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Molecular Basis of Inheritance Flashcards

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1
Q

Length of DNA is defined as

A

The number of nucleotides or base pairs present in it.

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2
Q

Nucleotide base pairs in Bacteriophage ɸ 174

A

5386

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3
Q

Nucleotide base pairs in Bacteriophage lambda

A

48502

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4
Q

Haploid content of the Human DNA

A

3.3 × 10⁹

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5
Q

Basic Components of a Nucleotide (3)

A
  1. Pentose sugar (ribose - RNA, deoxyribose - DNA), phosphate group and nitrogenous base.
  2. N base - purines - Adenine and Guanine and Pyrimidines - Cytosine, Thymine and Uracil.
  3. DNA - thymine (5 methyl uracil) and RNA - Uracil
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6
Q

Linkages in a Nucleotide

A
  1. N base is linked to the OH of 1’ carbon of the pentose sugar by a N- glycosidic bond to form a nucleoside.
  2. The phosphate is then linked to the OH of the 5’ carbon of the nucleoside to form a nucleotide.
  3. Two nucleotides can be linked by a 3’ - 5’ phosphodiester linkage to form a dinucleotide.
  4. Multiple nucleotides can be linked in this way to form a polynucleotide.
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7
Q

What is the 5’ end of a polynucleotide?

A

Free phosphate moiety on the 5’ end of the sugar , at the end of a polynucleotide is called the 5’ end.

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8
Q

What is the 3’ end of a polynucleotide?

A

Free OH group on the 3’ carbon at the end of a polynucleotide is called the 3’ end.

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9
Q

What is the backbone of a polynucleotide?

A

Sugar and phosphates are the backbone, N bases project inside from the backbone.

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10
Q

How is RNA diff from DNA? (2)

A
  1. It is a ribose sugar, - additional OH’ at the 2’ carbon.

2. Uracil instead of thymine (thymine - 5 methyl uracil)

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11
Q

Who discovered DNA?

A
  1. Fredrich Meisher discovered an acidic substance in the nucleus which he named “nuclein” in 1869.
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12
Q

Why did it take so long to elucidate the structure of DNA?

A

Due to technical limitations in isolating such a long polymer.

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13
Q

Who came up with the double helix structure of DNA?

A

James Watson and Francis Crick, based on the X ray diffraction studies provided by Maurice Wilkins and Rosalind Franklin came up with the double helix structure in 1953. Main proposition was the base pairing between the two strands, this was based on Chargaff’s rule.

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14
Q

What is Chargaff’s rule?

A

Observation provided by Erwin Chargaff that for double stranded DNA, the ratio between Adenine and Thymine and Cytosine and Guanine is constant and equal to one. A + T = C + G.

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15
Q

What unique properties did base pairing confer to the polynucleotide chains?

A
  1. If the sequence of base pairs on one strand is known, the sequence on the other strand can easily be predicted.
  2. During DNA synthesis, the daughter DNA produced would be exactly identical to the parent strands because the parent strand would act as a template.
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16
Q

What is the polarity of the strands in a DNA molecule?

A

The two strands are anti parallel to each other i.e. if one strand is 3’ to 5’, then the other is 5’ to 3’.

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17
Q

How are the bases in the two strands paired?

A

The bases on the opposite strands are paired by hydrogen bonds. Adenine pairs with Thymine by 2 H bonds. Cytosine pairs with Guanine by 3 H bonds. This ensures a purine is always opposite to a pyrimidine, and there is approximately uniform distance between the two strands.

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18
Q

How are the two strands in the DNA coiled?

A

They are coiled in a right handed fashion.

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19
Q

What is the pitch of the DNA?

A

3.4nm

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20
Q

How many base pairs in each turn of the DNA?

A

10bp

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21
Q

What is the distance between bp in a DNA helix?

A

0.34nm (pitch = 3.4/ base pairs in one turn = 10)

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22
Q

What confers stability to the double helix?

