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Biochemistry 300L > BCH 313 Protein Synthesis > Flashcards

BCH 313 Protein Synthesis Flashcards

1
Q

What is DNA replication?

A

It is a biological process that occurs in living organisms and copies their DNA it is the basis for biological inheritance

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2
Q

Why is DNA replication considered semi conservative?

A

Because each replicated duplex daughter DNA molecule contains one parent strand and one newly synthesized strand
Each strand of the original DNA molecule acts as a template for the production of the complementary strand

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3
Q

What is the origin of a DNA molecule

A

Specific locations on the genome where DNA replication begins

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4
Q

What is the replication fork

A

A replication DNA fork is defined as a Y-shaped region where the parent DNA double helix splits into two strands

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5
Q

DNA polymerase function

A

Builds a new duplex DNA strand by adding nucleotides in the 5’ to 3’ direction also performs proofreading and error correction

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6
Q

Function of DNA Helicase

A

Unwinds DNA double helix at the replication fork

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7
Q

Function of DNA clamp

A

a protein which prevents DNA polymerase three from dissociation from the DNA parent strand

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8
Q

Function of single-strand binding proteins

A

bind to ssDNA and prevent the DNA double helix from re-annealing after DNA Helicase unwinds it, thus maintaining the strand separation

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9
Q

Function of topoisomerase

A

Relax in the DNA from a strand
Prevents supercoiling

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10
Q

Function of DNA gyrase

A

released strain of unwinding by DNA Helicase

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11
Q

Function of DNA ligase

A

Re-anneals the semiconservative strand and joins the Okazaki fragments of the lagging strand

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12
Q

Function of primase

A

provides a starting point of DNA for DNA polymerase to begin synthesis of the new DNA strand by producing a primer,
Primers may consistently be added to the lagging strand

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13
Q

Function of telomerase

A

lengthens telomerase DNA by adding repetitive nucleotide sequences to the ends of the eukaryotic chromosomes

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14
Q

What are okazaki fragments?

A

Okazaki fragments are short sequences of DNA nucleotides which are synthesized discontinuously and later linked together by the enzyme DNA ligase to create the lagging strand during DNA replication.

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15
Q

What is a genetic mutation

A

a mutation is an alteration in the nucleic acid sequence of the genome of an organism
Genetic mutation alter the regulation or
expression of gene and results in dysfunctional or nonfunctional protein synthesis.

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16
Q

What is a mutagen?

A

a chemical or physical agent capable of inducing changes in DNA called mutations

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17
Q

Describe the dna structureand the bond between nucleotides?

A

These are molecules composed of a deoxyribose sugar, with a phosphate and a base (or nucleobase) attached to it. These nucleotides are attached to each other in strands via phosphodiester bonds to form a ‘sugar-phosphate backbone’.

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18
Q

DNA polymerase can only add nucleotides in what direction

A

5’to 3’ direction

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19
Q

What is DNA transcription and its stages?

A

A process where genetic information of DNA is transcribed (copied) to mRNA, transcription is the synthesis of RNA from DNA resulting in the transfer of the information stored in double stranded DNA into single stranded RNA which is used to direct the synthesis of its proteins.

Transcription has 4 stages: initiation, elongation, termination and post-transcriptional processing

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20
Q

What is DNA translation and where does it occur?

A
  • Translation is the process by which ribosomes convert the information carried by mRNA in the form of genetic code to the synthesis of new protein.
    Translation occurs in cytosol on ribosomes and is guided by mRNA.
    Eukaryotic protein synthesis involves more protein components with more intricate steps.
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20
Q

What are the requirements for translation?

A
  • mRNA
  • tRNAs
  • Ribosomes
  • Energy in the form of ATP and GTP
  • Enzymes and specific protein factors, e.g. initiation factors, elongation factors, etc.
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21
Q

What are the stages of Eukaryotic Translation

A
  1. Activation of amino acids
  2. Initiation
  3. Elongation
  4. Termination and Release
  5. Folding and post-translational processing
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22
Q

Describe translation Stage 1: Activation of Amino Acids

A

For the synthesis of a polypeptide with a defined sequence, two fundamental chemical requirements must be met:

(1) the carboxyl group of each amino acid must be activated to facilitate the formation of a peptide bond, and

(2) a link must be established between each new amino acid and the information in the mRNA that encodes it.

These two requirements are met by attaching the amino acid to a tRNA in the first stage of protein synthesis.

Attaching the right amino acid to the right tRNA is critical and This reaction takes place in the cytosol, not on the ribosome.

