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Genetics and evolution > Lecture 13 > Flashcards

Lecture 13 Flashcards

1
Q

the double helix

A

two nucleotide polymers
bound by complementary base pairing
phosphodiester bonds have 5’-3’ polarity
antiparallel strands
helical structure with specific features

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2
Q

Structural challenges of replication

A

how the new and old stands generate two new double helices

how are the nucleotides exposed and read

how to ensure fidelity

how to organise the enzymatic activities for multiple long DNA molecules

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3
Q

Conservative replication

A

histone scaffold

breaks hydrogen bonds

new nucleotide pairing without unwinding

need an isomerase activity to turn bases respect deoxyribose

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4
Q

Semi-conservative replication

A

breaks hydrogen bonds

new double helices are assembled as replication progresses

needs extensive unwinding: potential need for helicases

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5
Q

dispersive replication

A

no unwinding required
swapping templates strands
but requires multiple single-stranded breaks
assumes polymerisation is in both directions

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6
Q

Taylor, Woods and Hughes

A

growing root tip of vicia faba

they supplied thymidine labelled with tritium and after labelling the chromosomes incubate in cold medium with colchicine (blocks cytokinesis)

they looked at cells in metaphase and observed distribution of thymidine in chromosomes

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7
Q

Meselson and Stahl

A

reasoned that DNA with 15N in its nitrogens bases would be denser than with 14N so should lead to different sedimentations behaviour under CsCl gradient centrifugation

conc of Cs ions would be different along the test tube so DNA would sit at different heights based on density

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8
Q

Features of semi-conservative replication

A

fast
- human genome replicated in around 8 hours
accurate
-error rates of 10-5 in human polymerases but DNA repair mechanisms lower the overall rates to estimate 10-8
not perfect
- some errors , 7-0 new mutations per generation
Highly regulated
- dozens of enzymes participate in the process

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9
Q

Replication initiated at origins

A

replication proceeds in both ends of the bubble (forks)
however polymerisation is in one direction only

the replication bubble requires unwinding the strands of DNA

a eukaryotic chromomse may have 100s-1000s of origins of replications, this increases speed

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10
Q

replication forks

A

a very active area where DNA replication takes place

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11
Q

How do replication forks advance

A
  1. requires finding the 3-5 DNA polymerase
    or
  2. requires finding the small fragments and explaining how they are joined
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12
Q

Okazakis experiments

A

they used the replication of a phage in E.coli

they used sedimentation to separate DNA in this case of different lengths

they labelled the newly synthesised DNA with thymidine and then measured radioactivity in the distribution of sizes

then they used mutant phages with temperature-sensitive ligase (to join small fragments)

they did many other experiments using 5-3 exonuclease to prove that newly added nucleotides were only incorporated at the 3’ ends

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13
Q

leading strands

A

can be synthesised in a continuous process

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14
Q

lagging strands

A

the lagging strand requires additional steps

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15
Q

Essential enzymatic activities

A

DNA polymerase
can only synthesise in 5-3 orientation
needs a primer that provides 3 end

DNA primases
synthesise 5-10nt RNA primers from scratch using DNA as a template

Helices
unwind the double helix at the replication forks

Topoisomerases
relieve the strain of twisting of the double helix by breaking, swivelling and rejoining DNA strands ahead of the fork

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16
Q

Synthesis of the leading strands

A

as the fork advances the leading strand keeps elongating

17
Q

Synthesis of the lagging strand

A

like with the leading strand first the primase synthesises the RNA primer

DNA polymerase III elongates the primer, until it bumps into the 5’ end of another primer and falls off the DNA

the DNA polymerase I removes the ribonucleotides at the 3’ end of the Okazaki fragment

finally the ligase joins the adjacent nucleotides between two fragments closing the single strand DNA break

18
Q

The trombone

A

both strands together

the two polymerases at the fork advance in the same direction

19
Q

What happens at the end of the DNA

A

in prokaryotes circular chromosomes, the last active DNA polymerase III will always encounter an RNA primer and DNA polymerase I and ligase will finish the job

but the linear eukaryotic chromosomes have a problem: the last RNA primer of the lagging strand cannot be made into DNA

20
Q

Telomeres

A

telomeres are special, repetitive sequences at the end of the eukaryotic chromosomal DNA

Stabilised by proteins

They can be extended after every cell cycle, if the cell has telomerase

They do not prevent but postpone after every cell cycle, if the cell has telomerase

They do not prevent but postpone the erosion of genes near the ends of DNA molecules as they provide a buffer

21
Q

Effects of telomeres on the germline

A

loss of telomeres would lead to offspring inheriting chromosomes lacking some genes

22
Q

Effects of telomeres on somatic cells

A

progressive loss of terminal genes would impact tissue renewal

23
Q

Effects of telomeres on medicine

A

cancer cells need to express telomerase and somatic cells lacking telomerase may be related to aging

24
Q

Cyclic activity of telomerase

A

elongation- reverse transcription (RNA to DNA)

translocation- RNA template ends have the same sequence

25
Q

Replication errors

A

minimised by proofreading capacity of enzymes but polymerase mistakes and DNA damage due to tension ahead of forks can cause changes in the sequence

26
Q

If there are errors in the germline:

A

sequence changes can pass to the next generation
depending on the phenotypic effect the changes may be maintained in the population
this allows evolution

27
Q

If there’s errors in somatic cells:

A

cells can become defective and be removed or accumulate as senescent cells decreasing performance of tissue

potential development of cancer (the incidence of tumours is related to the frequency of cell division )

28
Q

Mutagens

A

ionising radiation (X and gamma rays)

chemical

29
Q

Chemical mutagens

A

Alkylating agents: modify the bases in the DNA making the polymerases pair them with the wrong base during replication

nucleotide analogues: mimic the structure of a nucleotide to fool DNA polymerase to introduce them in DNA

intercalating agents: insert themselves between he nitrobases and can create frame shifts during replication

30
Q

DNA repair mechanisms

A

DNA polymerase proofread newly made DNA replacing incorrect nucleotides

in mismatch repair of DNA repair enzymes correct error in base pairing

in nucleotide excision repair, DNA is scanned for structural deformations, a nuclease cuts out and replaces damaged stretched of DNA and the gap is closed by polymerase/ligase activities

Genetics and evolution (15 decks)

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