molecular basis of inheritance Flashcards
1
Q
2 types of nucleic acids found in the living world
A
deoxyribonucleic acid (DNA) & ribonucleic acid (RNA)
2
Q
nucleotide
A
A nucleotide consists of a sugar molecule (either ribose in RNA or deoxyribose in DNA) attached to a phosphate group and a nitrogen-containing base.
3
Q
DNA
A
its a long chain of deoxyribonucleotides, its length is defined by the no. of nucleotides present in it
4
Q
nitrogenous bases
A
purines(Adenine and guanine) and pyrimidines(Thymine, Cytosine, Uracil)
5
Q
cytosine -
uracil -
thymine-
A
both in DNA and RNA
RNA
DNA
6
Q
Nucleoside
A
sugar+base
7
Q
A bonds with-
G bond with-
A
T
C
8
Q
phosphate group linkage
A
it will come and link with the 5’c or 3’c of the sugar through a phosphodiester bond to form nucleotides
9
Q
pyrimidines
rings
purines
A
1 ring structure
2 rings
10
Q
DNA direction
A
DNA is composed of two strands of nucleotides held together by hydrogen bonding. The strands each run from 5’ to 3’ and run in antiparallel, or opposite, directions from one another.
11
Q
the bases are paired through
A
A-T 2H BONDS
G-C 3H BONDS
12
Q
stability of the DNA helix structure
A
Two factors are mainly responsible for the stability of the DNA double helix: base pairing between complementary strands and stacking between adjacent bases
13
Q
why the distance between the 2 chains always constant
A
because there are hydrogen bonds between the 2 nitrogen bases of the 2 strands which avoid movement
14
Q
e coli bacteria DNA packing
A
they don’t have a defined nucleus but the DNA is not scattered throughout the cell , it’s held by proteins inside the nucleoid
15
Q
eukaryotic cell packaging protein
A
set of a + charge basic proteins called histones
16
Q
how does a protein acquire its charge
A
depending upon the number of amino acids residual with charged side chains, histones are rich in the basic amino acid lysine and arginine
17
Q
nucleosome
A
the - charged DNA will wrap around the + 8 unit molecule called histone octamer to form a nucleosome
18
Q
chromatin repeating unit
A
nucleosome
19
Q
packing of a nucleosome in chromatin
A
they are packed to form chromatin fibres that are further coiled and condensed at the metaphase stage of cell division to form chromosomes
20
Q
higher level of packing of chromatins
A
we need an additional set of proteins called NON-HISTONE CHROMOSOMAL PROTEINS
21
Q
different regions of chromatin in the nucleus
A
- loosely packed and stain light called euchromatin, it’s the active chromatin
- dense and stain dark called heterochromatin, inactive
22
Q
transforming principle
A
The transforming principle was an early name for DNA
23
Q
Frederick Griffith worked with
A
streptococcus pneumoniae (bacteria responsible for pneumonia)
24
Q
2 types of streptococcus pneumoniae
A
S strain - produce smooth shiny colonies, have a polysaccharide coat, infectious (virulent)
R strain- produce rough colonies, no coating, (inverulent)
25
injection of S strain and R strain
when S is injected into mice, they get pneumonia and die, when R is injected they remain alive
26
what did Griffith do to the s strain
he heated the s strain to extreme temperatures which killed the virulent strain and made it ineffective
27
injecting heat-killed s strain
the mice survived
28
what happens when the heat-killed s strain is mixed with R strain and why?
the mice die and he recovered living S strain from the dead mice, this happened because he assumed that the genetic material of s stain was passed into the r strain which enabled the r strain to synthesis smooth polysaccharide coat and become virulent
29
Avery, MacLeod and McCarty
they worked to determine the transforming principle in Griffiths exp
30
what did the 3 scientists do
they mixed every component of the S strain with the R strain to see which one could transfer R cells into a virulent one (s cells), only one of those components could turn the r cells which had the same composition as that of DNA, diff enzymes like protease and RNAases did not affect transformation , and only DNAases could react and give us DNA
31
the conclusion from the transforming principle experiment
the transforming principle is the DNA
32
the Hershey chase exp requirements
```
culture medium
2 radioactive isotopes
mixer blender
centrifuge tube
bacteriophage
bacterial strain
```
33
bacteriophage
the virus which affects bacteria by inserting their genetic material into them and replicate themselves using the replication machinery of the host bacterium
34
radiolabelled bacteriophage
one strain had the radiolabelled proteins
| and the other had radiolabeled DNA
35
radiolabelled bacteriophages infect the bacteria ...............