A

The plane of the base pairs stacks one over the other and the H bonds confer stability to DNA.

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23
Q

What is the Central Dogma in molecular biology?

A

The Central Dogma in molecular biology was proposed by Francis Crick, according to which genetic information flows from DNA -> RNA -> proteins.

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24
Q

Length of DNA double helix in a typical mammalian cell is

A

2.2m

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25
Q

Length of DNA in E.coli is

A

1.36mm

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26
Q

Arrangement of DNA in prokaryotes is (4)

A
  1. Prokaryotes don’t have a well defined nucleus.
  2. But DNA is not scattered throughout the cell.
  3. Negatively charged DNA is held by positively charged proteins in a region called the nucleoid.
  4. DNA in the nucleoid is organized in large loops by proteins.
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27
Q

How do proteins acquire charge?

A

By an abundance of amino acid residues with charged side chains.

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28
Q

What are histones? (4)

A
  1. Histones positively charged basic proteins.
  2. They are rich in basic amino acid residues like lysine and arginine.
  3. They are organized to form a unit of 8 molecules called histone octamer.
  4. Negatively charged DNA wraps around the positively charged histone octamer to form a nucleosome.
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29
Q

What is a nucleosome? (3)

A
  1. A nucleosome - -vely charged DNA wraps around +vely charged histone octamer.
  2. 200bp of DNA in a nucleosome.
  3. Repeating units of nucleosomes form a structure in the nucleus called chromatin.
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30
Q

What is chromatin? (3)

A
  1. Chromatin is made up of nucleosomes.
  2. Chromatin is a thread like stained bodies seen in the nucleus.
  3. Under an electron microscope, nucleosomes in the chromatin are seen as a beads on a string structure.
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31
Q

Beads on a string structure of chromatin in packaged to form

A

Chromatin fibres.

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32
Q

Chromatin fibres are further condensed to form what?

A

During the metaphase of the cell cycle, chromatin fibres condense and coil to form chromosomes.

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33
Q

Packaging of chromatin at a higher level requires

A

An additional set of proteins called Non Histone Chromosomal (NHC) proteins.

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34
Q

Euchromatin

A

Loosely packed, lightly stained transcriptionally active chromatin.

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35
Q

Heterochromatin

A

Densely packed, darkly stained transcriptionally inactive chromatin.

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36
Q

By 1926, the quest to determine the genetic material had reached the

A

molecular level. (chromosomes located in most cells)

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37
Q

What year did Fredrick Griffith perform his transforming principle experiments?

A

1928

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38
Q

Which bacteria did Griffith use for his experiments?

A

Streptococcus pneumoniae (pneumococcus)

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39
Q

How do S train and R strain of Streptococcus pneumoniae differ?

A

S (smooth strain) is the virulent strain because it has a mucous (polysaccharide coat).
R (rough) strain does not have a polysaccharide coat and is not virulent.

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40
Q

What happens to the mice in Griffith’s experiments?

A

Mice infected with S strain die.
Mice infected with R strain survive.
Mice infected with heat killed S strain survive.
Mice infected with heat killed S strain + R strain die.

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41
Q

What was the conclusion of Griffith’s experiment?

A

The R strain had been transformed by the heat killed S strain. The transforming principle had enabled the R strain to produce a polysaccharide coat and become virulent. (Transforming principle was stable enough to resist the heat that killed the bacteria.)

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42
Q

Drawback of Griffith’s experiment?

A

Did not reveal the biochemical nature of the genetic material.

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43
Q

Who worked to determine the biochemical nature of the transforming principle of Griffith’s experiments?

A

Oswald Avery, Colin MacLeod and Maclyn McCarty

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44
Q

By 1926, the quest to determine the genetic material had reached the

A

molecular level. (chromosomes located in most cells)

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45
Q

What year did Fredrick Griffith perform his transforming principle experiments?