Each of the 20 amino acids is covalently attached to a specific tRNA at the expense of ATP energy, using Mg2+-dependent activating enzymes known as aminoacyl-tRNA synthetases.

When attached to their amino acid, the tRNAs are said to be “charged.”

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22
Q

Describe translation Stage 2: Initiation

A

The mRNA bearing the code for the polypeptide to be made binds to the smaller of two ribosomal subunits and the initiating aminoacyl-tRNA. Carrying the initiator tRNA.

The large ribosomal subunit then binds to form an initiation complex. The initiating aminoacyl-tRNA base pairs with the mRNA codon AUG which signals the beginning of the polypeptide.

This process, which requires GTP, is promoted by cytosolic proteins called initiation factors (which are released after the binding of the large ribosomal sub-unit).

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23
Q

Describe translation Stage 3: Elongation

A

The nascent polypeptide is lengthened by covalent (peptide bond) attachment of successive amino acid units, each carried to the ribosome and correctly positioned by its tRNA, which base-pairs to its corresponding codon in the mRNA.
Elongation requires cytosolic proteins known as elongation factors. The binding of each incoming aminoacyl-tRNA and the movement of the ribosome along the mRNA are facilitated by the hydrolysis of GTP as each residue is added to the growing polypeptide.

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23
Q

Describe translation Stage 4: Termination and Release

A

Completion of the polypeptide chain is signalled by a termination codon in the mRNA.
The new polypeptide is released from the ribosome, aided by proteins called release factors.

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24
Q

Describe Stage 5: Folding and Posttranslational Processing

A

To achieve its biologically active form, the new polypeptide must fold into its proper three-dimensional conformation.
Before or after folding, the new polypeptide may undergo enzymatic processing, including removal of one or more amino acids (usually from the amino terminus); addition of acetyl, phosphoryl, methyl, carboxyl, or other groups to certain amino acid residues; proteolytic cleavage; and/or attachment of oligosaccharides or prosthetic groups.

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24
Q

What is the A site on the ribosome?

A

The entry site with new tRNA charges with amino acid (aka aminoacyl tRNA)

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24
Q

What is the E site on the ribosome?

A

The exit site for the tRNA after it is done delivering the amino acid

24
Q

What is the P site on the ribosome?

A

Occupied by peptidyl tRNA, the tRNA that carries the growing polypeptide chain

25
Q

Where is the initiator tRNA positioned?

A

the P site

26
Q

What are posttranslational modifications?

A

Before or, after folding, the polypeptide may undergo processing by enzymatic action. Collectively these alterations are known as posttranslational modifications. These modifications may include removal of part of the translated sequence or the covalent addition of one or more chemical groups required for protein activity.

26
Q

What are chaperones?

A

To achieve native biologically active form of the polypeptide, it must undergo folding into its proper three-dimensional conformation. The chaperones (a group of specialized proteins) ensure the folding of a protein into its native form.

27
Q

What are some types of ptms?

A

Amino Terminal Modifications
Loss of Signal Sequence
Covalent Modification of Proteins

28
Q

What are some Covalent Modification of Proteins?

A

Glycosylation
Phosphorylation
* Carboxylation
* Hydroxylation
* Methylation
* Addition of prosthetic group

29
Q

What are the start codons and stop codons?

A

Start: AUG
Stop: UGA, UAG, UAA

30
Q

Describe the useful of inhibitors of protein synthesis

A

Several commonly used variety of antibiotics, act by inhibiting selectively the process of prokaryotic protein biosynthesis. The most useful antibiotics do not interact with eukaryotic protein synthesis and thus are not toxic to eukaryotes.
Such antibiotics can be used as therapeutic drugs.

31
Q

Importance of Puromycin and cycloheximide

A

Puromycin and cycloheximide are not clinically useful because they inhibit protein biosynthesis in eukaryotes also but are used for research purposes.