analyze the bacteria to know which radiolabelled component made its way into it
36
obtaining the radiolabelled bacteriophage strains
in one nutrient medium, they added the radioactive isotope sulphur-35
in the other, they added phosphorus-32
37
why sulphur-35 & phosphorus-32?
sulphur is an integral part of protein and phosphorus is present in DNA
38
bacteria were made to grow in both the mediums
S-35 AND P-32 WERE INCORPORATED INTO THEIR BODIES
39
Radioactive bacteria + bacteriophage
S-35 AND P-32 WERE INCORPORATED INTO THEIR protein coats and radioactive DNA respectively
40
the new bacteriophages infected new bacteria
and then agitated in a mixer blender to release the empty phage coats attached to the bacterial cells
41
centrifugation of the mix from the blender
we obtain 2 layers; supernatant (lighter particles) pellet (heavy)
42
case 1 - if the radioactive component is present in the supernatant
it means that the protein component S-35 has not been transformed into the bacteria as a result the protein is not the genetic material
43
case2- if the radioactive component is present in the pellet
it means that all the new phage particles have got the p-32 in them, DNA got transferred and DNA IS THE GENETIC MATERIAL
44
example of viruses who have RNA as genetic material
Tobacco mosaic viruses, QB bacteriophage
45
A molecule that acts as a genetic material must fulfil
REPLICATION
chemically and structurally stable
mutation
express itself in the form of mendelian characters
46
DNA stability eg:
in Griffiths exp, they heat-killed the strain but at least did not destroy the genetic material, which shows how stable it is
47
2'-OH group at every nucleotide
its a very reactive group which makes RNA very labile and easily degraded
48
reactivity of DNA and RNA
RNA - catalytic; reactive but not stable
| DNA -more stable because of the presence of thymine in place of uracil, less reactive
49
RNA and DNA mutation
RNA being unstable can mutate faster, viruses with RNA have a short life span so they mutate and evolve faster
50
synthesis of proteins
RNA can directly code for the synthesis of proteins and can easily express characters .DNA is dependent on RNA for protein synthesis
51
function of rna
translation, the transmission of genetic material, metabolism, act as a catalyst
52
replication of DNA
semiconservative DNA replication
53
semiconservative DNA replication
2 strands would separate and act as a template for the synthesis of a new complementary strand, after replication, each DNA would have one parental and one newly synthesised strand
54
semiconservative DNA replication was first show in
Escherichia coli
55
meselson and stahl exp
| 1st step
They began by growing E. coli in a medium containing a "heavy" isotope of nitrogen (15N), the result was 15N was incorporated into newly synthesised DNA, which produced a single band after centrifugation in cesium chloride. This result made sense because the DNA should have contained only heavy 15 N at that time.
