moves in the direction of 5’ → 3’)

<p>DNA polymerase 1 (DNA pol I) removes RNA primers &amp; fills in the gap w/ DNA</p><ul><li><p>(aka simultaneously removes the primer from the 5’ end of newly synthesized strand &amp; replaces it w/ DNA)</p></li></ul><div data-type="horizontalRule"><hr></div><p>(<strong><u>DNA pol binds</u></strong> at the <strong><u>3’ end of the RNA primer</u></strong> &amp; moves in the direction of 5’ → 3’)</p>
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function of DNA ligase

seals the gap by forming a phosphodiester bond

  • (the gap is b/w where DNA pol filled in the gap & part of synthesized strand)

  • (aka when the primer is replaced by DNA, a new piece of DNA is made BUT that piece is not attached to the neighboring DNA, which creates a gap b/w the 3’ & 5’ ends of neighboring pieces)

<p><strong>seals the gap by forming a <u>phosphodiester bond</u></strong></p><ul><li><p>(the gap is b/w where DNA pol filled in the gap &amp; part of synthesized strand)</p></li><li><p>(aka when the primer is replaced by DNA, a new piece of DNA is made BUT that piece is not attached to the neighboring DNA, which creates a <strong><u>gap b/w the 3’ &amp; 5’ ends of neighboring pieces)</u></strong></p></li></ul><p></p>
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name these 2 synthesized daughter strands that: (& def. ← 2 each)

during genome replication, the 2 synthesized daughter strands (that are anti-parallel) in the 5’ → 3’ direction (b/c only direction DNA pol. moves in)

^ these 2 daughter stranded are synthesized __

leading strand (aka leading daughter strand)

  • is made in 1 piece, using only 1 (RNA) primer

  • made in the same direction that the replication fork moves

lagging strand (aka lagging daughter strand)

  • is made in pieces (Okazaki fragments), using multiple (RNA) primers

  • made in the opposite direction of the replication fork

__

simultaneously

<p>leading strand (aka leading daughter strand)</p><ul><li><p>is made in 1 piece, using only 1 (RNA) primer</p></li><li><p>made in the same direction that the replication fork moves</p></li></ul><p>lagging strand (aka lagging daughter strand)</p><ul><li><p>is made in pieces (<strong>Okazaki fragments</strong>), using multiple (RNA) primers</p></li><li><p>made in the opposite direction of the replication fork</p></li></ul><p>__</p><p>simultaneously</p>
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<p>also, which part would be the chromosome end (aka __)</p>

also, which part would be the chromosome end (aka __)

telomere (aka chromosome end) is the end where the 2 parent strands haven’t unwound/separated/denatured yet (aka where parent strands are still together/intact)


for the problem in pic:

you can identify the 5’ and 3’ end from the direction of the synthesized DNA daughter strands that move in 5’ → 3’ direction, SO the ends of parent strands would the opposite of the synthesized strands (b/c anti-parallel)

you can identify the leading vs. lagging strand by splitting the diagram in half vertically

  • the blue move in same direction as fork

  • the red move in opp. direction from fork

<p><strong><u>telomere </u></strong><u>(aka chromosome end) </u>is the end where the 2 parent strands haven’t unwound/separated/denatured yet (aka where parent strands are still together/intact)</p><div data-type="horizontalRule"><hr></div><p>for the problem in pic:</p><p>you can identify the 5’ and 3’ end from the direction of the synthesized DNA daughter strands that move in 5’ → 3’ direction, SO the ends of parent strands would the opposite of the synthesized strands (b/c anti-parallel)</p><p>you can identify the leading vs. lagging strand by splitting the diagram in half vertically</p><ul><li><p>the blue move in same direction as fork</p></li><li><p>the red move in opp. direction from fork</p></li></ul><p></p>
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at the ends of linear chromosomes (aka at telomeres), what happens if nothing after ligase happens & you continue DNA replication?

  • there is a gap (b/w 3’ and 5’ of neighboring ends), but (the leading strand has no gap)

  • for lagging strand synthesis:

    • primase adds an RNA primer at 5’ end of chromosome

    • when primer is removed, there’s a gap at the end of the new chromosome

  • ligase joins the 2 Okazaki fragments w/ phosphodiester bonds

there is a gap at the end of the telomere WHERE chromosomes get shorter each time IF you don’t fill in the gap and just continue genome/DNA replication

<ul><li><p>there is a gap (b/w 3’ and 5’ of neighboring ends), but (the leading strand has no gap)</p></li></ul><ul><li><p>for lagging strand synthesis:</p><ul><li><p><strong><u>primase adds an RNA primer</u></strong> <strong><u>at 5’ end</u></strong> of chromosome</p></li><li><p>when primer is removed, there’s a gap at the end of the new chromosome</p></li></ul></li><li><p>ligase joins the 2 Okazaki fragments w/ phosphodiester bonds</p></li></ul><p><u>there is a gap at the end of the telomere WHERE chromosomes get shorter each time IF you don’t fill in the gap and just continue genome/DNA replication</u></p><p></p>
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function of telomere (1), which is a __ polymerase

telomerase is a __ __ (← term relates to transcription)

_

after telomerase performs its role, what happens to the lagging daughter strand?

telomerase (a DNA polymerase)

  • extends the 3’ end of the parent strand, using a RNA template, to maintain chromosome length

telomerase is a reverse transcriptase

  • (b/c makes DNA based on an RNA template, where central dogma is DNA → RNA → protein)

_

then, the lagging daughter strand extends via primase + DNA pol + ligase using this newly synthesized parent strand as the template

<p>telomerase (a DNA polymerase)</p><ul><li><p>extends the<span style="color: blue;"><strong> <u><mark data-color="blue" style="background-color: blue; color: inherit;">3’ end</mark></u></strong></span> of the <strong><u>parent </u></strong>strand, using a RNA template, to maintain chromosome length</p></li></ul><p>telomerase is a <strong><u>reverse transcriptase</u></strong></p><ul><li><p>(b/c <u>makes DNA based on an RNA template</u>, where central dogma is DNA → RNA → protein)</p></li></ul><p>_</p><p>then, the <u>lagging daughter strand extends</u> via <u>primase + DNA pol + ligase</u> using this <strong><u>newly synthesized parent strand as the template</u></strong></p><p></p>
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multiple ori allows for fast genome replication, but this can lead to (1)

you want …

fast replication can lead to mistakes in DNA replication, like mismatches

instead, you want a balance b/w replicating a genome quickly & accurately (i.e. making only 2-3 mistakes)

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def. mutation

  • can be caused by (1) or (1)

  • can be either (3)

ANY change in DNA sequence of a cell

  • caused by mistakes during cell division (replication) OR exposure to DNA-damaging agents in the environment

  • can be harmful, beneficial, or have no effect (neutral)

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just state the 2 mechanisms of repairing mismatches during DNA replication (1 of them also has a specific term)

which is immediate vs. not

  • exonuclease activity by DNA polymerase (“proofreading”)

    • immediate / during replication

  • mismatch repair

    • not immediate / after replication is done

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t/f: different forms of nucleotide bases can result in mismatches (“errors”) during DNA synthesis

  • explain generally

  • then, explain w/ ex. of cytosine

how do these spontaneous ““ shifts in the base occur? they can cause more __ nucleotides

true

  • nucleotide bases can exist in 2 forms called tautomers (aka shifting of protons), where 1 form is more stable than the other: predominant, major vs. minor form

