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bio12 > Unit 3: Molecular Genetics > Flashcards

Unit 3: Molecular Genetics Flashcards

1
Q

Who were credited with deducing the structure of DNA

A

James Watson and Francis Crick

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2
Q

What year was the DNA structure deduced

A

1953

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3
Q

3 main components of DNA

A
  1. A deoxyribose sugar
  2. A phosphate group
  3. A nitrogenous base
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4
Q

DNA is a what

A

Nucleotide polymer

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5
Q

What is DNA structure like

A
  • Thread like
  • it is twisted in a clockwise direction to form a double helix
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6
Q

What direction does DNA strands run in

A

Consists of 2 anti parallel strands of nucleotide( run side by side in opposite direction)

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7
Q

Each DNA strand has what

A

5’ and 3’ end which run in opposite directions to one another

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8
Q

The 5’ end terminates with what

A

Phosphate group

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9
Q

The 3’ end terminates with what

A

Hydroxyl (OH) group of the sugar

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10
Q

What are the complementary nitrogenous base pairings

A

Adenine and thymine
Cytosine and guanine

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11
Q

What nitrogenous bases are purines

A

adenine and Guanine

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12
Q

What nitrogenous bases are pyrimidines

A

Cytosine and thymine

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13
Q

How long would s stretched out long double helix be

A

1.8m but an individual núcleos would be 5nano meter

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14
Q

Histones

A

8 stabilizing proteins

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15
Q

For every 200 nucleotides , DNA is what

A

Coiled around a group of 8 stabilizing proteins

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16
Q

What charge are histones

A

Positively

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17
Q

Nucleosomes

A

The complex of histones supporting coiled DNA

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18
Q

Chromatin fibre

A

A series of nucleosomes coiled

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19
Q

How is final chromosome structure obtained

A

Supercoiling of chromatin fibre

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20
Q

What are the 3 forces contributing to DNA molecule

A
  1. Phosphodiester Linkages
  2. Hydrogen bonding
  3. Hydrophilic/ hydrophobic reactions
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21
Q

Phosphodiester linkages

A

Link the sugar / phosphate of one nucleotide to another ( stabilizes the handrail)

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22
Q

Hydrogen bonding

A

Keeps the 2 strands together as well as stabilizing the inner core (rungs)

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23
Q

Hydrophilic/ hydrophobic reactions

A

Cause the bases (hydrophobic) to remain inside the structure while the phosphate group and group (hydrophilic ) face out into the watery nucleus of the cell
(Similar to the cell membrane )

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24
Q

Senescent cell

A

larger than young cells, excrete proteins at a different rate and no longer divide.

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25
Q

telomerase

A

enzyme that appears to stabilize telomere length and keep cells useful

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26
Q

telomeres

A

chromosome tips and they consist of thousands of identical secions of DNA strung together like beads of a neckalce.

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27
Q

which type of cells seem to have a supply of telomerase

A

cancer cells and stem cells

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28
Q

the link between telomerees and aging has been supproted by what evidence

A

by studying people with progeria who are born with short telomeres and have an average lifespan of 12,7 yrs

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29
Q

why might cell senescence have evolved

A

to keep us tumor free during our reproductive years

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30
Q

Watson and crick discovery

A

credited with deducing the structure of DNA (double helix) in 1953

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31
Q

Rosalind Franklin

A
  • suggested that the sugar phosphate backbone of DNA faced the outside of the molecule and not the inside
  • suggested DNA was a double helix
  • determined the diameter of DNA moleucle
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32
Q

Alfred Hershey and Matha chase

A
  • confirmed that the DNA of a bacteriophage transformed the DNA of an infected bacterium
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33
Q

Avery, Mccarty, and mcleod

A

showed that DNA rather than RNA or protein, was the transforming substance

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34
Q

Chargaff

A

discovered that the nitrogenous bases in DNA always occur in exact ratios, with the number of adenine matching the number of thymine, and the number of cytosine matching the number of guanine

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35
Q

Hammerling

A

the nucleus of a cell contains hereditary info that controls the development of organisms

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36
Q

Griffith

A

discovered a substance derived from infectious bacteria that could turn non-infectious bacteria into infectous bacteria

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37
Q

Miescher

A

“nuclein” - DNA

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38
Q

what did melseson and stahl discver and what is it known as

A

proof of semiconservative replication aka “the msot beautiful experiment on earth”