56
why cesium chloride
because it created a density gradient
57
meselson and stahl exp
| 1 GEN
Then, the bacteria were switched to a medium containing a "light"14 N isotope and allowed to grow for several generations. whuch produced a single band when centrifuged. However, this band was higher, intermediate in density between the heavy DNA and the light DNA
58
meselson and stahl exp
| 2 GEN
When second-generation DNA was centrifuged, it produced two bands. One was in the same position as the intermediate band from the first generation, while the second was higher (appeared to be labelled only with 14N
59
meselson and stahl exp
| 3 AND 4 GEN
over the third and fourth generations, we'd expect the hybrid band to become progressively fainter (because it would represent a smaller fraction of the total DNA) and the light band to become progressively stronger (because it would represent a larger fraction)
60
DNA polymerase
"the builder", replicates DNA molecules to build a new strand of DNA, it uses DNA as a template to catalyse the polymerisation of deoxynucleotides
61
the place where the replication originates
origin of replication
62
deoxyribonucleoside triphosphate
act as substrates and serve as the energy source for the reaction, the terminal phos[ates breakdown to provide energy as ATP
63
replication fork
for long DNA molecules, the 2 strands of DNA cannot separate in their entire length, the replication occurs within a small opening in the helix
64
polymerase catalyses polymerisation in only one direction :)(5to3)
on one strand (3to5) replication is continuous
the other strand (5to3) replication is discontinous, which forms fragments that are later joined by the enzyme DNA ligase
65
transcription
The process of copying genetic information from one
strand of the DNA into RNA is termed as
transcription
66
complementary base pairing in transcription
adenosine pairs with uracil
67
Why both the strands are not copied during transcription
1. if both strands act as templates 2 types of proteins will be produced, one with the correct sequence and the other with the reverse one
2. the 2 rna molecules produced would be complementary to each other and form a double-stranded rna, it will not be translated to protein and the process of transcription will get nullified
68
transcription unit in DNA
it has 3 regions 1.promoter, 2.the structural gene, 3. terminator
69
template strand
Since the two strands have the opposite polarity and the DNA-dependent RNA polymerase also catalyse the
polymerisation in only one direction, that is, 5'→3', the strand that has the polarity 3'→5' acts as a template
70
coding strand
has 5'→3 polarity which is the same as RNA except for uracil in place of thymine
71
promoter
starting site of transcription provides a binding site for rna polymerase which runs along the DNA
72
location of promoter and terminator with respect to the coding strand
the promoter is present towards 5' of the coding strand (upstream) and the terminator towards 3' of the coding strand (downstream)
73
structural gene
it can be defined in terms of a cistron or a segment of DNA that codes for a polypeptide
74
2 types of the structural gene
monocistronic and polycistronic
75
monocistronic
contains genetic material to translate only a single protein and its found in eukaryotes
76
polycistronic
caries information for several genes and translate into several proteins, prokaryotes
77
split gene arrangement
it is present in monocistronic genes, where sequences like exons and introns are present
78
exons
expressed sequences, they appear in mature rna, they are interrupted by introns
79
introns
do not appear in mature rna, by the process of transcription they are removed and joins exons as mrna
80
function of exons
it is the exons in the structural gene along with the promoter that determines the inheritance of a character
81
regulatory sequences in s.g
they regulate the functions of other genes but do not code for any rna or proteins they are also known as regulatory genes
82
terminator
defines the end of the process of transcription
83
transcription in prokaryotes
in prokaryotes, there are no cytoplasmic separations, both transcription and translation can happen at the same time
84
rna polymerase enzyme in bacteria
there's only 1 DNA dependent rna polymerase enzyme that catalyzes all 3 types of rna
85
3 types of rna
mRNA (messenger RNA),
| tRNA (transfer RNA), and rRNA (ribosomal RNA).
86
mrna
it carries the genetic information to the sites of protein synthesis, it helps to put together amino acids to make protein
87
trna
it caries each amino acid to the ribosome according to the coded msg in mrna
88
rRNA
provides a mechanism to decode mRNA into amino acids
89
3 stages of transcription of prokaryotes
1. intiation
2. chain elongation
3. termination
90
initiation
rna polymerase along with the initiation factor(sigma )
| binds to the promoter which opens the DNA helix and separates the 2 strands
91
chain elongation
here the rna polymerase builds a rna strand by using DNA as a template
92
rna polymerase uses _____
uses nucleoside triphosphates as substrates and polymerises using the law of complementarity
93
what happens after chain elongation
the sigma factor dissociates from the rna polymerase and can me reused
94
termination
here the polymerase along with the termination factor RHO reaches the termination region and the newly created mrna falls off along with the enzyme
95
how many rna polymerases are required in prokaryotic and eukaryotic transcription
p-1
| E-3
96
RNA POLYMERASE 1
it transcribes various rRNAs: 28s,18s, 5.8s
97
RNA POLYMERASE 2
transcribes precursor of mRNA, the
| heterogeneous nuclear RNA (hnRNA).
98
RNA POLYMERASE 3
transcription of tRNA, 5srRNA, and snRNAs (small nuclear RNAs).
99
mrna in E and P
in P it doesn't require any processing, in E the hRNA is inactive
100
hRNA
it has split gene arrangement with exons and introns it has to undergo different processes like splicing, capping and tailing
101
capping
In capping an unusual nucleotide (methyl guanosine triphosphate) is added to the 5'-end of hnRNA. it helps mrna to bind to small ribosomal units during protein synthesis
102
splicing
where the introns are removed
| and exons are joined in a defined order.