  • the minor and major forms have different patterns of H bond donors and acceptors

__

  • cytosine has a major & minor form

  • the minor form of C can pair w/ A (adenine), but the major form cannot & only binds to G (guanine)

    • this C/A can lead to a change in the DNA sequence

_

these spontaneous proton shifts in the base can cause more mismatched nucleotides

  • occur by:

    • minor form is paired to the template (w/ the complementary nucleotide base)

    • before DNA polymerase continues to add nucleotides after this, the minor base changes back to the major form B/C the minor form is unstable

    • this would be a spontaneous mismatch

<p>true</p><ul><li><p>nucleotide bases can exist in 2 forms called <strong><u>tautomers</u></strong><span style="color: blue;"> (aka shifting of protons)</span>, where 1 form is more stable than the other: predominant, major vs. minor form</p></li><li><p>the minor and major forms have different patterns of H bond donors and acceptors</p></li></ul><p>__</p><ul><li><p>cytosine has a major &amp; minor form</p></li><li><p>the minor form of C can pair w/ A (adenine), but the major form cannot &amp; only binds to G (guanine)</p><ul><li><p>this C/A can lead to a change in the DNA sequence</p></li></ul></li></ul><p>_</p><p>these spontaneous <u>proton</u> shifts in the base can cause more <u>mismatched</u> nucleotides</p><ul><li><p>occur by:</p><ul><li><p>minor form is paired to the template (w/ the complementary nucleotide base)</p></li><li><p>before DNA polymerase continues to add nucleotides after this, the minor base <u>changes back to the major form</u> B/C the <u>minor form is unstable</u></p></li><li><p>this would be a <u>spontaneous mismatch</u></p></li></ul></li></ul><p></p>
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what are the 2 enzyme activities of DNA pol? what bond is affected?

  • specific term for 1 of them

(^^ this question is not completely referring to how to repair mismatch)

  • polymerization, where DNA pol adds nucleotides via polymerase domain (Pol)

  • exonuclease activity, where DNA pol removes mismatched nucleotides via exonuclease domain (Exo)

    • “proofreading”

^ both affect phosphodiester bonds (make vs. break, respectively)


(can refer to past flashcard for pic of polymerization, via phosphoryl-transfer rxn)

<ul><li><p><u>polymerization</u>, where DNA pol <u>adds nucleotides</u> via <strong><u>polymerase domain (Pol)</u></strong></p></li><li><p><u>exonuclease activity</u>, where DNA pol <u>removes mismatched nucleotides</u> via <strong><u>exonuclease domain (Exo)</u></strong></p><ul><li><p><strong><u>“proofreading”</u></strong></p></li></ul></li></ul><p>^ both affect phosphodiester bonds (make vs. break, respectively)</p><div data-type="horizontalRule"><hr></div><p>(can refer to past flashcard for pic of polymerization, via phosphoryl-transfer rxn)</p>
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DNA pol adds/removes nucleotides to/from the __ end

3’ end

(i.e. DNA pol adds nucleotide at the 3’ OH of prior neighboring sequence, in the direction of 5’ to 3’)

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DNA polymerase is what type of exonuclease & what type of polymerase?

  • function of both

DNA polymerase I is what type of exonuclease & what type of polymerase?

  • function of both

DNA pol

  • 5’-to-3’ polymerase (adds DNA)

    • start at 3’ end

  • 3’-to-5’ exonuclease / 3’ exonuclease (“proofreading” — remove incorrectly added bases, like mismatch and insertion/deletion)

    • starts acting at the 3’ end

DNA pol I

  • 5’-to-3’ polymerase (adds DNA)

    • start at 3’ end

  • 5’-to-3’ exonuclease / 5’ exonuclease (remove RNA primer)

    • starts acting at the 5’ end


(the green and purple ovals are the Exo / exonuclease domain)

<p>DNA pol</p><ul><li><p><mark data-color="purple" style="background-color: purple; color: inherit;">5’-to-3’</mark> <u>poly</u>merase (adds DNA)</p><ul><li><p>start at 3’ end</p></li></ul></li><li><p><span style="color: blue;"><strong><u><mark data-color="blue" style="background-color: blue; color: inherit;">3’-to-5’</mark> exonuclease / <mark data-color="blue" style="background-color: blue; color: inherit;">3’</mark> exonuclease (“proofreading”</u></strong></span><strong><u> — remove incorrectly added bases, like mismatch</u></strong> <span style="color: rgb(194, 188, 188);">and insertion/deletion</span>)</p><ul><li><p><u>starts acting at the 3’ end</u></p></li></ul></li></ul><p>DNA pol I</p><ul><li><p><mark data-color="purple" style="background-color: purple; color: inherit;">5’-to-3’</mark> <u>poly</u>merase (adds DNA)</p><ul><li><p>start at 3’ end</p></li></ul></li><li><p><span style="color: purple;"><strong><u><mark data-color="purple" style="background-color: purple; color: inherit;">5’-to-3’</mark> exonuclease / <mark data-color="purple" style="background-color: purple; color: inherit;">5’</mark> exonuclease (remove RNA primer)</u></strong></span></p><ul><li><p><u>starts acting at the 5’ end</u></p></li></ul></li></ul><div data-type="horizontalRule"><hr></div><p>(the green and purple ovals are the Exo / exonuclease domain)</p>
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<p>how would DNA pol recognize &amp; remove mismatched nucleotides, based on the picture</p>

how would DNA pol recognize & remove mismatched nucleotides, based on the picture

DNA pol recognize mismatched base b/c 3’ OH of cytosine is not positioned properly

DNA pol then uses its 3’-to-5’ exonuclease activity to remove the mismatched base

<p>DNA pol recognize mismatched base b/c <u>3’ OH of cytosine is not positioned properly</u></p><p>DNA pol then uses its <strong><u>3’-to-5’ exonuclease activity</u></strong> to remove the mismatched base</p>
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exonuclease activity by DNA pol starts when there’s a spontaneous tautomeric shift that stops continuation of DNA synthesis,

so why doesn’t mismatch repair stop DNA synthesis when there’s a mismatch?

DNA synthesis continues b/c the mismatched nucleotide was not recognized

  • (mismatch repair occurs after DNA synthesis is done b/c there’s no proofreading)

while exonucl. activity by DNA pol is proofreading that occurs during DNA synthesis


exonuclease activity/proofreading occurs during DNA synthesis

mismatch repair (MMR) occurs after end of DNA synthesis

BER and NER occurs after the end of DNA replication

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name & def. 2 mismatch repair enzymes

  • 1 of them can act __ or __

_

review(ish):

another name for each (& function of each):

  • 3’-to-5’ exonuclease

  • 5’-to-3’ exonuclease

endonuclease — cleaves phosphodiester bonds w/in a strand

  • can act EITHER upstream (at 5’ end) OR downstream (at 3’ end) of the mismatch (cleaving ONE phosphodiester bond in mismatch repair)

exonuclease — cleaves phosphodiester bonds at the end of a DNA strand

  • EITHER upstream OR downstream

(^^ both cleave phosphodiester bonds, just at different positions in a DNA strand)

__

3’-to-5’ exonuclease (aka 3’ exonuclease) (aka 3’-5’ ““)

  • proofreading / remove mismatch

5’-to-3’ exonuclease (aka 5’ exonuclease) (aka 5’-3’ ““)

  • remove RNA primer

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<p>which DNA strand do endonucleases cleave?</p><ul><li><p>how do they know?</p><ul><li><p>why is it like this?</p></li></ul></li></ul><p></p>

which DNA strand do endonucleases cleave?

