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CELS191 Molecular Biology, Chromosomes, Genes > Cell and DNA replication > Flashcards

Cell and DNA replication Flashcards

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1
Q

2 significant findings of Chargaff’s work

A
  1. Within a species, the amount of adenine equals the amount of thymine, and the amount of guanine equals the amount of cytosine.
  2. The composition of DNA varies between species
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2
Q

Griffiths experimental conclusion

A

The information that determines a bacteria’s strain and virulence must be encoded in a nonliving chemical, as this information can be transferred from dead to living bacteria: a chemical substance is the bearer of genetic information

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3
Q

Avery’s conclusion

A

DNA is the chemical substance that acts as genetic material

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4
Q

main components of DNA

A
  • The DNA has a double stranded helical structure.
  • The sugar phosphate backbone is on the outside.
  • The bases are on the inside.
  • Stabilised by hydrogen bonds
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5
Q

bonds between nucleotide monomers and between strands

A

phosphodiester bond, hydrogen bonds

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6
Q

Formation of the phosphodiester bond

A

The hydroxyl group (OH) on the 3rd carbon of one nucleus reacts with the phosphate group attached to the 5th carbon on another nucleotide

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7
Q

features of prokayrotic DNA replication

A
  • single circular chromosome
  • bidirectional
  • single origin of replication
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8
Q

semi-conservative replication

A

Each DNA strand of the double helix is used as a template strand for the synthesis of two new strands

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9
Q

direction of DNA synthesis

A

DNA or RNA is always synthesised in 5 to 3 direction (therefore parental template strands are run in 3 to 5 direction)

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10
Q

molecules needed for DNA replication

A
  1. primase
  2. DNA polymerase III
  3. helicase
  4. topoisomerase
  5. DNA polymerase I
  6. DNA ligase
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11
Q

function of primase

A

Enzyme (RNA polymerase) that makes an RNA primer = starting point for DNA polymerisation

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12
Q

function of DNA polymerase III

A

Progressive addition of new nucleotides (A, C, T or G)

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13
Q

function of DNA polymerase I

A

removes RNA primers (RNase H) and fills the gap between okazaki fragments with DNA nucleotides (DNA polymerase)

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14
Q

function of DNA ligase

A

joins newly synthesised Okazaki fragments together (creates phosphodiester bonds)

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15
Q

function of helicase

A

Release tension generated by unwinding the DNA helix

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16
Q

major differences between eukaryotic and prokaryotic DNA replication

A
  • Multiple large linear (vs single small circular) chromosomes
  • Multiple (vs single) origin of replication (ori)
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17
Q

2 types of DNA error repair

A

exonuclease (during replication) and endonuclease (after replication)

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18
Q

features of exonuclease repair (during replication)

A
  • DNA pol III has a proof-reading mechanism - checks newly inserted nucleotide bases against the template.
  • These types of incorrect bases are removed by a 3’ to 5’ exonuclease activity of DNA pol III
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19
Q

features of endonuclease repair (after replication)

A
  • damaged region is removed by endonuclease.
  • DNA pol makes new DNA. DNA ligase joins DNA to existing DNA
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20
Q

cause of error during replication

A

DNA pol III makes very few mistakes - high fidelity

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21
Q

causes of error after replication

A
  • Incorrectly inserted bases are not corrected by DNA pol III.
  • Radiation damage (e.g. UV).
  • Chemical modifications of bases (natural and chemical causes)
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22
Q

importance of correcting DNA errors

A

if not corrected, DNA becomes part of DNA template which causes permanent DNA change and mutation

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23
Q

features of polymerase chain reaction

A
  • In vitro method of making multiple DNA copies.
  • Only ‘targeted’ DNA region will be copied.
  • Rapid exponential increase of DNA molecules.
  • Method utilises cycles of heating and cooling
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24
Q

PCR components

A
  • DNA template
  • primers
  • heat stable DNA polymerase
  • dNTPs (deoxynuclotide triphosphates = free nucleotides)
  • Buffer solution
  • Divalent cations (Mg2+)
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25
Q

Process of PCR

A
  1. Denaturation - double stranded DNA is denatured by heat into single strands; takes place at 94-98 degrees.
  2. Annealing - primers anneal (bind) to DNA template strand and provide OH group to start forming phosphodiester bonds; takes place at 45-70 degrees C.
  3. Extension/Elongation - DNA polymerase adds free nucleotides (dNTP’s); takes place at 72 degrees C
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26
Q

function of primer in PCR

A

Provides a free ‘3 OH group, the chemical group that is essential to initiate DNA synthesis. Defines the region of the DNA molecule to be replicated

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27
Q

function of heat stable DNA polymerase in PCR

A

Enzyme which adds nucleotides, (complementary to the DNA template), and joins them together forming a phosphodiester bond (e.g. Taq DNA polymerase but there are many others)