A

1928

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46
Q

Which bacteria did Griffith use for his experiments?

A

Streptococcus pneumoniae (pneumococcus)

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47
Q

How do S train and R strain of Streptococcus pneumoniae differ?

A

S (smooth strain) is the virulent strain because it has a mucous (polysaccharide coat).
R (rough) strain does not have a polysaccharide coat and is not virulent.

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48
Q

What happens to the mice in Griffith’s experiments?

A

Mice infected with S strain die.
Mice infected with R strain survive.
Mice infected with heat killed S strain survive.
Mice infected with heat killed S strain + R strain die.

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49
Q

What was the conclusion of Griffith’s experiment?

A

The R strain had been transformed by the heat killed S strain. The transforming principle had enabled the R strain to produce a polysaccharide coat and become virulent. (Transforming principle was stable enough to resist the heat that killed the bacteria.)

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50
Q

Drawback of Griffith’s experiment?

A

Did not reveal the biochemical nature of the genetic material.

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51
Q

Who worked to determine the biochemical nature of the transforming principle of Griffith’s experiments?

A

Oswald Avery, Colin MacLeod and Maclyn McCarty

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52
Q

What did Avery, MacLeod and McCarty do?

A

They worked to determine the biochemical nature of Griffith’s transforming principle. They isolated biochemicals (protein, DNA, RNA) from the heat killed S strain to observe which transformed the R strain, only DNA did.
They also discovered that protein digesting enzyme (Protease), RNA digesting (RNAase) did not affect the transformation, only DNAase inhibited the transformation. They concluded that DNA is the hereditary material but not all biologists were convinced.

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53
Q

Where did the unequivocal proof of DNA being the genetic material come from?

A

Alfred Hershey and Martha Chase experiments of 1952.

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54
Q

What did Hershey and Chase work with?

A

Bacteriophages, viruses that infect bacteria,

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55
Q

What was the objective of the Hershey and Chase experiment?

A

To check if the genetic material was DNA or protein.
They worked with bacteriophages, which attach to bacteria and their genetic material enters the bacterial cell, the bacteria treats the virulent material as it’s own and manufactures more virus particles.

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56
Q

What was the Hershey and Chase experiment?

A

Some bacteriophages were grown on radioactive phosphorus, and therefore had radioactive DNA because DNA has phosphorus and did not have radioactive protein because protein doesn’t have P. Similarly phages grown on radioactive Sulphur had radioactive protein and not radioactive DNA.
Radioactive phages were allowed to attack E coli bacteria.
After the infection occurred, the viral coats were removed by agitating them in a blender.
Virus particles were separated from the bacteria by spinning them in s centrifuge.
Bacteria infected with phages with radioactive DNA were radioactive while those with radioactive protein were not.
Therefore DNA was the genetic material that passed from virus to bacteria and not protein.

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57
Q

Organisms in which RNA is the genetic material?

A

QB bacteriophage and Tobacco Mosaic Virus

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58
Q

Characters required for a molecule to act as a genetic material?

A
  1. Should be able to generate it’s replica. (Replication)
  2. Should be structurally and chemically stable.
  3. Should provide scope for slow changes (mutation) as required for evolution.
  4. Should be able to express itself in the form of Mendelian characters.
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59
Q

Stability of genetic material.

A

Genetic material should be stable enough to not change with different stages of a life cycle, age or physiology of an organism.

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60
Q

Why did heat not kill the DNA in Griffith’s experiment?

A

Because the two strands were complementary, even if they got separated by the heat they would come together when appropriate conditions were provided.

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61
Q

Why is DNA the better genetic material?

A

2’ OH group of RNA is a reactive group which makes RNA liable and easily degradable. Further RNA is a catalytic molecule hence it is reactive. DNA is chemically less reactive and structurally more stable, presence of thymine instead of uracil increases stability, which is why it is the better genetic material.

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62
Q

Which mutates faster, RNA or DNA?