  • how do they know?

    • why is it like this?

mismatch repair machinery cleaves the non-methylated DNA

  • the mismatch repair machinery can identify the parent strand b/c it is methylated (CH3), while the daughter strand is not methylated

    • over time, genomic DNA accumulates methylation (parent strand is old, while daughter strand is new)

__

picture:

No b/c methylation at this position doesn’t disrupt the hydrogen bonds that form b/w adenine & thymine (A-T base pair is intact)

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how do they both work together in mismatch repair:

  • endonuclease, exonuclease, mismatch repair machinery, helicase, ligase, DNA pol (← is stated out of order)

^^ aka steps of mismatch repair (5)

_

& include which specific exonuclease is used depending on relative positions of two specific things (← 2)

  • repair machinery/proteins recognize the mismatch & identify the unmethylated daughter strand

  • endonuc. makes a nick”/cut ONE phosphodiester bond near the mismatch

    • (nicks forms “near” aka EITHER upstream OR downstream of the mismatch, but not “at” the mismatch)

  • helicase unwinds the DNA for exonuclease to access the DNA

  • exonuc. removes nucleotides past the mismatch (aka including and after the mismatch), EITHER - OR -:

    • 5’ (5’ to 3’) exonuclease IF “nick” is 5’ / upstream from the mismatch

      • so moving from nick to mismatch is downstream (5’→3’)

    • 3’ (3’ to 5’) exonuclease IF “nick” is 3’ / downstream from the mismatch

      • so moving from nick to mismatch is upstream (3’→5’)

  • DNA pol synthesizes new NDA

  • ligase joins the 3’ OH & 5’ phosphate of neighboring nucleotide w/in the same strand (fills in the gap) in newly synthesized daughter strand

^^ endonuclease can cleave either upstream (5’ → 3’) or downstream (3’ → 5’)

<ul><li><p><strong><u>repair machinery/proteins</u></strong> <u>recognize </u>the mismatch &amp; <u>identify the unmethylated daughter strand</u></p></li><li><p><strong><u>endonuc</u></strong>. makes a <u>“<mark data-color="blue" style="background-color: blue; color: inherit;">nick</mark>”/cut </u><strong><u>ONE </u></strong><u>phosphodiester bond near the mismatch</u></p><ul><li><p>(nicks forms “<span style="color: blue;">near</span>” aka <strong><u>EITHER upstream OR downstream</u></strong> of the mismatch, but<span style="color: blue;"> not “at” the mismatch)</span></p></li></ul></li><li><p><strong><u>helicase</u> </strong><u>unwinds</u> the DNA <u>for exonuclease to access the DNA</u></p></li><li><p><strong><u>exonuc</u></strong>. <u>removes nucleotides</u> <u>past the mismatch</u> (aka including and after the mismatch), <strong><u>EITHER</u></strong> - OR -:</p><ul><li><p><strong><u>5’ (5’ to 3’) exonuclease</u></strong> IF “nick” is 5’ / upstream from the mismatch</p><ul><li><p>so moving from nick to mismatch is downstream (5’→3’)</p></li></ul></li><li><p><strong><u>3’ (3’ to 5’) exonuclease</u></strong> IF “nick” is 3’ / downstream from the mismatch</p><ul><li><p>so moving from nick to mismatch is upstream (3’→5’)</p></li></ul></li></ul></li><li><p><strong><u>DNA pol</u></strong> synthesizes new NDA</p></li><li><p><strong><u>ligase</u> </strong>joins the 3’ OH &amp; <u>5’ phosphate of neighboring nucleotide w/in the same strand</u> (fills in the gap) in newly synthesized daughter strand</p></li></ul><p>^^ endonuclease can cleave either upstream (5’ → 3’) or downstream (3’ → 5’)</p>
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name 2 of various ways that double-stranded breaks (DSBs) occur

_

name 2 ways that double-stranded DNA breaks (DSBs) can be repaired

  • which is error-prone vs. precise repair? why?

  • exposure to environmental mutagens

  • deliberate breaks caused during gene editing (aka gene editing machinery) (will go over in later unit)

__

double-stranded DNA breaks can be repaired via:

  • end-joining (EJ)

    • error-prone (b/c mutations are introduced a lot at the break)

  • homology-directed repair (HDR), such as homologous recombination

    • precise repair (b/c homologous sequences act as the template for precise repair of the break)

<ul><li><p>exposure to <u>environmental mutagens</u></p></li><li><p>deliberate breaks caused during gene editing (aka gene editing machinery) <span style="color: rgb(198, 193, 193);">(will go over in later unit)</span></p></li></ul><p>__</p><p>double-stranded DNA breaks can be repaired via:</p><ul><li><p><strong><u>end-joining</u></strong><u> (EJ)</u></p><ul><li><p><u>error-prone</u> (b/c <u>mutations are introduced a lot at the break</u>)</p></li></ul></li><li><p><u>homology-directed repair (HDR)</u>, such as <strong><u>homologous recombination</u></strong> </p><ul><li><p><u>precise repair</u> (b/c <u>homologous sequences act as the template</u> for precise repair of the break)</p></li></ul></li></ul><p></p>
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briefly state what happens in end-joining vs. homologous recombination (1 each)

& again, explain why it’s error-prone or precise repair

end-joining

  • broken ends are joined together

^^ during EJ, nucleotides might be added or removed around the break → changing the DNA sequence

__

homologous recombination

  • new DNA is made to repair the break

^^ a similar/homologous sequence acts as the template sequence for repairing the break

<p><u>end-joining </u></p><ul><li><p><strong>broken ends are joined together</strong></p></li></ul><p>^^ during EJ, nucleotides might be added or removed around the break → changing the DNA sequence</p><p>__</p><p><u>homologous recombination</u></p><ul><li><p><strong>new DNA is made to repair the break</strong></p></li></ul><p>^^ a similar/homologous sequence acts as the template sequence for repairing the break</p>
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explain the process of end-joining (for double-stranded break repair) (3)

  • a protein complex (Ku + DNA-PK) binds to the broken ends & bring the ends together

  • broken ends are modified (in varied ways) that change the DNA sequence

    • (aka usually have mutations at the site of the break)

  • DNA ligase makes sure that the broken DNA is re-sealed

__

^^ the modifications are variable (i.e. exonuclease often removes a variable # of nucleotides from the ends) → SO variations in end-repair

<ul><li><p>a <u>protein complex <mark data-color="blue" style="background-color: blue; color: inherit;">(Ku + DNA-PK)</mark> binds to</u> the broken ends &amp; <u>bring the ends together</u></p></li><li><p>broken ends are <u>modified (in varied ways)</u> that <u>change the DNA sequence</u></p><ul><li><p>(aka <span style="color: blue;">usually have mutations at the site of the break</span>)</p></li></ul></li><li><p>DNA <strong><u>ligase </u></strong>makes sure that the broken DNA is <u>re-sealed</u></p></li></ul><p>__</p><p>^^ the modifications are variable (i.e. exonuclease often removes a variable # of nucleotides from the ends) → SO variations in end-repair</p>
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briefly describe the 2 DNA helices involved in homologous recombination

