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CYTOGENETICS - SECOND SEM > LESSON 2 - PRELIM > Flashcards

LESSON 2 - PRELIM Flashcards

1
Q

heredity

A

Gregor Mendel

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2
Q

flies, linkage

A

Thomas Hunt Morgan

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3
Q

1928: transformation and mice

A

Frederick Griffith

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4
Q

1944: DNA as the transforming agent

A

Oswald Theodore Avery, Colin MacLeod and Maclyn
McCarty

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5
Q

late 40’s-early 50’s: base pairing=AT
CG

A

Erwin Chargaff

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6
Q

(1952: DNA is not a
protein

A

Alfred Hershey-Martha Chase

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7
Q

1953: chemical structure of DNA –
secondary structure: double-helix

A

Watson and Crick

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8
Q

mid 1950’s: DNA Replication details:
semi-conservative replication model

A

Meselson-Stahl

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9
Q

__________ in 1869
 Isolated what he called nuclein from the nuclei of pus
cells
 Nuclein was shown to have acidic properties, hence
it became called nucleic acid

A

Friedrich Miescher

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10
Q

structure in the cell nucleus which is the visible
carrier of genetic information

A

Chromosomes

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11
Q

 portion of a chromosome that controlled a specific
inheritable trait

A

Genes

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12
Q

 carries information which directs the process of
protein synthesis
 within the nucleic acids are the codes needed for
transcription and translation of proteins

A

Nucleic Acids

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13
Q

________ (entire set of genes of an organism) size is based
on number of nucleotide pairs present

A

Genome

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14
Q

Among eukaryotes, there is no
consistent relationship on the C-value (DNA content of
the genome) and the metabolic, developmental, or
behavioural complexity of the organism

A

C-value paradox

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15
Q

Within the nucleus, __________are located (as pairs)

A

chromosomes

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16
Q

___________ are tied together (by protein centromere)

A

Chromosomes

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17
Q

(Telomeres)

A

Ends of the chromosome

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18
Q

(genes are specific
portions of chromosome coding for a protein which
functions in various phase)

A

Within the chromosome are genes

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19
Q

(DNA: double-helix
molecule containing base pair)

A

In the genes are nucleic acids

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20
Q

Composition of Nucleic Acids

Nucleic Acids (repeating series of nucleotide)

A

Polymers (polynucleotides)

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21
Q

 Parts of Nucleotide

A

 A five-membered ring monosaccharide
 A nitrogen-containing cyclic compound (nitrogenous
bases)
 A phosphate group

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22
Q

 Types of Nucleic Acids

A

 DNA (genetic material – doesn’t function without
RNA)
 RNA

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23
Q

Sugars

A

2-deoxyribose (for DNA)
o 5th carbon – phosphate group
o 3rd – next nucleotide attached
 Ribose (for RNA)
o 2nd carbon - oxygen

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24
Q

 Nitrogenous Base

A

Purines (2)
o Contains two-fused N-containing ring
 Adenine (A)
 Guanine (G)

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25
Q

 Pyrimidines (3)

A

o Has one nitrogen-containing ring
 Cytosine (C)
 Thymine (T)
 Uracil (U)

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26
Q

 Sugar + Base =

A

Nucleoside

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27
Q

Adenine + (deoxy)ribose =

A

(deoxy)adenosine

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28
Q

Guanine + (deoxy)ribose =

A

(deoxy)guanosine

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29
Q

Cytosine + (deoxy)ribose =

A

(deoxy)cytidine

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30
Q

Thymine + deoxyribose =

A

deoxythymidine

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31
Q

Uracil + ribose

A

uridine

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32
Q

nucleoside formed after combination of
adenine with ribose

A

Adenosine

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33
Q

NUCLEOSIDE + Phosphate =

A

NUCLEOTIDE (Tide labada)

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34
Q

Are the building blocks of nucleic acids
 Monomers of the DNA and RNA polymers
 Is a 5’-monophoshpate ester of a nucleoside
 Are named by adding 5’-monophosphate at the end
of the name of the nucleoside

A

NUCLEOTIDE

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35
Q

Can add additional phosphate groups to form
diphosphate or triphosphate esters (necessary to
produce energy needed in transcription, translation
and repli)