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28
Q

function or divalent cations in PCR

A

act as cofactors - essential for DNA pol to work

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29
Q

central dogma of molecular biology

A

the coded genetic information in DNA can be transcribed into transportable molecules (mRNA) and these contain the programme for synthesis of a particular protein

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30
Q

gene expression

A

process by which information from a gene is used in the synthesis of a functional gene product

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31
Q

process of transcription

A
  1. catalysed by RNA polymerase.
  2. RNA polymerase synthesises mRNA by catalysing the formation of covalent phosphodiester bonds between ribonucleotides.
  3. RNA polymerase selects the correct nucleotides to be incorporated with the mRNA based on the sequence of DNA that is being transcribed.
  4. mRNA is made from template strand.
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32
Q

what is a gene

A

defined region (sequence) of DNA that produces a diffusible product (RNA)that has some function

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33
Q

steps of transcription

A
  1. initiation
  2. elongation
  3. termination
34
Q

process of initiation

A
  1. Transcription factors bind to the TATA box and other regions of the promoter
  2. RNA pol II binds, forming a transcriptional initiation complex together with the transcription factor
  3. The two DNA strands separate and RNA pol II starts mRNA synthesis without the need of a prime
35
Q

process of elongation

A

RNA pol II uses the template strand, which runs in the 3’ -> 5’ direction, as a template, and inserts complementary RNA nucleotides in the 5’ -> 3 direction

36
Q

anatomy of prokaryotic gene

A
37
Q

Anatomy of eukaryotic genes

A
38
Q

standard splicing process

A

one gene codes for one protein

39
Q

Coding sequence (exon)

A

Portion of a gene’s DNA that is translated into a protein

40
Q

Promoter

A

DNA segment recognised by RNA polymerase to initiate transcription

41
Q

5’ UTR and 3’ UTR (UnTranslated Region)

A
  • Contain regulatory elements that influence on gene expression at the transcriptional and/or translational level (e.g. by influencing on mRNA stability, translation efficiency or localisation)
  • Transcribed but (usually) not translated
42
Q

5’ G cap

A

Prevents mRNA degradation, regulates translation (by providing a ribosome recognition and binding site), regulation of nuclear export, promote intron excision

43
Q

Poly-A tail

A

Prevents mRNA degradation, regulates translation and nuclear export

44
Q

Alternative splicing

A

one gene codes for several proteins

45
Q
A
46
Q

Post-translational Chemical modification

A
  1. Phosphorylation (addition of a phosphate group) – regulation of cellular processes
  2. Glycosylation (addition of sugar groups) – affect protein folding, conformation, distribution and activity)
  3. Methylation (addition of methyl group(s)) – change hydrophobicity
47
Q

difference between prokaryotic and eukaryotic transcription

A

In prokaryotes:

  • Transcription and translation in cytoplasm
  • Transcription and translation are coupled

In eukaryotes:

  • Transcription in nucleus
  • Translation in cytoplasm
  • Transcription and translation are NOT coupled
48
Q

Key features of the genetic code

A
  • bases must come in triplets so there are enough combinations to code for 20 amino acids
  • 61 of a possible 64 codons specify an amino acid
  • most amino acids have more than one codon
  • three codons specify stop (UAA, UAG and UGA)
  • one codon specifies start (AUG - this codon also specifies Methionine)
  • genetic codon table must be read in 5’ to 3’ direction
49
Q

three stop codons

A

UAA, UAG, UGA

50
Q

start codon

A

AUG

51
Q

role and features of tRNA

A
  • functions as adaptor molecule between amino and mRNA
  • single-strand of RNA
  • 70 - 80 nucleotides in length
  • at least one tRNA for each amino acid
  • Each tRNA has a region which can bind an amino acid AND a region which can interact with mRNA
52
Q

Steps involved in ‘charging’ a tRNA

A
  • An enzyme (aminoacyl tRNA synthetase), recognises both a specific amino acid and the correct tRNA for this amino acid and joins them together
  • There are 20 different (aminoacyl tRNA synthetase) enzymes, one for each amino acid
  • charging = attaching amino acid
53
Q

what is translation

A

synthesis of proteins by ribosomes using mRNA as a set of instructions

54
Q

process of translation

A
  1. initiation
  2. elongation
  3. termination
55
Q

process of translational initiation

A
  1. The small ribosomal subunits finds the initiation AUG codon on the mRNA. The AUG codon is positioned in the P site of the small ribosomal subunit.
  2. A tRNA ‘charged’ with the amino acid Methionine (Met, M) binds to the P site
  3. The large ribosomal sub unit attaches
  4. Energy is required
56
Q