__

where can a cell find a homologous sequence to repair the DSB? (2 w/ explanation)

  • a DNA helix w/ 2 broken strands

  • a DNA helix w/ a homologous sequence to the double-stranded broken helix

__

find homo- sequence via:

  • sister chromatid — human cells are diploid, meaning they have 2 copies of each chromosome (1 from each parent) SO chromosome pairs are homologous chromosomes

  • exogenous DNA — DNA w/ homologous sequence can be injected into cells during gene editing

<ul><li><p>a DNA helix <u>w/ 2 broken strands</u></p></li><li><p>a DNA helix <u>w/ a homologous sequence</u> to the double-stranded broken helix</p></li></ul><p>__</p><p>find homo- sequence via:</p><ul><li><p><strong><u>sister chromatid</u></strong> — human cells are <u>diploid</u>, meaning they have 2 copies of each chromosome (1 from each parent) SO <u>chromosome pairs are homologous chromosomes</u></p></li><li><p><strong><u>exogenous DNA</u></strong> — DNA w/ homologous sequence can be <u>injected into cells</u> during gene editing</p></li></ul><p></p>
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<p>answer the Q</p><p>&amp; what does it result in/create? (1)</p><p>draw the 3’/5’ ends &amp; the 5’ exonuclease with arrows of direction its moving in</p>

answer the Q

& what does it result in/create? (1)

draw the 3’/5’ ends & the 5’ exonuclease with arrows of direction its moving in

5’ exonuclease — removes nucleotides from the 5’ end of one strand

  • SO that the 3’ end of the opposite strand is single-stranded → 3’ overhang


(red shows 3’ overhangs)

<p><u>5’ exonuclease</u> — removes nucleotides from the 5’ end of one strand</p><ul><li><p>SO that the 3’ end of the opposite strand is single-stranded → <u>3’ overhang</u></p></li></ul><div data-type="horizontalRule"><hr></div><p>(red shows 3’ overhangs)</p>
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explain the process/steps of homologous recombination (for double-stranded break repair) (6)

  • in the helix w/ DSB, 5’ exonuclease cleaves the nucleotides at the 5’ end of one strand → make 3’ single-stranded ends on the opposite strand aka 3’ overhang

  • 3’ overhang invades the helix w/ the homologous sequence

  • DNA pol. extends the 3’ end of this invading strand UNTIL the displaced strand (of homo- sequence) base pairs w/ the OTHER 3’ overhang (to cover the break)

    • (3’ overhang acts a primer for DNA pol., except it is part of the DNA sequence)

  • DNA pol. extends this other 3’ end (to cover the break)

  • DNA ligase seals the two gaps → to form Holliday structures (“X”)

  • the 2 Holliday structures are resolved → get repaired DNA

    • (in 2 ways: cut → non-crossover & cut → crossover) ← don’t have to know

<ul><li><p>in the <mark data-color="blue" style="background-color: blue; color: inherit;">helix w/ DSB</mark>, <strong><u>5’ exonuclease</u></strong> cleaves the nucleotides at the 5’ end of one strand → make 3’ single-stranded ends on the opposite strand aka <span style="color: blue;"><strong><u>3’ overhang</u></strong></span></p></li></ul><ul><li><p>3’ overhang invades the <mark data-color="blue" style="background-color: blue; color: inherit;">helix w/ the homologous sequence</mark></p></li><li><p><strong><u>DNA pol.</u></strong> extends the 3’ end of this invading strand UNTIL the displaced strand (of homo- sequence) base pairs w/ the OTHER 3’ overhang (to cover the break)</p><ul><li><p>(<span style="color: rgb(221, 132, 229);">3’ overhang acts a primer for DNA pol., except it is part of the DNA sequence</span>)</p></li></ul></li><li><p><strong><u>DNA pol.</u></strong> extends this other 3’ end (to cover the break)</p></li><li><p><strong><u>DNA ligase</u></strong> seals the two gaps → to form <strong><u>Holliday structures</u></strong><u> (“X”)</u></p></li><li><p>the 2 <u>Holliday structures are resolved </u>→ get <u>repaired DNA</u></p><ul><li><p><span style="color: rgb(205, 200, 200);">(in 2 ways: </span><span style="color: rgb(227, 142, 218);">cut → non-crossover &amp; cut → crossover</span><span style="color: rgb(205, 200, 200);">) ← don’t have to know</span></p></li></ul></li></ul><p></p>
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why is it important to form the 3’ overhangs (1) & explain

the 3’ overhang brings in DNA polymerase

  • b/c the 3’ overhang has a free 3’ OH, that DNA pol can add nucleotides to (b/c DNA pol starts adding nucleotides at the 3’ end)

^ this is also why you don’t need RNA primase in homologous recombination

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review:

which specific exonuclease is used depending on relative positions of nick & mismatch (← 2)

__

review:

which removes single mismatch vs. large nucleotide sequence that can contain multiple mismatches: exonuclease activity via DNA pol OR mismatch repair

  • 5’ (5’ to 3’) exonuclease IF “nick” is 5’ / upstream from the mismatch

    • so moving from nick to mismatch is downstream (5’→3’)

  • 3’ (3’ to 5’) exonuclease IF “nick” is 3’ / downstream from the mismatch

    • so moving from nick to mismatch is upstream (3’→5’)

^^ just depends on relative position of nick and mismatch

can act EITHER upstream or downstream of the nick, but not in both directions at the same time

__

removes single mismatch — exonuclease activity via DNA poly

  • (b/c acts once it recognize a mismatch during replication)

large nucleotide sequence that can contain multiple mismatches — mismatch repair

  • (b/c acts after replication is over, so can possibly have multiple mismatches that are removed by the exonuclease)

<ul><li><p><strong><u>5’ (5’ to 3’) exonuclease</u></strong> IF “nick” is 5’ / upstream from the mismatch</p><ul><li><p>so moving from nick to mismatch is downstream (5’→3’)</p></li></ul></li><li><p><strong><u>3’ (3’ to 5’) exonuclease</u></strong> IF “nick” is 3’ / downstream from the mismatch</p><ul><li><p>so moving from nick to mismatch is upstream (3’→5’)</p></li></ul></li></ul><p>^^ just depends on relative position of nick and mismatch</p><p>can act  <strong><u>EITHER </u></strong>upstream or downstream of the nick, but not in both directions at the same time</p><p>__</p><p>removes single mismatch — exonuclease activity via DNA poly</p><ul><li><p>(b/c acts once it recognize a mismatch during replication)</p></li></ul><p>large nucleotide sequence that can contain multiple mismatches — mismatch repair</p><ul><li><p>(b/c acts after replication is over, so can possibly have multiple mismatches that are removed by the <u>exo</u>nuclease)</p></li></ul><p></p>
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what 2 enzymes catalyze the formation of phosphodiester bonds?