A

Nucleotides

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36
Q

Bases Deoxyribonucleosides Deoxyribonucleotides

Adenine (A)

A

Deoxyadenosine Deoxyadenosine 5’-

Monophosphate
(dAMP)

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37
Q

Bases Deoxyribonucleosides Deoxyribonucleotides

Guanine
(G)

A

Deoxyguanosine Deoxyguanosine 5’-
Monophosphate
(dGMP)

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38
Q

Bases Deoxyribonucleosides Deoxyribonucleotides

Cytosine
(C)

A

Deoxycytidine Deoxycytidine 5’-
Monophosphate
(dCMP)

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39
Q

Bases Deoxyribonucleosides Deoxyribonucleotides

Thymine
(T)

A

Deoxythymidine Deoxythymidine 5’-
Monophosphate
(dTMP)

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40
Q

Bases
Ribonucleosides Ribonucleotides

Adenine (A)

A

Adenine (A) Adenosine Adenosine 5’-
Monophosphate
(AMP)

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41
Q

Bases
Ribonucleosides Ribonucleotides

Guanine (G)

A

Guanine (G) Guanosine Guanosine 5’-
Monophosphate
(GMP)

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42
Q

Bases
Ribonucleosides Ribonucleotides

Cytosine (C)

A

Cytosine (C) Cytidne Cytidine 5’-
Monophosphate
(CMP)

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43
Q

Bases
Ribonucleosides Ribonucleotides

Uracil (U)

A

Uracil (U) Uridine Uridine 5’-
Monophosphate
(UMP)

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44
Q

the repeating sequence of nucleotides form its
primary structure (forming alternating ribose and
phosphate backbone – providing structural stability)

A

Primary Structure (from polymerization of monomers)

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45
Q

Based on:
o Chargaff rule

A

 Secondary Structure

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46
Q

Obtained by Rosalind Franklin and Maurice
Wilkins
 Diagonal image: helical structure of DNA

A

X-ray diffraction photographs

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47
Q

A, T, G, and C (complimentary) are present
in equimolar quantities (refers to similarity in
molar concentration in DNA hydrolysis;
equal concentration)
 If this 2 molecules are placed together, they
form hydrogen bonds
 Similar molar concentration upon DNA
hydrolysis

A

Chargaff rule

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48
Q

o The 2 (single strand of DNA) polynucleotide
chains run in opposite directions
o One 5’ – OH and one 3’ – OH terminal
o Bases are hydrophobic (non polar, tucked
inside)
o Sugar phosphate backbone is hydrophilic (polar,
exposed to environment)

A

Double helix

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49
Q

Basic protein to w/c the DNA is coiled around

A

 Higher Structure

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50
Q

Further arrangement of DNA in order to organize
them in the chromosomes

A

Nucleosome

51
Q

11 base pairs before helix rotates

A

A form

52
Q

10 base pairs before helix rotates

A

B form

53
Q

_____ is rarest and is only obtained in experiments

A

Z form

54
Q

________ was in
B form (common structure of DNA, followed by A)

A

X-ray photograph (Franklin and Wilkins) of DNA

55
Q

3 forms of the helical structure of DNA: _____, _____, _______

A

A, B, Z

56
Q

12 base pairs before helix rotates (glycosidic
bonds are anti and syn)

A

Z form

57
Q

 Single strand nucleic acids
 Usually located outside the nucleus (but made inside the
nucleus and transported to cytoplasm to do its function)
 The important intermediary player in the central dogma
 The only genetic material of viruses (viruses are
classified as non-living things because it only has one
genetic material)

A

Ribonucleic Acids

58
Q

Three Types of Ribonucleic Acids

A

Messenger RNA (mRNA)
Ribosomal RNA (rRNA)
Transfer RNA (tRNA)

59
Q

Codes for protein

A

Messenger RNA (mRNA)

60
Q

Forms the core of the ribosomes
 Machinery for making proteins

A

Ribosomal RNA (rRNA)

61
Q

Matches code for amino acid on mRNA and position
the right amino acid in place during protein synthesis
 Transfers free amino acids to the polypeptide chain

A

Transfer RNA (tRNA)