103
tailing
```
adenylate residues (200-300) are
added at 3'-end in a template-independent giving rise to a POLY A TAIL, this process is also known as POLYADENLATION
```
104
what happens after splicing, capping and tailing
It is the fully processed hnRNA, now called mRNA, that is transported out of the nucleus for translation
105
George Gamow
he said that there are only 4 bases and if they have to code for 20 amino acids, the code should constitute a combination of bases
106
codons
combination of 3 bases in the mRNA
107
stop codons
do not code for any amino acid: UAA, UAG, UGA
108
degenerate
Some amino acids are coded by more than one codon, hence
| the code is degenerate
109
codons are universal
from bacteria to human
| UUU would code for Phenylalanine (phe).
110
AUG
. It codes for Methionine (met), and it
| also, act as initiator codon
111
point mutation
point mutation is a change of single base pair in the gene for the beta-globin chain that results in the change of amino acid residue glutamate to valine.
It results into a diseased condition called sickle cell anaemia
112
frameshift mutation
insertion or deletion because they shift every codon
113
adapter molecule
trna has an anticodon loop that has bases complementary to the code, and it also has an amino acid acceptor end to which it binds to amino acids
114
initiator tRNA
For initiation, there is another specific tRNA
| There are no tRNAs for stop codons
115
translation
process of polymerisation of amino acids to form a polypeptide
116
use of mrna in translation
has the message that contains the order and sequence of amino acids
117
amino acids are joined by which bond?
peptide bond
118
activation of amino acids
binding of ATP to the amino acids results in the trna and amino acid binding together
119
binding of amino acid to tRNA
charging of tRNA or aminoacylation of tRNA
120
The cellular factory responsible for
| synthesising proteins
RIBOSOME, made of rRNAs, has 2 subunits: large subunit and small subunit
121
small subunit
it binds to the mRNA as a result, the process of
translation of the mRNA to protein
begins
122
large subunit
the tRNAs carrying amino acid reaches the large subunit and a peptide bond is formed between the amino acids
123
start codon
signals the ribosome to start synthesizing the protein and recognised only by the initiator tRNA
124
stop codon
signals the ribosome to stop synthesizing the protein
125
release factor
a release factor binds to the stop codon, terminating translation and releasing the complete polypeptide from the ribosome.
126
untranslated regions (UTR)
An mRNA also has some additional sequences that are not translated,The UTRs are present
at both 5' -end (before start codon) and at 3' -end (after stop codon). They
are required for efficient translation process
127
DNA fingerprinting
DNA fingerprinting involves identifying differences in some specific
regions in DNA sequence called as repetitive DNA, because in these
sequences, a small stretch of DNA is repeated many times
128
bulk dna
its the 99.9 % of the dna in which all nucleotide bases are exactly the same in all people
129
satellite DNA
it is the repetitive DNA, non-coding sequences, that have a high degree of polymorphism and form
the basis of DNA fingerprinting
130
polymorphism
gentetic varaition
131
uses of DNA fingerprinting
fronesic applications and paternity testing, in case of disputes.
132
VNTR or loci
(i) isolation of DNA,
(ii) digestion of DNA by restriction endonucleases,
(iii) separation of DNA fragments by electrophoresis,
(iv) transferring of separated DNA fragments to synthetic
membranes, such as nitrocellulose or nylon,
(v) hybridisation using labelled VNTR probe(fragment), and
(vi) detection of hybridised DNA fragments by autoradiography
133
what is hybridisation using labelled VNTR probe
a hybridization probe is a fragment of DNA which can be radioactively labelled. It can then be used in DNA samples to detect the presence of nucleotide substances that are complementary to the sequence in the probe
134
what happens after autoradiography
it results in an autoradiogram that gives many bands of differing sizes. These bands give a
characteristic pattern for an individual DNA It differs from individual to individual except in the case of twins
135
PCR
many copies of dna can be made - polymerase chain reaction
136
gene regulation
regulation of the gene to perform a particular or a set of functions and which portion of the DNA should be used