  • difference b/w the 2 things that the bond forms between

DNA ligase

  • b/w 3’ OH and 5’ phosphate w/in a strand (that already exist, which is why they’re w/in the strand) of 2 different nucleotides w/in a strand

DNA polymerase

  • b/w 3’ OH & 5’ phosphate of incoming dNTP of 2 different nucleotides w/in a strand

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t/f: primase (a type of RNA polymerase) can initiate DNA synthesis de novo

  • what does “de novo” mean?

true

  • “de novo” — from the beginning

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<p>just notice that:</p><ul><li><p>one side of ori is the synthesized leading strand, other side of ori is the synthesized lagging strand</p></li><li><p>DNA always synthesized from 5’ → 3’</p></li><li><p>leading or lagging strand is deter. by its synthesis relative to the direction the replication fork moves in</p></li></ul><p></p>

just notice that:

  • one side of ori is the synthesized leading strand, other side of ori is the synthesized lagging strand

  • DNA always synthesized from 5’ → 3’

  • leading or lagging strand is deter. by its synthesis relative to the direction the replication fork moves in

knowt flashcard image
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<p>just notice that:</p><p></p>

just notice that:

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name & def. 3 types of mutagenic chemical modifications that affect the DNA base

(^ aka changes to nucleotide bases can cause mutations if not repaired via DNA repair mechanisms)

alkylation — addition of methyl or ethyl that disrupts a H-bond from forming (can’t form H bond aka disrupts base pairing)

  • (methylation w/ CH3 or ethylation w/ CH2CH3)

deamination — remove amine group & replace w/ carbonyl group

  • (replace “—NH2” w/ “=O”)

depurination — remove the entire purine base

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is methylation of guanine O6 mutagenic? (may help to draw picture)

  • what happens to base pairing if methylated guanine O6?

when is methylation mutagenic?

_

diff. b/w methylation & alkylation (revew-ish)

yes b/c prevents it from being a H-bond acceptor

  • instead, O6-methylguanine pairs with thymine (G6-me-T, instead of G-C)

__

methylation is mutagenic if it occurs at a position that disrupts hydrogen bonds b/w standard base pairs

__

methylation — only adds a methyl group

alkylation — can add either a methyl or ethyl group

<p>yes b/c prevents it from being a H-bond acceptor</p><ul><li><p>instead, O6-methylguanine pairs with thymine (<strong><u>G<sup>6-me</sup>-T</u></strong>, instead of G-C)</p></li></ul><p>__</p><p>methylation is mutagenic if it occurs at a position that <u>disrupts hydrogen bonds b/w </u><strong><u>standard </u></strong><u>base pairs</u></p><p>__</p><p>methylation — <u>only </u>adds a methyl group</p><p>alkylation — can add <u>either </u>a methyl or ethyl group</p>
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<p><strong><u>try to draw out adenosine w/o looking at picture &amp; figure it out</u></strong></p><p>for <u>(deoxy)adenosine</u>, is methylation at the circled positions N1, N6, and N7 mutagenic or not? explain</p><p></p>

try to draw out adenosine w/o looking at picture & figure it out

for (deoxy)adenosine, is methylation at the circled positions N1, N6, and N7 mutagenic or not? explain

at N1: mutagenic

  • N1 is acceptor in a base pair w/ T (thymine)

at N6: not mutagenic

  • N6 is donor in a base pair w/ T, BUT N6 has 2 H’s (b/c NH2)

at N7: not mutagenic

  • even though N7 is a hydrogen bond acceptor, it is NOT involved in base pairing

  • aka N7 is not a H-bond acceptor in a standard base pair

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(review that deamination is removing amine group & replacing it w/ carbonyl group)

t/f: deamination of a base can change the base’s identity

__

deamination occurs __, but can also be catalyzed by __

__

deamination of C will convert C into __

  • which makes the standard base pair of (1) into (1) mismatch, which can lead to mutations ← usually the most common mutation from this specific mismatch is called (1)

what may explain why DNA lacks uracil (U)?

true

__

deamination occurs spontaneously, but can also be catalyzed by enzymes

_

deamination of C: regularC (cytosine) → U (uracil)

  • standard base pair of C:G into U:G mismatch

  • U:G mismatch can cause mutations, most commonly the C>T mutation (aka C → T mutation)

    • (C>T mutation goes from standard base pair of C:G → T:A)

__

DNA may lack uracil b/c of spontaneous deamination of cytosine

<p>true</p><p>__</p><p>deamination occurs <strong><u>spontaneously</u></strong>, but can also be catalyzed by <u>enzymes</u></p><p>_</p><p>deamination of C: <strong><sub>regular</sub></strong>C (cytosine) → U (uracil)</p><ul><li><p>standard base pair of <u>C:G</u> into <u>U:G mismatch</u></p></li><li><p>U:G mismatch can cause <u>mutations</u>, most commonly the <strong><u>C&gt;T mutation</u></strong> (aka C → T mutation)</p><ul><li><p>(C&gt;T mutation goes from standard base pair of <u>C:G → T:A</u>)</p></li></ul></li></ul><p>__</p><p>DNA may lack uracil b/c of <u>spontaneous deamination of cytosine</u></p>
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what base cannot be deaminated?

__

base-deamination changes these bases into what:

  • cytosine

  • 5-methylcytosine

  • adenine

  • guanine

& know the structure of 5-methylcytosine (draw it)

thymine can’t be deaminated b/c thymine has no amine group to replace w/ carbonyl group

__

C → U

C5-me → thymine

A → hypoxanthine

G → xanthine

<p>thymine can’t be deaminated b/c <u>thymine has no amine group</u> to replace w/ carbonyl group</p><p>__</p><p><strong><u>C → U</u></strong></p><p><strong><u>C<sup>5-me </sup>→ thymine</u></strong></p><p><strong><u>A → hypoxanthine</u></strong></p><p><strong><u>G → xanthine</u></strong></p><p></p>
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review:

def. of depurination

  • what bases are affected

  • def. apurinic

  • what bond is cleaved/removed?

remove entire purine base (adenine, guanine)

  • cleaves the glycosidic bond

_

apurinic — no purine base b/c it has been removed

  • (NOT apurinic if it has a pyrimidine base)

<p>remove entire purine base (adenine, guanine)</p><ul><li><p><strong>cleaves the <u>glycosidic bond</u></strong></p></li></ul><p>_</p><p><strong><u>apurinic </u></strong>— no purine base b/c it has been removed</p><ul><li><p>(NOT apurinic if it has a pyrimidine base)</p></li></ul><p></p>
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alkylation, deamination, depurination were all mutagenic CHEMICAL modifications, but

what is the mutagenic STRUCTURAL modification (1 ← name & def.)