62
Q

 RNA with enzymatic properties
 Functions in mRNA splicing

A

 Ribozyme

63
Q

Types of nucleic acid

A

DNA and RNA

64
Q

 Carry the genetic information from the DNA in the
nucleus directly to the cytoplasm

A

Messenger RNA (mRNA)

65
Q

Contains 73 to 93 nucleotides per chain
 Can carry a single type of amino acid
 Every amino acid have one tRNA carrier
 There is at least one different tRNA for each of the
20 amino acids
 Transports amino acids to the site of protein
synthesis in the ribosomes

A

 Transfer RNA (tRNA)

66
Q

 Complementary to the codon present on the mRNA
(GCC)
 Corresponds to the amino acid alanine
 Amino acid is connected to the 3’ end of transfer RNA

A

Antiocodon: CGG

67
Q

Structural formula of tRNA:
Yello and Blue

A
67
Q

Structural formula of tRNA:

A

Yellow and Blue

68
Q

Structural formula of tRNA:

nitrogenous
bases

A

Yellow

69
Q

Structural formula of tRNA:

sugar
phosphate backbone

A

Blue

70
Q

_________
complementary base pairs of mRNA
 3 nitrogenous base pairs present at the end
of tRNA is complementary to the codon
present in mRNA
 Each anticodon corresponds to a specific
type of amino acid carried by the tRNA

A

Anticodon arm

71
Q

 RNA that is complexed with proteins in ribosomes
 Main component of ribosome
 Complex machinery that is the site of protein
synthesis

A

 Ribosomal RNA (rRNA)

72
Q

 2 subunits
of rRNA
________ catalyzes the peptide bond formation

___________ – binds mRNA and tRNA

A

1 large and 1 small

73
Q

_________RNA with enzymatic capabilities
 Catalytic RNA – they can catalyse different reactions
or initiate different reactions specifically splicing of
mRNA
 Catalyse the splicing of mRNA – refers to the
process of removing unnecessary parts of the
mRNA for it to become more efficient in protein
synthesis

A

Ribozyme

74
Q

_______ where mRNA will bind

A

Small subunit

75
Q

_______ amino acids will combine in order to
form primary structure of proteins

A

Large subunit

76
Q

_______ (unit used to determine the sedimentation
rate of the different molecules or compound

A

s = Svedberg unit

77
Q

 Coding sequence
 Expressed sequence
 Portion in the mRNA that codes for a specific amino
acid to make protein

A

 Exons

78
Q

Most important parts of tRNA

A

anticodon arm and acceptor arm

79
Q

________ they can catalyse different reactions
or initiate different reactions specifically splicing of
mRNA

A

Catalytic RNA

80
Q

________ refers to the
process of removing unnecessary parts of the
mRNA for it to become more efficient in protein
synthesis

A

Catalyse the splicing of mRNA

81
Q

Noncoding sequence
 Part in the RNA that has no purpose; do not code for
proteins
 Intervening sequence
 Where DNA analysis happens
- Identifies specific identity in a person using DNA
(to single out an individual they use introns)
- Because every individual have their own unique
sequence of introns

A

Introns

82
Q

Occurs in the nucleus (to protect mRNA)
 Information encoded in a DNA molecule is copied
into an mRNA molecule

A

Transcription

83
Q

Information encoded in an mRNA molecule is used
to assemble a specific protein

A

Translation

84
Q

acts as a “manager” in the process of making
proteins

A

DNA

85
Q

Means to produce molecules that have the same base
seuquence
 To distribute the DNA of the parent cells to its daughter
calls (cells die eventually – It needs to be passed to code
proteins)
 Process that ensures the stability of an organism
 Producing two identical replicas of DNA

A

DNA Replication

86
Q

cell is metabolically active

A

G1 Phase

87
Q

DNA Replication (8 hours for normal
somatic cells: body cells

A

S Phase

88
Q

cell growth continues

A

G2 Phase

89
Q

where the cell will divide

A

Mitotic phase

90
Q

– phosphate is connected

A

5th carbon

91
Q

2 H bonds

A

Adenine and Thymine

92
Q

3 H bonds

A

Guanine and Cytosine

93
Q

DNA Replication Models

A

 Semiconservative Replication
 Conservative Replication
 Dispersive Replication

94
Q

 DNA Replication would create two molecules
 Each of them would be a complex of an old
(parental) and a daughter strand
 Newly formed molecules is composed of 1 strand
from parent and 1 strand from daughter.