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39
Q

what did meselson and stahl do to discover what they did

A
  1. grew Ecoli bacteria in a nutrient medum rich in 15N for several generations to ensure that the 15N would be encorporated in the DNA.
  2. bacteria were then transfered to a culture medium containing the light isotope 14N
  3. centrifugation was used to isolate the DNA strands
  4. cetrifugation was centrifugal force and boyant forces to seperate materials of diffeent densities
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40
Q

what was in test tube 1 , 2, 3 , 4 and what did they prove

A

TEST TUBE 1: containing heavy DNA - produce one discrete bond in a low region
TEST TUBE 2: Containin regular DNA- produce one discrete bond in a higher region
TEST TUBE 3: Bacteria from heavy N medium allowed to replicate once in the normal N medium - produced an discrete bond intermediate between the bonds in tube 1 and 2
- PROVED THAT DNA WAS A HYBRID(combined one heavy and one regular strand)
TEST TUBE 4: Bacteria from hevy n medium allowed to replciate twide in the normal nitogen medium - produced 2 discrete bonds: one at at the intermediate level and one at regular N level
- PROVED THAT THE INITIAL DNA WAS CONTINUALLY USED AS A TEMPLATE

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41
Q

what type of replication is DNA

A

semi-conservative

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42
Q

what is each new DNA strand composed of

A

an existing parent half and newly synthesized half combined together

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43
Q

what do proteins bind to on DNA

A

replicated origins

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44
Q

DNA helicase

A

breaks the hydrogen bond holding the two complementary parents together. Resulting in an unzipped helix that terminates at the replication fork

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45
Q

Single stranded binding proteins

A
  • keep the DNA strands seperated by blocking reformation of hydrogen bonds
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46
Q

DNA gyrase and Topoisomerase

A

relieves tension and unwinds double helix

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47
Q

what does DNA polymerase III do

A

builds the complementary strand

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48
Q

what direction does DNA polymerase synthesize DNA in

A

5’ to 3’ direction

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49
Q

leading strand

A

strand which uses the 3’ to 5’ direction template strand and synthesized continuosly(5’to3’) in the direction of the replication fork

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50
Q

lagging strand

A

strand which uses the 5’ to 3’ template strand and is synthesized discontinuosly (5’ to 3’) in short fragments called okazaki fragments in the opposite direction of the replication fork

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51
Q

DNA polymerase 1

A

removes RNA primers from the leading strand and fragments of the lagging strand and replaces them with the correct deoxribonucleotide

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52
Q

DNA ligase

A

joins the okazaki fragments together via phosphodiester bonds.

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53
Q

termination of DNA rep

A

as strands are built, they automatically rewind to form a double helix

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54
Q

how does DNA rep ensure accuracy

A
  • DNA polymerase III and I can recognize whether hydrogen bonding has taken place between base pairs, if not , a mismatch has occurred and the polymerase excises the incorrect base and inserts the correct one
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55
Q

telomeres

A
  • long sequences of repetitive non coding DNA
  • buffer zones at the ends of chromosomes that help guard against the loss of valuable genetic material during replication.
56
Q

DNA in comparison to RNA

A
  • SUGAR: DNA has deoxyribose and RNA has ribose
  • STRUCTURE: DNA has double helix and RNA has single helix
  • SIZE: DNA is a larger molecule that can be thousands of nucleotides long while RNA is a smaller molecule that transcribed only a section of DNA
    -AMOUNT: DNA is few molecule in the cell and RNA is many molecules in the cell
  • WHERE: DNA in nucleus and RNA in both nucleus and cytoplasm
    -KINDS: DNA has one type and RNA has three types (mRNA,tRNA and rRNA)
    -NITROGENOUS BASES: DNA- A,C,G,T while RNA- A,C,G,U
57
Q

genes

A

a sequence of nucleotides in DNA or RNA that encodes the synthesis of a protein

58
Q

genes are not what

A

spaced regularly along chromosomes

59
Q

size of genome and number of genes is?

A

Not directly related

60
Q

Exons

A

coding regions

61
Q

Introns

A

noncoding regions

62
Q

In general, the frequency and length of introns is loosely related to

A

the development complexity of the organism

63
Q

In vertebrates what percentage of introns we have

A

95% non coding regions

64
Q

multigene family

A
  • contain from a few hundred up to hundred of thousands of copies of the same or very similar genes
  • in some cases they code for a high in demand protein
65
Q

triplet hypothesis

A

genetic code codons are made up of nucleotide triplets - Francis crick

66
Q

genetic code central dogma

A

the genetic code reveals the amino acid that each codon (nucleotide triplets/3 letters) codes for

67
Q

characteristics of genetic code

A
  1. continuity - reads as a set of 3 letter codons with no spaces and has a correct reading frame
  2. redundancy- 64 possible codons but only 20 amino acids
    - offers some protection against mutation
    - not random
  3. Universality- genetic code is essentially the same in all living organisms(ie. same mRNA codons correspond to the same amino acids)
    - provides evidence that most organisms share a common ancestor
68
Q