process of translational elongation

A
  1. A ‘charged’ tRNA, with an anticodon complementary to the A site codon, lands in the A site.
  2. Then two things happen at the same time:
    1. i. The ribosome will break the bond that binds the amino acid to the tRNA in the P site, transfer the amino acid to the newly arrived amino acid (attached to the tRNA in the A site) and form a peptide bond between them
    2. ii. While the tRNAs are bound to the mRNA (in the P and A sites), the ribosome moves three nucleotides down the mRNA.
  3. In the E site, the ‘uncharged’ tRNA detaches from its anticodon and is expelled. 4. Energy is required 5. A new ‘charged’ tRNA with an anticodon complementary to the next A site codon enters the ribosome at the A site and the elongation process repeats itself.
57
Q

process of translational termination

A
  1. When the ribosome reaches a stop codon, a protein called release factor enters the A site
  2. The release factor uses water to break the bond between the P site tRNA and the lastly added amino acid. This causes the polypeptide chain to detach from its tRNA and the newly made polypeptide (protein – well, it will the functional protein one the long chain of amino acids/polypeptide has folded into a 3D structure) is released
  3. The small and large ribosomal sub units dissociate from the mRNA and each other.
  4. Energy is required
58
Q

impacts of application genetics on society

A

advancements in medicine, law, sociology, philosophy, ecology and agriculture

59
Q

steps of mitosis

A
  1. prophase
    1. prometaphase
  2. metaphase
  3. anaphase
  4. telophase
    1. cytokinesis
60
Q

features of prophase

A
  • centrosomes separate, but are connected by tubular spindles that reach out from the asters around the centrosomes
  • chromosomes inside the nuclear envelope condense and duplicate to become sister chromatids connected by centromere
  • each chromatid contains 1 double stranded copy of DNA
61
Q

features of prometaphase

A
  • nuclear envelope dissolves
  • the kinetochore on the chromosomes connect to the microtubular spindles called kinetochore microtubules
  • the microtubular spindles without chromosomes attached are called non-kinetochore microtubules
62
Q

features of metaphase

A

microtubules stretch to the edge of the cell and chromosomes aligned at the equator of the cell as the metaphase plate

63
Q

features of anaphase

A
  • metaphase plate is split and the microtubules get shorter so the chromatids are pulled apart by their kinetochores
  • kinetochore microtubules shorten
  • non-kinetochore microtubules lengthen
64
Q

Telophase and cytokinesis

A
  • cleavage furrow forms to create two daughter cells
  • the nuclear envelope forms around the DNA
65
Q

what happens in G1 phase of cell replication

A

organelles replicate

66
Q

what happens in S phase of cell replication

A

DNA replication

67
Q

what happens in G2 phase of cell replication

A

enzymes used in mitosis produced

68
Q

what is the sexual cycle

A
69
Q

steps of meiosis I

A
  1. prophase I - homologous chromosomes made up of sister chromatids, line up into tetrad pair and exchange segments (crossing over - occurs at chiasmata); spindle fibres form
  2. metaphase I - chromosomes line up on the metaphase plate through the chiasmata whilst attached to the kinetochore microtubules through the centromeres
  3. anaphase I - pairs of homologous chromosomes split up and sister chromatids remain unaffected; kinetochore microtubules get shorter and the non-kinetochore microtubules get longer
  4. Telophase I and cytokinesis- two cells form through the cleavage furrow
70
Q

functions of mitosis

A

growth and repair

71
Q

function of meiosis

A

make gametes for sexual reproduction

  • halves number of chromosomes in cell
72
Q

steps of meiosis II

A
  1. Prophase II - Spindle apparatus forms
  2. Metaphase II - centromeres line up on the metaphase plate
  3. Anaphase II - sister chromatid pulled apart as the non-kinetochore microtubules get longer and the kinetochore microtubules get shorter
  4. Telophase II and cytokinesis- haploid daughter cells form through the cleavage furrow
73
Q

Segregation and assortment

A

two equally probably arrangements of chromosomes at metaphase I

74
Q

mitosis vs meiosis

A

Mitosis:

  • chromosomes align independently
  • no chiasmata
  • centromeres on metaphase plate
  • chromatids disjoin
  • 2n -> 2n

Meiosis:

  • homologous chromosomes synapse
  • chiasmata
  • chiasmata on metaphase plate
  • chromosomes disjoin
  • 2n -> n
75
Q

function of cell division in multicellular organisms

A
  • development from a fertilised egg
  • growth to adult
  • repair
76
Q

function of mitosis in context of cell cycle

A

two copies of chromosomes and organelles replicated in interphase are halved to produce cells with normal numbers of each

77
Q

ovum production

A
78
Q

structures of meiosis I

A
79
Q

structures of meiosis II

A
80
Q

how does meiosis lead to gametic and zygotic diversity

A

crossing over in prophase I and segregation and assortment result in genes of parental chromosomes being mixed up and swapped around in gametes, causing variation

CELS191 Molecular Biology, Chromosomes, Genes (2 decks)

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