31
Q

Action of streptomycin

A

It binds to 30 S subunits of prokaryotes at the A site thereby inhibits chain elongation by preventing the binding of additional aminoacyl tRNA

32
Q

Action of tetracycline

A

Binds to the 30S subunits and inhibits binding of aminoacyl tRNA to mRNA in prokaryotes

33
Q

Action of Chloramphenicol

A

Binds to 50 S ribosomal subunits and blocks peptidyl transferase reaction in prokaryotes

34
Q

Action of erythromycin

A

Bind to the 50 S ribosomal subunit that inhibits the translocation reaction in prokaryotes

35
Q

Action of lincomycin and clindamycin

A

Binds to the 50 S subunits and inhibits peptidyl transferase thereby preventing the peptide bond formation in prokaryotes

36
Q

Action of puromycin

A

Causes premature chain termination in both prokaryotes and eukaryotes

37
Q

Action of cycloheximide

A

Inhibits peptidyl transferase activity in the 60s ribosomes subunits in eukaryotes

38
Q

Initiation stage of transcription

A

RNA polymerase binds to the promoter region of DNA
The promoter region is a short sequence of DNA (in eukaryotes TATAAA)
RNA polymerase also causes the DNA strands to separate

39
Q

Elongation stage of transcription

A

RNA polymerase begins to add nucleotides to the growing mRNA strand (antisense) to synthesize mRNA

39
Q

What is the anti-sense strand of DNA?

A

Antisense is the non-coding DNA strand of a gene. In a cell, antisense DNA serves as the template for producing messenger RNA (mRNA).

40
Q

What is the sense strand of DNA?

A

a sense strand, or coding strand, is the segment within double-stranded DNA that carries the translatable code in the 5′ to 3′ direction

40
Q

Termination stage of transcription

A

a transcribing RNA polymerase releases the DNA template and the nascent RNA

40
Q

Post-transcriptional processing stage of transcription

A

During post-transcriptional processing, portions of the RNA chain that are not supposed to be translated into proteins are cut out of the sequence

41
Q

What are retroviruses?

A

a subgroup of RNA viruses e.g. HIV which causes aids
Here the RNA acts as a template for the synthesis of new DNA molecules with the help of reverse transcriptase

42
Q

Reverse transcription

A

Reverse transcription is the synthesis of DNA from an RNA template. This process is driven by RNA-dependent DNA polymerases, also known as reverse transcriptases.

43
Q

What is PCR?

A

PCR is an in vitro DNA amplification procedure in which millions of copies of a particular sequence of DNA can be produced within a few hours.
PCR allows scientists to take a very small amount sample of DNA and amplify it, to a large enough amount to study it in detail

44
Q

Discuss PCR methods?

A

They rely on thermal cycling which exposes DNA samples to repeated cycles of heating and cooling to permit different temperature-dependent reactions, employing 2 main reagents known as the ‘primer’ and a DNA polymerase

45
Q

Step 1 of PCR

A

DNA strands are first separated (melted) by heating at 95 degree for 15s - 2 minutes

46
Q

Step 2 of PCR

A

The primers are annealed by cooling to 50 degrees and the primers hybridise with their complementary single-stranded DNA produced in the 1st step

47
Q

Step 3 of PCR

A

New DNA strands are synthesized by Taq polymerase enzyme derived from the bacteria, Thermus acquaticus found in hot springs therefore the enzyme does not denature at high temp.

The PCR is allowed to take place at 72 degrees for 30s in the presence of deoxyribonucleotide triphosphate and at this time, both strands of DNA are now duplicated.

48
Q

Step 4 of PCR

A

Steps 1,2 and 3 are repeated, in each cycle the DNA strands are doubled. Thus 20 cycles provide 1 million times amplification.
The cycles are generally repeated by automated instruments called temp cycler/ thermocycler.

49
Q

Features of genetic code

A
  1. Triplet codon
  2. Non-overlapping
  3. Non-punctuated
  4. Degenerate
  5. Unambiguous
  6. Universal
  7. Wobbling phenomenon
  8. Terminator codons/non-codons’
  9. Initiator codon
49
Q

Triplet codon

A

the codes are on the mRNA consisting of 3 bases of RNA

49
Q

Degenerate

A

One amino acid has more than one codon

49
Q

Non-overlapping

A

the codes are read one after the other in a continuous manner

49
Q

Non-punctuated

A

There is no punctuation between codons, it is consecutive or continuous

49
Q

Unambiguous

A

one codon stands for one amino acid

50
Q

Universal

A

codons are the same for the same amino acids in all species

50
Q

Wobbling phenomenon

A
51
Q

Terminator codon/ non-codons

A

They do not code for any amino acid
they put a full stop to protein synthesis
They are UAA, UGA, UAG
UGA is a stop codon but in certain circumstances, it stand for seleno-cysteine (the 21st amino acid)

52
Q

Initiator codon

A

AUG acts as the initiator and also codes for methionine

53
Q

What is the genetic code?

A

the system of the nucleotide sequence of mRNA that determines the sequence of amino acids in protein

Biochemistry 300L (11 decks)

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