A

Both RNA and DNA can mutate, but RNA mutates faster because it is unstable, which is why viruses with RNA as genetic material have shorter lifespans and mutate and evolve faster.

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63
Q

Protein synthesis machinery has evolved around

A

RNA.
RNA can directly code for the synthesis of proteins and easily express the characters. DNA is dependent on RNA for the synthesis of proteins.
DNA is preferred for the storage of genetic information and RNA for it’s transfer.

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64
Q

Which was the first genetic material?

A

RNA was the first genetic material. Essential life processes- metabolism, translation and splicing evolved around RNA. It was both a genetic material and a catalyst. Being a catalyst it was reactive and unstable. So DNA evolved from RNA, with chemical modifications ( double strands with complementary strands - resists change + easy to repair) and became the genetic material.

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65
Q

What was semiconservative replication?

A

The two strands of DNA could separate and each strand would act as a template for the formation of the new strand. After the completion of replication, each DNA molecule would have one parental strand and one new strand.

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66
Q

Whose experiment proved that DNA replicates semi conservatively?

A

Matthew Meselson and Franklin Stahl’s 1958 experiment proved that DNA replicates semi conservatively.

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67
Q

Is N15 radioactive?

A

No, N15 is not radioactive, it can be separated from N14 only based on densities.

68
Q

How could heavy DNA with N15 be separated by normal DNA with N14?

A

By centrifugation in a CsCl (Cesium Chloride) density gradient.

69
Q

E coli divide in what time?

A

20 minutes. (or 18 minutes)

70
Q

How was the DNA extracted after one generation ie 20 minutes after transfer from N15 to N14?

A

It was of intermediate density, that is half heavy and half light.

71
Q

How was the DNA extracted after one generation ie 40 minutes after transfer from N15 to N14?

A

It was composed of equal amounts of hybrid and light (N14) DNA.

72
Q

Ratio between normal (light) to hybrid DNA is

A

2 ^ n - 2 / 2 where n is the number of generations.
20 min - 1 generation
40 min - 2 generations
60 min - 3 generations and so on

73
Q

What experiment did Taylor and his colleagues perform?

A

An experiment to prove semiconservative replication of DNA using radioactive thymidine in the chromosomes of vicia faba beans in 1958.

74
Q

Main enzyme for DNA replication is

A

DNA dependent DNA polymerase.

It uses a DNA template to catalyze the polymerization of deoxynucleotides.

75
Q

Average rate of polymerization of E coli is

A

2000 bp per second

76
Q

Characteristics of DNA polymerases

A
  1. Highly efficient - catalyze the polymerization of a large number of nucleotides in a very short amount of time. ( E coli - polymerization of 2000 bp per second)
  2. Highly accurate - Enzymes have to be both efficient and accurate because any mistake could result in mutations.
77
Q

What role do deoxyribonucleotide triphosphates play during DNA replication?

A

Deoxyribonucleotide triphosphates serve a dual purpose. They act as substrates for polymerization and they also provide energy for polymerization because two terminal phosphates are high energy phosphates same as ATP.

78
Q

Is DNA polymerase the only enzyme required for DNA polymerization?

A 
No, other enzymes are also required.  (NCERT) 
{Outside NCERT} They are :
1. SSBP - Single Cell Binding Protein
2. DNA helicase
3. Topoisomerases
4. DNA primase
5. DNA ligase. 
Total : 6 enzymes.
79
Q

What is the replication fork?

A

In very long DNA molecules, the two strands cannot be separated along its entire length due to very high energy requirements so replication occurs within a small opening in the DNA helix called the replication fork.

80
Q

Along what direction does DNA dependent DNA polymerase catalyze polymerization?

A

5’ to 3’ direction.

Therefore polarity of the parent strand is 3’ to 5’.

81
Q

How is the replication of the strand with template polarity 5’ to 3’?