1 protein has

A

hundreds of amino acids

69
Q

the sequene of amino acids is determined by

A

the sequene of nucleotides in an organism

70
Q

a protein is composed of combinations of

A

20 amino acids

71
Q

3 types of RNA

A

mRNA- messenger RNA
tRNA- transfer RNA
rRNA- ribosomal RNA

72
Q

rRNA

A

provide the site on the ribosome where the polypeptide is assembled

73
Q

tRNA

A
  • smaller than rRNA
  • there are a max number of the kinds of tRNA, though may vary from organism to organism
  • float freely in cytoplasm
  • transport amino acids to the ribosome and position amino acids on the chain
74
Q

mRNA

A
  • long single strand of RNA
  • passes from nucleus to cytoplasm through nuclear pores
  • brings info from chromosomes to ribosomes to direct protein synthesis
    -spliced, capped and tailed before leaving nucleus
75
Q

2 main processes in protein syntheis

A

Transcription
Translation

76
Q

transcription

A
  • RNA copy of a gene is produced (mRNA) which is transcribed from DNA
  • ## this is initiated when the enzyme RNA polymerase binds to a sequence of nucleotides on one DNA strand
77
Q

translation

A
  • synthesis of polypeptides by a ribosome
  • information of mRNA directs the choice of amino acids
  • ribosomes moves along in step 3(codons)
78
Q

DNA to mRNA

A

transciption

79
Q

4 stages of transcription in DNA to mRNA

A
  1. Initiation
  2. Elongation
  3. Termination
  4. Processing of mRNA transcript
    a)capping and tailing
    b) splicing
80
Q

Initiation in transcription

A
  • RNA polymerase contains a binding site that only recognizes a promoter sequence on the DNA
81
Q

promoter sequence

A
  • rich in A and T nucleotides
  • sequences are not palindromic so RNA polymerase will bind in only one orientation
  • this sequence does not get transcribed only serves as the attachment site for RNA polymerase
82
Q
  1. Elongation in transcription
A
  • RNA polymerase works in the 5’ to 3’ direction
  • Does not require a primer to begin
  • transcribes only one strand( template strand)
  • RNA polymerase opens the helix one section at a time and arranges the correct RNA nucleotides according to the DNA template
  • Once the RNA polymerase passes a given point the DNA helix reforms and the mRNA strand seperates from the DNA
  • RNA polymerase no proofreading capability
83
Q
  1. Termination in transcription
A
  • transcription stops when RNA polymerase encounters a terminator sequence (highly specific)
84
Q
  1. processing of mRNA transcript a)capping and tailing
A
  • 5’ cap- made of modified guanine nucleotides which is added to the start of the primary transcript(pre-cursor mRNA)
  • Poly A tail- string of 200 adenine ribonucleotides is added to the 3’end by poly A polymerase
  • both addition protect the mRNA from digestion by enzymes and assist in the initiation of translation
85
Q
  1. processing of mRNA transcript b) splicing
A
  • eukaryotic genes contain exons and introns
  • spliceosome cut out introns and join exons together to create a continuous coding regions on the mRNA transcript
86
Q

From mRNA to protein

A

translation

87
Q

5 sidesteps of translation

A
  1. ribosomes
  2. transfer RNA
  3. initiation
  4. elongation(of polypeptide)
    5.termination
88
Q

ribosomes in translation

A
  • ribosomes consists of 2 subunits (1 large and 1 small) and rRNA
  • 2 subunits bind mRNA between them
  • moves along the mRNA transcript in the 5’ to 3’ direction
89
Q

tRNA in translation

A
  • tRNA is a small single stranded RNA with a 3 lobed stercutre resembling a clover leaf
  • anti codonregion- contains a nucleotide triplets complementary to a codon on mRNA
  • amino acid attachment site - across from anticodon region where the appropriate amino acid bonds
  • aminoacyl tRNA- a tRNA bound to its particular amino acid
90
Q

3.initiation in translation

A
  • 1st tRNA is brought into the P(peptide) site carrying methionine since it corresponds to the start codon AUG
  • 2nd tRNA enters the A(acceptor) site
91
Q
  1. Elongation of polypeptide in translation
A
  • A peptide bond is formed between the 2 adj amino acids
  • the ribosomes translocates (shifts) one codon over to the 5’ to 3’ direction and the next tRNA enters the A site
  • process is repeated until a stop codon is reached ( a codon that does not code for an amino acid)
92
Q
  1. termination in translation
A
  • a release factor protein aids in dismantling the ribosome- mRNA complex releasing the polypeptide chain
  • polypeptide chain may undergo further processing
    • sugar (glycosylation) or phosphate(phosphorylation) may be added
    • polypeptyide may be cleaved
    • 2 or more polypeptides may combine together
93
Q

mutations

A

errors made in the DNA sequence that can be inheritied or caused by the environment