  • 2 affected bases

  • def. kink

  • usually forms b/w …

pyrimidine dimers

  • UV light causes a kink to form B/C of covalent bond that is made b/w pyrimidines on the same strand

    • kink — structural change to DNA helix

  • affects pyrimidine bases: thymine, cytosine

    • NOT usually uracil

    • SO like covalent bond b/w T-T on same strand

  • usually forms b/w T-T

<p>pyrimidine dimers</p><ul><li><p><strong><u>UV </u></strong><u>light </u>causes a <strong><u>kink </u></strong>to form B/C of <strong><u>covalent </u></strong><u>bond that is made b/w pyrimidines on the </u><strong><u>same </u></strong><u>strand</u></p><ul><li><p><strong><u>kink </u></strong>— structural change to DNA helix</p></li></ul></li><li><p>affects pyrimidine bases: <u>thymine, cytosine</u></p><ul><li><p><span style="color: red;"><strong>NOT usually uracil</strong></span></p></li><li><p>SO like covalent bond b/w T-T on same strand</p></li></ul></li><li><p>usually forms b/w T-T</p></li></ul><p></p>
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just name 3 DNA repair mechanisms (that repair damaged base(s) that occur independently from replication)

base excision repair (BER)

nucleotide excision repair (NER)

direct repair

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def. direct repair

is not a pathway for DNA repair, but is instead, a class of different enzymes that change/modify a base

(i.e. methyltransferase does demethylation/remove methyl group to revert the mutagenic methylation of O6 → into normal/unmethylated guanine)

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how do you prevent a mistake from becoming a mutation?

repair machinery must act before the cell divides

aka

DNA repair must happen before replication happens, or else the mistake will become a PERMANENT mutation (after replication, mismatch becomes a mutation)

<p>repair machinery must act before the cell divides</p><p>aka</p><p>DNA repair must happen before replication happens, or else the mistake will become a <u>PERMANENT </u>mutation (after replication, mismatch becomes a mutation)</p>
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def. base excision repair (BER)

  • explain w/ example of: BER steps for deaminated 5-methylcytosine aka guanine (4)

repairs that change DNA at ONE or few nucleotides & do NOT cause a major distortion of DNA

(remove 1 or few damaged/mismatched bases before replication occurs)

__

BER w/ deaminated 5-methylcytosine aka guanine:

  • DNA glycosylase cleaves the glycosidic bond (b/w the sugar ring & base) —→ to make abasic nucleotide (aka nucleotide w/o base)

    • glycosylase is NOT specific in which strand’s mismatched base is changed (aka recognizes the mismatch/damage, but not the strand)

  • AP endonuclease cleaves the phosphodiester bonds to remove the abasic nucleotide w/in the strand → forms a gap

    • (AP endonuc. is either apurinic or apyrimidinic endonuclease)

  • DNA polymerase synthesizes new DNA to fill in gap

  • DNA ligase seals the remaining nick

<p>repairs that change DNA at <strong><u>ONE </u></strong><u>or few nucleotides</u> &amp; do <strong><u>NOT </u></strong><u>cause a major distortion of DNA</u></p><p><span style="color: rgb(175, 165, 165);">(remove 1 or few damaged/mismatched bases before replication occurs)</span></p><p>__</p><p>BER w/ deaminated 5-methylcytosine aka guanine:</p><ul><li><p><strong><u>DNA glycosylase</u></strong> cleaves the <u>glycosidic bond</u> (b/w the sugar ring &amp; base) —→ to make <u>abasic nucleotide</u> (aka nucleotide w/o base)</p><ul><li><p><span style="color: blue;"><strong><u>glycosylase is NOT specific in which strand’s mismatched base is changed</u></strong><u> </u></span><u>(aka recognizes the mismatch/damage, but not the strand)</u></p></li></ul></li><li><p><strong><u>AP endonuclease</u></strong> cleaves the <u>phosphodiester bonds</u> to remove the abasic nucleotide w/in the strand → forms a <u>gap</u></p><ul><li><p>(AP endonuc. is either apurinic or apyrimidinic endonuclease)</p></li></ul></li><li><p><strong><u>DNA polymerase</u></strong> synthesizes new DNA to fill in gap</p></li><li><p><strong><u>DNA ligase</u></strong> seals the remaining nick</p></li></ul><p></p>
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is DNA glycosylase need to repair depurinated bases? why? what would happen as the first step?

no

  • b/c depurinated bases don’t have a base, meaning that there is no glycosidic bond (which usually forms b/w sugar ring + base)

  • so glycosylase (which cleaves glycosidic bonds) is not needed for depurinated bases

SO BER’s the first step would be AP endonuclease (skip the glycosylase step)

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def. nucleotide excision repair (NER)

  • name example of NER

explain steps of NER using pyrimidine dimers

repairs that change DNA at multiple nucleotides & causes a major distortion of DNA

  • like pyrimidine dimers

__

NER using pyrimidine dimers:

  • recognize damage (aka pyrimidine dimers, like T-T)

  • helicase opens/unwinds the helix

  • endonucleases remove multiple nucleotides surrounding the damaged region (on the same strand)

    • specifically, excision endonucleases aka excinucleases, like XP-F and XP-G

  • DNA polymerase synthesizes new DNA to fill in the gap

  • DNA ligase seals the remaining nick

<p>repairs that change DNA at <strong><u>multiple </u></strong><u>nucleotides </u>&amp; causes a major distortion of DNA</p><ul><li><p>like <strong><u>pyrimidine dimers</u></strong></p></li></ul><p>__</p><p>NER using pyrimidine dimers:</p><ul><li><p>recognize damage (aka pyrimidine dimers, like T-T)</p></li><li><p><strong><u>helicase </u></strong>opens/unwinds the helix</p></li><li><p><strong><u>endonucleases </u></strong>remove <strong><u>multiple </u></strong>nucleotides surrounding the damaged region (on the same strand)</p><ul><li><p>specifically, <strong><u>excision endonucleases </u></strong><u>aka </u><span style="color: blue;"><strong><u>excinucleases</u></strong></span>, like <strong><u><mark data-color="blue" style="background-color: blue; color: inherit;">XP-F and XP-G</mark></u></strong></p></li></ul></li><li><p><strong><u>DNA polymerase</u></strong> synthesizes new DNA to fill in the gap</p></li><li><p><strong><u>DNA ligase </u></strong>seals the remaining nick</p></li></ul><p></p>
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explain how (regular/unmethylated) C deamination in dividing cells → results in base-pair mismatch & in a C>T mutation (3 steps)

  • & state transition of standard & non-standard/mismatched base pairs @ each step

  • (unmethylated) C is deaminated into U

    • so goes from C:G → U:G

  • cell division/replication occurs (dividing cells), where U is though to be part of the parent strand, SO DNA pol. pairs it w/ A in newly synthesized daughter strand

    • aka U:G —(cell div./repl. using U strand as the parent template strand)→ U:A

  • after cell division is done, BER occurs to replace U w/ T (B/C it recognizes that uracil is not apart of DNA)

    • so U:A → T:A

(C:G → U:G → U:AT:A ————— this is a C>T mutation)

^^^ b/c replication happened when mismatch was still present/not fixed, it caused a (permanent) mutation

<ul><li><p>(unmethylated) C is <strong><u>deaminated </u></strong>into U</p><ul><li><p>so goes from <span style="color: blue;"><strong>C:G → U:G</strong></span></p></li></ul></li><li><p><strong><u>cell division/replication</u></strong> occurs <u>(dividing cells)</u>, where U is though to be part of the parent strand, SO DNA pol. pairs it w/ A in newly synthesized daughter strand</p><ul><li><p>aka <span style="color: blue;"><strong>U:G</strong></span> —(cell div./repl. using U strand as the parent template strand)<span style="color: blue;"><strong>→ U:A</strong></span></p></li></ul></li><li><p>after cell division is done, <strong><u>BER</u></strong> occurs to replace U w/ T (<u><mark data-color="blue" style="background-color: blue; color: inherit;">B/C it recognizes that uracil is not apart of DNA</mark></u>)</p><ul><li><p>so <span style="color: blue;"><strong>U:A → T:A</strong></span></p></li></ul></li></ul><p>(<mark data-color="red" style="background-color: red; color: inherit;">C</mark>:G → <u>U</u>:G → U:<u>A</u> → <u><mark data-color="red" style="background-color: red; color: inherit;">T</mark></u>:A ————— this is a <span style="color: red;"><strong>C&gt;T mutation)</strong></span></p><p>^^^ b/c <strong>replication happened when mismatch was still present/not fixed, it caused a </strong>(permanent) <strong>mutation</strong></p><p></p>
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explain how 5-methycytosine deamination in dividing cells → results in mismatch & in mutation