A

Replication of the strand with template polarity 5’ to 3’ is discontinuous because DNA polymerase catalyzes along 5’ to 3’ so parent strand should be 3’ to 5’. The fragments of the discontinuous strand (Okazaki fragments) are later joined by DNA ligase.

82
Q

What are Okazaki fragments?

A

Okazaki fragments are the discontinuous strands formed due to polymerization along the parent strand with 5’ to 3’ polarity. They are later joined by DNA ligase.

83
Q

What is Origin of Replication?

A

Definite region in the (E coli) DNA where replication originates. In recombinant DNA technology, vectors provide the origin of replication.

84
Q

In eukaryotes when does the replication of DNA take place?

A

During the S phase of the cell cycle.

85
Q

Why should DNA replication and the cell cycle be highly coordinated?

A

Because a failure in cell division after DNA replication results in polyploidy which is a chromosomal anomaly.

86
Q

What is transcription? (def only)

A

The process of copying genetic information from one strand of DNA into RNA.

87
Q

How does transcription differ from the replication of DNA?

A
  1. Both transcription and replication are based on complementarity but in transcription, adenine pairs with uracil on the RNA instead of thymine.
  2. Replication once set in ensures that the total DNA of the organism is duplicated but in translation only a segment of one of the strands is copied into RNA.
88
Q

Why does only one strand act as a template during transcription?

A

If both strands acted as templates, they would code for RNA molecules with different sequences and in turn two different proteins. One segment of DNA would produce 2 diff proteins complicating the genetic information transfer machinery.
The two strands of RNA formed would be complementary to each other and would form double stranded RNA, preventing RNA from being translated into protein, rendering the process of transcription futile.

89
Q

Three regions of the transcription unit

A
  1. Promoter
  2. Structural gene
  3. Terminator
90
Q

Transcription is catalyzed by?

A

DNA dependent RNA polymerase (different types catalyze all three types of RNA- messenger, transfer and ribosomal)

91
Q

Along what direction does DNA dependent RNA polymerase catalyze transcription?

A

5’ to 3’ direction.

Therefore polarity of the template strand is 3’ to 5’.

92
Q

Which is the template strand during transcription?

A

The strand with the polarity 3’ to 5’ is the template strand during transcription.

93
Q

Which is the coding strand during transcription?

A

The strand with the polarity 5’ to 3’ is called the coding strand. It has the same sequence as RNA with thymine in the presence of uracil. It is displaced during transcription. All references of the transcription unit are made with respect to the coding strand.

94
Q

The promoter and the terminator flank the _________ in a transcription unit.

A

Structural gene

95
Q

What is a promoter? Where is it located on the transcription unit?

A

The promoter is a DNA sequence that provides binding site for RNA polymerase. It is located on the 5’ end of the coding strand of the structural gene. Presence of promoter defines coding and template strand. On switching the position of the promoter with the terminator, the definition of coding and template strand is reversed.

96
Q

What is the terminator in a transcription unit?

A

Terminator is located along the 3’ end of the coding strand. It defines the end of the process of transcription.

97
Q

________________ may be present further upstream or downstream to the promoter in a transcription unit.

A

Additional regulatory sequences.

98
Q

Why is it difficult to define a gene in terms of DNA sequences?

A

A gene is a functional unit of inheritance. The DNA sequence coding for tRNA or rRNA molecule, apart from those coding for a polypeptide can also be defined as a gene. Which is why it is difficult to define a gene in terms of DNA.

99
Q

What is a cistron?

A

A segment of DNA coding for a polypeptide. Eukaryotic genes are monocistronic and Prokaryotic genes are polycistronic.

100
Q

What is the split gene arrangement of Eukaryotes?

A

The monocistronic genes in a eukaryote have interrupted coding sequences - they are split. Exons- the coding sequences or expressed sequences which appear in mature or processed RNA. Introns- intervening sequences which are absent in mature RNA. This split gene arrangement further complicates the definition of genes in terms of DNA sequences. The split gene arrangement and the presence of introns - process of splicing represents the dominance of the RNA world.