94
Q

silent mutation

A

A mutation that does not alter the resulting sequence

95
Q

missense mutation

A

a mutation that changes a single amino acid in the coding sequence

96
Q

nonsense mutation

A

a mutation that results in a premature stop codon

97
Q

framshift mutation

A

a sifdt in the reading frame resulting in multiple missense and/or nonsense effect

98
Q

point mutation

A

A change in a single nucleotide within a gene

99
Q

substitution

A

the replacement of one base pair in a DNA sequence by another base pair

100
Q

deletion

A

the removal of a base pair (small scale mutation) or larger coding regions(large scale mutation ) to a DNA sequence

101
Q

translocation

A

the movement of entire genes or sequences of DNA from one chromosome to another

102
Q

inversions

A

chromosomal segment that has reversed its orentation in the genome

103
Q

transporable elements

A

jumping genes that can move from one location to another n the genome

104
Q

proteins are produced in your body on the basis of

A

supply and demand

105
Q

different proteins are needed at

A

different stages of your life, different time of the day and different cells of the body

106
Q

all cells in your body contain the information (gene) for producing hemoglobin but

A

not all cells activate this gene

107
Q

the LAC operon

A
  • Ecoli produces the enzyme B-galactoidse, responsible for breaking down lactose only when latose is present
  • the gene for B-galactosidase is part of an operon
  • if lactose is not avaliable for the E.coli a repressor protein binds to the operator region making it impossible for RNA polymerase to bind to the promoter
  • if lactose is present in some E.colis environment the cell will take up the lactose and some will bind to the repressor. xausin a conformational change in the repressor resulting in it detavching from the operator site- lactose acts as an inducer in this case
108
Q

operon

A

a cluster of genes under the control of one promoter and one operator

109
Q

operator

A

the region in the operon that regulatory factors

110
Q

corresporessor

A

a signal molecule that binds to reduce the expression of an operons gene

111
Q

repressor

A

a protein that binds to the operator to repress gene transcription

112
Q

inducer

A

a signal molecule that triggers the expression of an operons gene

113
Q

the TRP operon

A
  • Trpytophan is an amino acid that is used by E.coli cells for the production of protein
  • if concentration of tryptophan is high the genes for this amino acid production are no longer transcribed
  • with high levels of tryptophan it binds it to TRP repressor protein , altering it’s shape so that it can now bind to the TRP operator
  • Tryptophan is required to inactivate the trp-operon corepressor
114
Q

gene expression regulation in eukaryotic cells

A

pretranscriptional control
transcriptional control
posttranscriptional control
translational control
post translational control

115
Q

pre transcriptional control

A
  • cells control the extent for which DNA is exposed to transcriptional enzymes 9the more condensed regions of DNA are not transcribed)
116
Q

transcriptional control

A

the cells control whether or not exposed DNA is transcribed - involves regulatory proteins

117
Q

post transcriptional control

A

cells controls the rates of processing of pre-mRNA into finished mRNA
- accomplished by not adding cap or tail

118
Q

translational control

A

regulatory proteins can bind to 5’cap of the molecule preventing the small ribosomal subunit from altering to the mRNA

119
Q

post translational control

A

regulatory proteins may break down the polypeptide before ity is able to reach its destination

120
Q

during prophase of meiosis homologous chromosomes

A

pair up forming a tetrad

121
Q

crossing over is

A

the exchange of DNA between the 4 chromatids of a tetrad

122
Q

linkage

A
  • genes that are part of the same chromosome are said to be linked
  • the degree of crossing over between 2 genes on a single chromosome is proportionate to the distance between them
123
Q

mapping

A
  • the frequency of crossovers in offspring can be used to map the relative position of the genes on the chromosome
  • this method is called recombination mapping
  • distance is measured in map units (m.u.)
124
Q

the highest possible recombination frequency is

A

50%

125
Q

if recombination occurs twice between 2 genes

A

then there is no recombination in the offsprings(only odd num of crossovers will show up in offspring)

126
Q

HIV 1

A

first indentifed in 19812, evolved in chimps and jumped species to humans

127
Q

HIV 2

A

first identified in 1985, less prevalent, less virulent form

128
Q

a virus must be able to

A

make it past the bodys defense mechanisms

129
Q

HIV is unique as it

A

invades the cell whose job it is to protect the body

130
Q

with HIV you become

A

susbsceptible to infections our immune system could normally fight off

131
Q

HIV infects what cell

A

helper T-cell

132
Q

HIV is a what

A

Retrovirus - genetic information is in the form of RNA instrsad of DNA

133
Q

retrovirus encodes a special virus called

A

reverse transcriptase

134
Q

reverse transcriptase

A

coverts RNA into a complementary strand of DNA

135
Q

HIV is very good at

A

mutating to avoid immune detection, it does this by changing the protein on the outside of its memebrane

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