Ch. 6 Learning Objectives Flashcards

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1
Q

Explain how a DNA double helix provides a template for its own replication

A

because DNA has complimentary base pairs

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2
Q

describe the resulting daughter helices in terms of their sequence and the distribution of parental and
newly synthesized DNA strands.

A

the daughter DNA has one of the original (old) and one strand that is completely new. (semiconservative)

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3
Q

Recall where along a chromosome DNA synthesis begins (ori)

A

the initiator protein binds to specific DNA sequences called replication origins

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4
Q

function of initiator protein

A

The initiator protein pry the 2 DNA strand apart breaking

the hydrogen bonds between the bases.

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5
Q

explain what characterizes these nucleotide sequences (ori) in simple cells such as bacteria and yeast.

A
  • replication origins span approximately 100 nucleotide
    pair
  • composed of DNA (A-T basepairs) that attract initiator proteins
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6
Q

why do initiator proteins like to bind to DNA rich in A-T basepairs

A

they are easier to break because they only have 2 H bonds they have to break

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7
Q

Compare the direction in which replication forks move with the direction in which the new
DNA strands are synthesized.

A
  • The 2 forks moves away from the origin in opposite directions
  • DNA replication in both bacterial
    and eukaryotic is bidirectional.
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8
Q

Do prokaryotes or eukaryotes replicate faster

A

Prokaryotes because eukaryotes have a more complex chromatin structure

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9
Q

Compare the bonds that link together nucleotides in a DNA strand with the bonds that hold
together the two strands of DNA in a double helix.

A
  • Covalent bonds occur within each linear strand and strongly bond the bases, sugars, and phosphate groups
  • Hydrogen bonds occur between nucleotides
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10
Q

Explain how nucleoside triphosphates provide the energy for DNA synthesis

A

is provided by the incoming deoxyribonucleoside triphosphate itself

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11
Q

Explain why an asymmetrical replication fork poses a challenge for DNA polymerization

A
  • one strand is 5’ –3’ and the other 3’-5’ at each replication fork. This causes an issue with DNA replication as it makes it impossible to synthesized straight forward.
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12
Q

how DNA polymerase solves the asymmetrical replication fork problem to keep the replication fork moving forward

A
  • all DNA polymerases add new subunits only to the 3’ end of a DNA strand. A new DNA chain can be synthesized only in a 5’-3’ direction.\
  • for the (3’-5’) strand, DNA uses the “backstitching” maneuver.
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13
Q

the primers required for DNA replication leading strand

A
  • rna primer is needed only to start replication at the replication origin then DNA polymerase takes over.
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14
Q

the primers required for DNA replication lagging strand

A
  • new rna primers are continuously needed
    to keep polymerization going.
  • DNA polymerase then adds a deoxyribonucleotide to the Okazaki fragments until it runs into the previously synthesized RNAprimer.
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15
Q

what 3 additional enzymes are need to join these Okazaki fragments together:

A
  • nuclease
  • DNA repair polymerase
  • DNA ligase
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16
Q

nuclease function

A

degrade the RNA primer.

17
Q

DNA repair polymerase function

A

replace the RNA primers with

DNA

18
Q

DNA ligase function

A

join the 5’-phosphate to the 3’-hydrolxly

19
Q

Name 7 proteins that form part of the replication machine

A
  • DNA polymerase
  • DNA helicase
  • Single-stranded DNA- binding protein (SSB)
  • DNA topoisomerase
  • Sliding clamp
  • Clamp loader
  • DNA ligase
20
Q

DNA polymerase role in DNA replication

A

catalyzes the addition of nucleotides to the 3’ end of a growing strand of DNA strand as a template.

21
Q

DNA helicase role in DNA replication

A

uses the energy of ATP hydrolysis to unwind the DNA double helix ahead of the replication fork.

22
Q

SSB role in dna replication

A

Binds to single-stranded DNA, preventing base pairs from re-forming before the lagging strand can be replicated.

23
Q

DNA topoisomerase role in dna replication

A

produces transient nicks in the DNA backbone to relieve the tension built up by the unwinding of DNA ahead of the DNA helicase.

24
Q

sliding clamp role in dna replication

A

Keeps DNA polymerase attached to the template

25
Q

clamp loader role in dna replication

A

uses the energy of ATP hydrolysis to lock the sliding clamp onto DNA

26
Q

DNA ligase role in dna replication

A

uses the energy of ATP hydrolysis to join Okazaki fragments made on the lagging-strand template.

27
Q

Describe the problem created by a moving replication fork

A

when the end of a chromosome approaches, while leading strand can be replicated all the way to the end, lagging strand cannot.

28
Q

explain how DNA topoisomerases solve the moving replication fork problem

A
  • in bacteria they have circular DNA molecules as chromosomes.
  • Eukaryotes, DNA topoisomerase adds long repetitive nucleotide sequences to the ends of every chromosomes.
29
Q

Explain how DNA polymerase contributes to the accuracy of DNA replication.

A

DNA repair polymerases (I and III) removes the RNA, proofreads, and synthesizes DNA

30
Q

Thermal collisions

A
  • chemical changes in DNA.

- purine bases (A&G) will be lost by a spontaneous reaction called Depurination.

31
Q

Deamination

A

loss of amino group from a cytosine in DNA to produce the base uracil

32
Q

UV radiation dna damage

A

cause covalent linkage between two adjacent pyrimidine bases, forming thymine dimer.

33
Q

cell metabolism dna damage

A

incorrect base-pairing during replication or to deletion of one or more nucleotide pairs

34
Q

List the three main steps involved in repairing damage that affects only one strand of the DNA double helix

A

Step 1: The damaged DNA is recognized and removed.
Step 2: A repair DNA polymerase binds to the 3’-hydroxl end of the cute DNA strand. the enzyme fills in the gap
Step 3: This nick in the helix is sealed by DNA ligase.

35
Q

Explain how the mismatch repair system recognizes and corrects replication errors.

A
  • removed a portion of the DNA strand containing the error, and then resynthesize the missing DNA
36
Q

Non-homologous end joining

A

specialized group of enzymes that “clean” the broken ends and rejoin them by DNA ligation

37
Q

Homologous recombination

A

specific nuclease chews back the 5’ ends of the two broken strands at the break. Then, with the help of specialized enzymes one of the broken 3’ ends “invades” the unbroken homologous DNA duplex and searches for a complementary sequence through base pairing.

38
Q

where are the 3 places that require tons of atp

A
  • dna helicase
  • ## sliding clamp