101
Q

What are exons?

A

Exons- the coding sequences or expressed sequences which appear in mature or processed RNA. (from the monocistronic gene of a eukaryote)

102
Q

What are introns?

A

Introns- intervening sequences which are absent in mature RNA. (from the monocistronic gene of a eukaryote)

103
Q

What are regulatory genes (with reference to transcription unit)

A

Inheritance of a character can be affected by the promoter and regulatory sequences of a structural gene which is why regulatory sequences can be referred to as regulatory genes, even though they do not code for any RNA or proteins.

104
Q

What are the major types of RNA and what are their functions?

A

The major types of RNA are: mRNA (messenger RNA), tRNA (transfer RNA) and rRNA (ribosomal RNA). All three RNAs are required to synthesize a protein in a cell. mRNA provides the template, tRNA brings the amino acids as well reads the genetic code and rRNA acts as a structural and catalytic molecule.

105
Q

How does DNA dependent RNA polymerase catalyze the process of transcription? (How does transcription happen?)

A

RNA polymerase binds to the promoter and initiates transcription (initiation).
It uses nucleotide triphosphates as a substrate and polymerizes in a template dependent fashion following the rule of complementarity.
It also facilitates opening of the helix and continues elongation.
Only a small stretch of RNA remains bound to the enzyme.
Once the polymerase reaches the terminator region the nascent RNA falls off along with RNA polymerase (termination)

106
Q

What are the three steps of transcription?

A
  1. Initiation
  2. Elongation
  3. Termination
107
Q

RNA polymerase is capable of catalyzing which step of transcription?

A

Elongation

108
Q

What is initiation factor?

A

Initiation or SIGMA factor associates transiently with RNA polymerase to alter it’s specificity to help initiation in the process of transcription.

109
Q

What is termination factor?

A

Termination or RHO factor associates transiently with RNA polymerase to alter it’s specificity to help termination in the process of transcription.

110
Q

How can transcription and translation be coupled in bacteria (prokaryotes) ?

A

mRNA formed by transcription is not required to go under any processing to become active and transcription and translation take place in the same compartment (no separation between nucleus and cytosol) so sometimes translation can begin even before mRNA has been fully transcribed.

111
Q

How many RNA polymerases are present in eukaryotes?

A

3 in the nucleus and one in the organelles

112
Q

RNA polymerase I transcribes

A

rRNA - 28S, 18S and 5.8S

113
Q

RNA polymerase III transcribes

A

tRNA , 5srRNA and snRNA (small nuclear RNA)

114
Q

RNA polymerase II transcribes

A

hnRNA - heterogenous nuclear RNA - precursor to mRNA

115
Q

What is splicing?

A

Process by introns are removed and exons are joined in a defined order.

116
Q

What is capping?

A

An unusual nucleotide methyl guanosine triphosphate is added to the 5’ end of the hnRNA.

117
Q

What is tailing?

A

Adenylate residues (200-300) are added to the 3’ end in a template independent manner.

118
Q

When is hnRNA fully processed?

A

After splicing, capping and tailing, hnRNA is fully processed and called mRNA. mRNA is transported out of the nucleus for translation.

119
Q

What is translation?

A

Transfer of genetic information from a polymer of nucleotides to synthesize a polymer of amino acids.
This led to the proposition of a genetic code that could direct the sequence of amino acids during synthesis of proteins. (Translation is the process of polymerization of amino acids to form a polypeptide.)

120
Q

Who were involved in deciphering the genetic code?

A

Scientists from all disciplines - physicists , organic chemists, biochemists and geneticists.

121
Q

What was George Gamow’s contribution to cracking the genetic code?

A

George Gamow was a physicist. He suggested that since there are 4 bases that have to code for 20 amino acid the code should consist of a combination if bases.
Two bases - 4^2 = 16 would be insufficient so the code should be made up of 3 nucleotides. ( 4^3 = 64) This code would generate 64 codons.