A

 Semiconservative Replication

95
Q

DNA Replication process would create a brand new
DNA double helix made of two daughter strands
while the parental chains would stay together

A

 Conservative Replication

96
Q

Replication process would create two DNA doublechains, each of them with parts of both parent and
daughter molecules

A

 Dispersive Replication

97
Q

nitrogen weighing 15 amu

A

N15

98
Q

nitrogen weighing 14 amu

A

N14

99
Q

Meselsohn and Stahl Experiment 2 isotopes used

A

N15 and N14

100
Q

Only one replication origin is
needed that is because the
chromosomes of prokaryotes are
simple

A

Prokaryotes

101
Q

Multiple replication origins are
needed because the chromosomes
of eukaryotes are way more
complex than prokaryotes

A

Eukaryotic chromosomes have many
bubbles
-

102
Q

Prokaryote specifically bacteria contain
extrachromosomal DNA which are called _______.

A

plasmids

103
Q

Prokaryote specifically bacteria contain
extrachromosomal DNA

A

Rolling Circle Replication

104
Q

There are 2 so called origin of the plasmid replication

A

Single Stranded Origin and Double Stranded Origin

105
Q

Present on the separated DNA strand and
while the new DNA is being made for the
separated DNA strand the DNA template is
single stranded thus making it single
stranded

A

Single Stranded Origin

106
Q

At this point the DNA is double stranded
before a new DNA strand was made

A

Double Stranded Origin

107
Q

Unwinds the DNA double helix
- Helicase will cut the paired DNA strands.
Helicase is the equivalent of the UVR in the
Rolling Circle Replication

A

Helicase

108
Q

Prevents supercoiling; relaxes the part of
the DNA that is not yet separated.

A

Topoisomerase

109
Q

Breaks one DNA strand
and will connect another
strand to became a very
loose strand in a part of the
DNA.
- Although the DNA strand is
loose and can be coiled
again it will not be
supercoiled unlike before
- Prevents supercoiling in
the DNA strand

A

TYPE I Topoisomerase

110
Q

Breaks double strands of
the DNA and pass another
loop over it
- Prevent supercoiling within
2 DNA pairs

A

TYPE II Topoisomerase

111
Q

Synthesize short oligonucleotides (primers)

A

Primase

112
Q

A.k.a. Processivity clamps allows the
leading strand to be threaded through.
- At the same time makes the process in the
leading strand efficient. This clamp protein
helps keep the replication protein in place.

A

Clamp Protein (PCNA/Proliferating Cell Nuclear
Antigen for Eukaryote

113
Q

Joins the assembled nucleotides in order to
from nucleic acids

A

DNA polymerase

114
Q

DNA Polymerase enzymatic activity (3)

A

Polymerase
(2) Exonuclease
Endonuclease

115
Q

Polymerizing the new DNA
strand by adding
nucleotides

A

Polymerase

116
Q

Break the sugar-phosphate
backbone in the end of a
nucleotide strand
- Proof reading capacity of
DNA polymerase.
- Nucleotides removed from
the ends

A

Exonuclease

117
Q

Remove nucleotide from
the middle nucleotide
strand
- Internal cuts

A

Endonuclease

118
Q

Joins Okazaki fragments in the lagging
strand

A

Ligase

119
Q

The end-replication problem
Loss of DNA in each eukaryotic
replication cycle because of primer
overhangs
- The answer to this problem are
Telomeres

A

Termination

120
Q

Regions of repetitive DNA close to
the ends and help prevent loss of
genes due to this shortening
- G-C rich

A

Telomeres

121
Q

Binds and stabilizes the double -
stranded telomeric DNA
- Helps the overhangs to form
protective loops

A

Telomeric repeat-binding factor

122
Q

Enzyme normally present on stem cells
and germ cells which catalyzes the
formation of telomeres

A

Telomerase

123
Q

The number of cell division an
organism can make
- Caused by the limited and
consumable presence of telomeres
on somatic cells

A

Hayflick Limit

CYTOGENETICS - SECOND SEM (9 decks)

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