  • deamination of C5-me, where C5-me:G → T:G

  • cell division/replication occurs (dividing cells), where makes 2 daughter strands: 1 mutant T:A & 1 wild-type C:G strand

    • mutant strand is T:A (b/c has diff. base pair from original starting, standard base pair)

    • wild-type strand is C:G (b/c has same base pair as original starting base pair)

(C>T mutation in mutant strand)

<ul><li><p><strong><u>deamination </u></strong>of C<sup>5-me</sup>, where C<sup>5-me</sup>:G → T:G</p></li><li><p><strong><u>cell division/replication</u></strong> occurs (<u>dividing cells</u>), where makes 2 daughter strands: <span style="color: blue;"><strong><u>1 mutant T:A</u></strong></span><strong><u> &amp; </u></strong><span style="color: blue;"><strong><u>1 wild-type C:G strand</u></strong></span></p><ul><li><p>mutant strand is T:A (b/c has diff. base pair from original starting, standard base pair)</p></li><li><p>wild-type strand is C:G (b/c has same base pair as original starting base pair)</p></li></ul></li></ul><p>(<span style="color: red;"><strong>C&gt;T mutation</strong></span> in mutant strand)</p>
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explain how 5-methycytosine deamination in NON-DIVIDING cells → results in mutation

specifically results in C>T mutation

  • b/c deamination of C5-me, where C5-me:G → T:G

    • (aka C5-me → T)

<p>specifically results in <span style="color: red;"><strong>C&gt;T mutation</strong></span></p><ul><li><p>b/c <strong><u>deamination </u></strong>of C<sup>5-me</sup>, where C<sup>5-me</sup>:G → T:G</p><ul><li><p>(aka C<sup>5-me</sup> → T)</p></li></ul></li></ul><p></p>
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can a mutation (like a C>T mutation) occur from:

  • deamination of C in dividing cell

  • deamination of 5-methylcytosine in non-dividing cell

  • deamination of C in non-dividing cell

  • deamination of 5-methylcytosine in dividing cell

deamination of C in dividing cell

  • YES, b/c C:G (when deamination)→ U:GU:A (when cell division/replication) → T:A (when BER of T)

deamination of 5-methylcytosine in non-dividing cell

  • likely YES

    • b/c C5-me:G (when deamination) → T:G

      • aka C turns into T, so is C>T mutation

deamination of C in non-dividing cell

  • likely NOT

    • b/c C:G (deamination) → U:G (BER) → original C:G

      • b/c BER recognizes that U is not part of DNA

      • this happens despite the glycosylase step of BER doing non-specific cleaving of a strand, BER will recognize that U is usually part of RNA, not DNA

deamination of 5-methylcytosine in dividing cell

  • YES, b/c C5-me:G (when deamination) → T:G (when replication) → get two daughter strands of T:A & C:G

    • mutant strand of C>T mutation is T:A (b/c causes a change in/is different from the starting standard base pair)

    • wild-type DNA strand is C:G (b/c is the same base pair as the original standard base pair)

^^^

if mismatch/mutation is NOT repaired before a cell divides, it will cause a (permanent) mutation in the parent strand & in synthesized daughter strand

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term image

yes b/c there’s no 2’ OH (think RUH “ROH”), so that means it’s DNA

  • b/c DNA vs. RNA is deter. by the 2’ sugar ring, not the type of base (i.e. not U vs. T)

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<p>for 5-methylcytosine deamination in non-dividing cells (shown in picture),</p><p>how does the repair machinery deter. which nucleotide is the original vs. mutant?</p>

for 5-methylcytosine deamination in non-dividing cells (shown in picture),

how does the repair machinery deter. which nucleotide is the original vs. mutant?

it can’t

b/c the “mutant” deaminated 5-methylcytosine is T (thymine)

BER occurs as the next step after deamination (b/c non-dividing cells):

  • sometimes BER removes T → then restored to original base pair of C:G

  • sometimes BER removes G → then sequence will mutate/change into T:A

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t/f: spontaneous deamination of cytosine changes our genome & results in C>T mutations

  • # of cytosines spontaneously deaminated per day

  • % of human genome that is made of cytosine

__

t/f: C>T mutations are highly represented in “mutational signatures” associated w/ cancer

_

t/f: malfunctioning DNA repair machinery leads to predisposition to disease

true

  • 100-500 Cs deaminated / day

  • 20% of human genome that is made of cytosine

__

true

  • b/c cancers are actively dividing cells

__

true

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human genome has (#) base pairs

t/f: no two genomes have identical sequences b/c our genomes have many SNPs / single-nucleotide polymorphisms

t/f: a mutation may not even be detected b/c there may not be a “correct”/common nucleotide for that position in the genome

3.2 billion base pairs

_

true

true

<p>3.2 billion base pairs</p><p>_</p><p>true</p><p>true</p>
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incorrect base incorporation during DNA replication is fixed how?

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chemical changes to DNA bases (alkylation, deamination, depurination) is fixed by (← state for each of the 3)

structural changes to DNA (pyrimidine dimers) is fixed by (1)

fixed by DNA polymerase’s exonuclease activity OR mismatch repair

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chemical changes to DNA bases (alkylation, deamination, depurination) is fixed by:

  • alkylation — by BER or direct repair (methyltransferase)

  • deamination — by BER (need glycosylase)

  • depurination — by BER (don’t need glycosylase)

structural changes to DNA (pyrimidine dimers) is fixed by — NER or direct repair

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early discoveries from mapping genes (~4)

  • first genetic map of chromosome (Sturtevant)

  • DNA is the genetic molecule (…McCarty experiment)

  • pieces of DNA “hop” around the genome, which disrupts genes & changes traits

  • structure of DNA solved

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for study on location genes w/in a genome:

  • explain experiment w/ pigmentation of corn

kernel color (trait), corn/maize genome

  • genomic DNA (chromosomes) are treated w/ DNA stain to get patterns of light and dark bands → used to find parts of genome that contain genes correlating w/ a specific trait

    • when the genomic fragment Ds that jumps around is in a position → purple kernels

    • when genomic fragment Ds is in another position → colorless/yellow kernels

(the genomic fragment Ds is the transposon)

<p><span style="color: rgb(180, 171, 171);">kernel color (trait), corn/maize genome</span></p><ul><li><p>genomic DNA (chromosomes) are treated w/ DNA stain to get patterns of light and dark bands → used to find parts of genome that contain genes correlating w/ a specific trait</p><ul><li><p>when the genomic fragment Ds that jumps around is in a position → <strong><u>purple </u></strong>kernels</p></li><li><p>when genomic fragment Ds is in another position → <strong><u>colorless</u></strong>/yellow kernels</p></li></ul></li></ul><p>(the genomic fragment Ds is the transposon)</p>
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def. Mendelian inheritance

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def. transposons

  • name 2 broad classes of transposons & briefly def.