122
Q

What was Har Gobind Khorana’s contribution to cracking the genetic code?

A

Har Gobind Khorana developed a chemical method which was instrumental in synthesizing RNA molecules with defined combination of bases. (homopolymers and copolymers).

123
Q

What was Marshall Nirenberg’s contribution to cracking the genetic code?

A

Marshall Nirenberg’s cell free system helped the code be deciphered.

124
Q

What is Severo Ochoa enzyme?

A

Severo Ochoa enzynme - polynucleotide phosphorylase was helpful in polymerizing RNA with defined sequences in a template independent manner.

125
Q

Salient features of the genetic code? (6)

A
  1. The code is a triplet - 61 codons code for amino acids and 3 are stop codons( which do not code for anything)
  2. The code is degenerate - some amino acids are coded by more than one codon.
  3. The codon is read in a continuous fashion with no punctuation.
  4. The code is nearly universal - from bacteria to humans UUU codes for phenylalanine. Exceptions have been found in mitochondrial codons and some protozoans.
  5. AUG has dual function - codes for methionine and acts as the initiator codon.
  6. UAA, UAG and UGA are stop or terminator codons.
126
Q

Why is the genetic code degenerate?

A

Some amino acids are coded by more that one codon.

127
Q

Why is the genetic code universal?

A

From bacteria to humans the code is nearly the same - UUU codes for phenylalanine - exceptions are mitochondrial codons and some protozoans.

128
Q

Which is the initiator codon?

A

AUG which also codes for methionine.

129
Q

Which are the termination codons?

A

UAA, UAG, UGA - they do not code for any amino acids.

130
Q

Classical example of point mutation is

A

Change of a single base pair in the gene for beta globin results the change of amino acid from residue from glutamate to valine resulting in sickle cell anemia.

131
Q

Frameshift and deletion mutations

A

Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion .

132
Q

What did Francis Crick postulate after the proposition of the genetic code?

A

Because amino acids have no structural specialties to read the code uniquely, he postulated that there was an adapter molecule that would on one hand read the code and on the other hand bring specific amino acids.

133
Q

What was tRNA earlier known as?

A

sRNA or soluble RNA.

134
Q

Structure of tRNA

A

tRNA has an anticodon loop with bases complementary to the code and an amino acid acceptor end which binds to the amino acids.

135
Q

Specificity of the tRNA

A

tRNAs are specific for each amino acid and there is a specific tRNA for initiation - initiator tRNA.

136
Q

Shape of tRNAs

A

Looks like clover leaf. Actual structure is an compacted molecule looking like an inverted L.

137
Q

Order and sequence of amino acids (in translation) is defined by

A

Sequence of bases in the mRNA

138
Q

Amino acids are joined by which bond?

A

Peptide bond

139
Q

What is the first step of translation?

A

Formation of peptide bond requires energy so in the first step of translation the amino acids are activated by ATP and linked to their cognate tRNA. This step is called amino acylation of tRNA or charging of tRNA.

140
Q

What happens when two charged tRNA molecules are brought together?

A

Formation of peptide bond is favored energetically. The presence of catalyst would enhance the rate of formation of peptide bond.

141
Q

Cellular factory responsible for synthesizing protein is

A

Ribosome

142
Q

What does the ribosome consist of?

A

Ribosome consists of structural RNA and 80 different proteins.

143
Q

In inactive stage ribosome exists as

A

A large subunit and the small subunit.

144
Q

How does the ribosome facilitate translation? (Explain the process of translation) (6)

A
  1. When the small subunit encounters a mRNA, translation begins.
  2. The ribosome binds with the mRNA at the start codon, which is recognized only by the initiator tRNA.
  3. The ribosome proceeds to elongation. The large subunit provides two sites for two subsequent amino acids to be close enough to each other, to bind and form a peptide bond.
  4. The ribosome also acts as a catalyst. 23S rRNA is the bacterial enzyme ribozyme.
  5. The ribosome moves from codon to codon until the stop codon is reached.
  6. A release factor binds to the stop codon terminating translation and releasing the complete polypeptide from the ribosome.
145
Q

What is the translational unit?

A

Sequence of the mRNA flanked by a start codon and a stop codon that codes for a polypeptide.

146
Q

What are untranslated regions in the mRNA?

A

Untranslated regions (UTRs) are additional sequences that are not translated. They are present at 5’ end -before start codon and at the 3’ end - after the stop codon. They are required for an efficient translation process.

147
Q

Which ribosome acts as the bacterial enzyme ribozyme?

A

23S rRNA

148
Q

Regulation of gene expression in eukaryotes can be expressed at which levels?

A
  1. Transcriptional level - formation of primary transcript.
  2. Processing level - regulation of splicing
  3. Transport of mRNA from nucleus to cytoplasm
  4. Translational level
149
Q

Why does E. coli synthesize B galactosidase?

A

To catalyze the hydrolysis of disaccharide lactose into glucose and galactose. (Bacteria uses it as an energy source.)

150
Q

Which conditions regulate the expression of genes?

A

Metabolic, physiological and environmental conditions.

151
Q

What is the predominant site for control of gene expression in prokaryotes?

A

Control of transcriptional initiation.

152
Q

What are operators? ( With reference to control of gene expression in prokaryotes.)

A

Operators are protein sequences whose interactions regulate the accessibility of the promoter region of prokaryotic DNA. They can act positively (activators) and negatively repressors.

153
Q

Who elucidated on the working of the lac operon?

A

Geneticist - Francois Jacob and Biochemist Jacque Monod.

154
Q

In lac operon what does lac refer to?

A

lac refers to lactose

155
Q

What is an operon?

A

A polycistronic structural gene is regulated by common promoter and regulatory genes.

156
Q

Examples of operons

A

lac, val, ara. his and trp operons

157
Q

What does the lac operon consist of?

A

One regulatory gene - the i gene

and three structural genes - z, y and a genes.

158
Q

What is the i gene? What does it code for?

A

The i gene is the regulatory gene of the lac operon. The term i refers to inhibitor, not inducer. The i gene codes for the repressor of the lac operon.

159
Q

What is the z gene? What does it code for?

A

The z gene is one of three structural genes of the lac operon. It codes for B galactosidase.

160
Q

What is the y gene? What does it code for?

A

The y gene is one of three structural genes of the lac operon. It codes for permease.
Permease increases the permeability of the cell to B galactosidase.

161
Q

What does the enzyme permease do? ( With reference to lac operon)

A

Permease increases the permeability of the cell to B galactosidase - and also lets lactose into the cell.

162
Q

What is the a gene? What does it code for?

A

The a gene is one of three structural genes of the lac operon. It codes for transacetylase.

163
Q

What acts as an inducer for the lac operon?

A

Lactose acts as an inducer for the lac operon. It is the substrate for B galactosidase and it regulates the switching on and off of the operon.

164
Q

How does lactose enter the cell?

A

By the action of permease.

165
Q

Why does a very low level of the expression of the lac operon have to be present in the cell all the time?

A

A very low level of the expression of the lac operon have to be present in the cell all the time because otherwise lactose cannot enter the cell.

166
Q

How does the lac operon work?

A

A repressor protein is always synthesized from the i gene. The repressor binds to the operator and prevents RNA polymerase from transcribing the operon. (Operator is present after the promoter). In the presence of an inducer- lactose or allolactose, the repressor is inactivated by interaction with the inducer. This allows RNA polymerase to access the promoter and transcription proceeds.

167
Q

Regulation of lac operon by repressor is referred to as?

A

Negative regulation. (lac operon can also be positively regulated.)

Biology 🧬 (8 decks)

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