t/f: more than half the human genome s made of transposable elements

genes are transmitted in a constant, predictable way from parent-to-child

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transposons — pieces of genomic DNA that can “jump” around (aka can be pasted/inserted into different positions of a genome to affect the expressed traits)

  • DNA transposon — cut & paste

  • retrotransposon — copy & paste

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true

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describe how DNA transposons work (incl. enzymes involved) (2)

DNA transposons

  • transposase cut/excises DNA transposon out of the original position in genome/DNA sequence (& flanking DNA leaves) → make a DNA intermediate

  • transposase pastes DNA inter. into another part of the genome (the target DNA) → make a transposed mobile element, where original/same DNA transposon is in different position

<p>DNA transposons</p><ul><li><p><span style="color: blue;"><strong><u>transposase<mark data-color="blue" style="background-color: blue; color: inherit;"> </mark></u></strong></span><u><mark data-color="blue" style="background-color: blue; color: inherit;">cut/excises </mark></u>DNA transposon out of the original position in genome/DNA sequence (&amp; <u><mark data-color="purple" style="background-color: purple; color: inherit;">flanking DNA leave</mark></u><mark data-color="purple" style="background-color: purple; color: inherit;">s</mark>) → make a <strong><u>DNA intermediate</u></strong></p></li><li><p><span style="color: blue;"><strong><u>transposase</u></strong> </span><span><u><mark data-color="blue" style="background-color: blue; color: inherit;">pastes </mark></u></span>DNA inter. into another part of the genome (the <u>target </u>DNA) → make a transposed mobile element, where <strong><u>original/same DNA transposon is in different position</u></strong></p></li></ul><p></p>
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describe how retrotransposons work (incl. enzymes involved) (3)

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how many copies of the retrotransposon do you have at the end?

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def. reverse transcriptase (give an example)

retrotransposons

  • RNA polymerase copies retrotransposon into RNA (& original sequence w/ retrotransposon + flanking DNA leavesno excision of original retrotransposon) → make a RNA intermediate

  • via reverse transcriptase: RNA sequence is reverse transcribed from RNA into DNA → make a DNA intermediate

  • DNA inter. is pasted into another part of the genome (the target DNA) → make a transposed mobile element, where copy of retrotransposon is in different position

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2 copies of retrotransposon: original (“donor DNA”) & newly synthesized transposon (“transposed mobile element”)


reverse transcriptase is a DNA polymerase that uses an RNA template

  • like telomerase

<p>retrotransposons</p><ul><li><p><span style="color: blue;"><strong><u>RNA polymerase</u></strong></span> <u><mark data-color="blue" style="background-color: blue; color: inherit;">copies </mark></u>retrotransposon into RNA (&amp; <mark data-color="purple" style="background-color: purple; color: inherit;">original sequence w/ retrotransposon + flanking DNA leaves</mark> ← <u>no excision of original retrotransposon</u>) → make a <strong><u>RNA intermediate</u></strong></p></li><li><p>via <span style="color: blue;"><strong><u>reverse transcriptase</u></strong></span>: RNA sequence is <u><mark data-color="yellow" style="background-color: yellow; color: inherit;">reverse transcribed</mark></u> from RNA into DNA → make a <strong><u>DNA intermediate</u></strong></p></li><li><p>DNA inter. is <u><mark data-color="blue" style="background-color: blue; color: inherit;">pasted </mark></u>into another part of the genome (the <u>target </u>DNA) → make a transposed mobile element, where <strong><u>copy of retrotransposon is in different position</u></strong></p></li></ul><p>__</p><p>2 copies of retrotransposon: original (“donor DNA”) &amp; newly synthesized transposon (“transposed mobile element”)</p><div data-type="horizontalRule"><hr></div><p><u>reverse transcriptase</u> is a DNA polymerase that uses an RNA template</p><ul><li><p>like <strong><u>telomerase</u></strong></p></li></ul><p></p>
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cont. kernel pigmentation study

the movement of a transposon in the maize genome causes variation in the (1)


t/f: insertion of transposon can have the effects of decreasing OR increasing gene expression (i.e. pigment). ALSO, the presence and absence of transposons can also affect the expression of the trait

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t/f: transposons induce variations in traits, which can include intermediate expressions of the trait

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t/f: transposons induce variation in traits that contribute to evolution

kernel pigmentation

  • b/c insertion of transposons in new positions w/in genome causes variations in expressed traits

  • (aka transposons induce variations in traits)

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true

  • w/ inserted transposon → either decrease or increased expression of trait

  • w/ or w/o inserted transposon → either decrease or increased expression of trait

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true

  • i.e. no transposon is black grapes, w/ transposon is white grapes, partially removed/piece of transposon is red grapes

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true

  • i.e. rise in industry caused moths to evolve from peppered to black appearance

<p>kernel pigmentation</p><ul><li><p>b/c insertion of transposons in new positions w/in genome causes variations in expressed traits</p></li><li><p>(aka transposons induce variations in traits)</p></li></ul><p>__</p><p>true</p><ul><li><p>w/ inserted transposon → either decrease or increased expression of trait</p></li><li><p>w/ or w/o inserted transposon →  either decrease or increased expression of trait</p></li></ul><p>__</p><p>true</p><ul><li><p>i.e. no transposon is black grapes, w/ transposon is white grapes, partially removed/piece of transposon is red grapes</p></li></ul><p>_</p><p>true</p><ul><li><p>i.e. rise in industry caused moths to evolve from peppered to black appearance</p></li></ul><p></p>
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where are genes located in the genome (2 parts)

genes are present in consistent, predictable positions (fixed positions) w/in genome of an organism

BUT some pieces of DNA called transposons can jump in & out of genes (insertion + excision), which affects traits

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t/f: gene structure is sim. in all eukaryotic genomes

name & def. 3 common parts of a gene in eu-

  • what does 1 of them tell us (2)

true

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enhancer

  • part of DNA sequence that binds proteins, like TFs, that stabilize or destabilize the RNA polymerase bound to the promoter

promoter

  • part of DNA sequence that binds RNA polymerase, SO when stably bound, starts RNA transcription

transcription unit

  • part of DNA sequence that encodes for RNA (aka DNA template for RNA synthesis / DNA → RNA)

    • where if RNA codes for → proteins → trait, then that transcription unit is part of the “blueprint”

    • but can also have RNA encoding for→ trait

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enhancer tells us where a gene is expressed & the amount of RNA transcribed

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name & def. two elements/parts of mRNAs / of transcription unit

exons — part of mRNA that is transcribed & kept in mature mRNA

introns — is removed from mRNA

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what happens if transposon is inserted/pasted into an enhancer?

  • gene is expressed in different areas/different tissues (← pattern)

  • amount of RNA expression (from RNA transcription) & amount of protein changes

BUT

  • protein made is normal (just the amount of protein changes) b/c has no effect on transcription units, which code for RNA

^ basically functions of enhancers

<ul><li><p>gene is <u>expressed in different areas</u>/different tissues (← pattern)</p></li><li><p><u>amount of RNA</u> expression (from RNA transcription) &amp; amount of <u>protein </u>changes </p></li></ul><p>BUT </p><ul><li><p><strong><u>protein made is normal</u></strong> (just the amount of protein changes) <span style="color: rgb(218, 114, 223);">b/c has no effect on transcription units, which code for RNA</span></p></li></ul><p>^ basically functions of enhancers</p>