Gene expression/Protein synthesis Flashcards

1
Q

Explain why DNA needs to make a single stranded copy

A

DNA cannot leave the nucleus as it is too large
cellular structures that build proteins are outside the nucleus

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2
Q

Compare structure of mRNA and DNA

A

mRNA: single stranded, Uracil, ribose sugar, more oxygen
DNA: double stranded, Thymine, deoxyribose sugar, less oxygen
Both contain phosphate and nitrogenous bases

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3
Q

Outline the process of transcription

A

Initiation:
- RNA polymerase binds to promoter region of the DNA template strand.
- DNA strands are separated (by DNA helicases?) to expose nitrogenous bases
Elongation:
- RNA polymerase moves along the DNA template strand in a 3’ to 5’ direction, catalysing the joining of complementary RNA nucleotides which are added to the 3’ end of mRNA and attach to each other via condensation reactions
Termination:
- once RNA polymerase transcribes terminator region, it detaches from DNA, releasing the pre-mRNA.
- DNA strands pair via complementary base pairing and recoil (due to hydrogen bonds)

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4
Q

What are monomers?

A

atoms or mall moleucles that bond together to form more complex structures (nucleotide, amino acid)

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5
Q

What are polymers?

A

large molecule made up of repeating subunits (monomers)

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6
Q

DNA polymers:

A

DNA
- double helix
- strands run antiparallel to each other
- two strands are connected through hydrogen bonds between complementary base pairs

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7
Q

RNA polymer (structure and types):

A
  • single stranded
  • messenger RNA
  • transfer RNA
  • ribosomal RNA
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8
Q

Protein monomer:

A
  • amino acid
  • central carbon with hydrogen attached
  • amine group (NH2)
  • carboxyl group (COOH)
  • variable R group
  • 20 different amino acids (20 R groups)
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9
Q

Name and summarise the reaction that creates bonds that join nucleotides together to create mRNA (and which bond)

A
  • condensation reactions
  • pentose sugar of one nucleotide interacts with phohsphate of another nucleotide
  • creates a phosphodiester bond (covalent), eliminating a water
  • nucelotides added to 3’ end
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10
Q

Condensation polymerisation for amino acids (protein monomer)

A
  • carobxyl group of one amino acid and amine group of another amino acid interact, creating a peptide link
  • a water molecule is released
  • ATP is needed
  • synthesised from amino to carboxyl terminus (always added to carboxyl end)
    (mRNA is read 5’ to 3’
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11
Q

Primary structure (which bonds)

A
  • order of amino acids
  • joined together by peptide bonds (covalent- strong)
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12
Q

Secondary structure

A
  • Formation of alpha helices (e.g. wool)
  • beta pleated sheets (e.g. silk)
  • random coils
  • hydrogen bonding between amino acids (folding of polypeptide backbone (everything except R group))
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13
Q

Tertiary structure

A
  • formation of irregular 3D structure (single poplypeptide chain)
  • many proteins are functional at this level
  • bonding between the variable R groups
  • hydrogen bonds, ionic bonds, disulphide bonds (only type of covalent in tertiary), hydrophobic interactions, van der waals forces
  • e..g myoglobin
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14
Q

Quaternary structure

A
  • two or more polypeptide chains combining (it may be multiple of the same polypeptide chain) to form one functional protein.
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15
Q

Prokaryote vs Eukaryote DNA

A

Prokaryote: single circular chromsome and plasmid DNA
Eukaryote: linear chromosomes (humans- 46)

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16
Q

Eukaryote gene structure

A

Promoter:
- positioned at the front of a gene
- particular nucleotide sequence (TATAAA…)
- location where enzyme RNA polymerase binds and initiates beginning of gene expression

Exons- regions of DNA that contain information for making a protein

Introns- sections of DNA (no info for making a protein), removed in the process of gene expression

Terminator:
- positioned at the end of a gene
- RNA polymerase detaches at this region
- represents specific nucleotide sequence

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17
Q

Prokaryotic gene structure

A

Organised into operons:
- genes with similar functions are grouped together for efficiency
- under the control of a single promoter

  • Promoter
  • Operator: after promoter, before genes, represents specific nucleotide sequence, repressor proteins can bind to regulate gene expression
  • no introns
  • terminator
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18
Q

RNA processing- what is it, which cells does it occur in, process

A
  • production of mature mRNA from pre mRNA
  • occurs in the nucleus
  • only occurs in eukaryotes

Process:
1. Capping: Addition of the 5’ methyl cap (modified guanine nucleotide added to 5’ end)
- protection from degradation and enzyme attack to the 5’ end
- contributes to its stability
- assist in ribosome binding to mRNA (as ribosome first binds to 5’cap)

  1. Addition of the poly-A tail to the 3’ end (AAAAA…)
    - contributes to stability
    - facilitate mRNA export from the nucleus
    - protects mRNA from degradation
  2. Splicing (removing introns and joining exons)
    - done by spliceosomes

https://www.youtube.com/watch?v=m2lOf3ker9M&ab_channel=PeterCavnar

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19
Q

What is the point of the nucleus?

A

To protect the DNA as if the DNA is damaged, every cell produced could be affected

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20
Q

Why is thymine replaced with Uracil?

A
  • Thymine has a methyl group which makes it more stable in DNA
  • RNA is more transient (while DNA stores genetic information over a long period of time), doesn’t need the stability
  • Uracil is synthesised and broken down more easily, more energy efficient
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21
Q

Describe a codon and where they are found

A

3 nucleotides found on the m-RNA

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22
Q

Explain role of ribosomal RNA

A
  • rRNA in the small ribosomal subunit helps to bind mRNA
  • rRNA in the small subunit helps align the tRNA anticodons with the appropriate mRNA codons
  • catalyses protein synthesis by forming peptide bonds between amino acids
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23
Q

Outline process of translation

A
  1. Initiation
    - small subunit of the ribosome attaches to mRNA (at 5’ cap in eukaryotes)
    - ribosome moves along mRNA until it reaches a start codon (AUG- methionine)
    - initiator tRNA binds to start codon
    - large subunit joins
  2. ELongation
    - tRNA molecule with an anticodon complementary to the codon of mRNA will bind via complementary base pairing, bringing a specific amino acid with them.
    - Amino acids are attached to each other via condensation reactions.
    Termination:
    - When a stop codon is reached, the ribosome detaches, releasing the mRNA and polypeptide chain.
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24
Q

why do tRNA come to ribosomes?

A

attracted by hydrogen bonds of complementary base pairs

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25
Q

Describe structure of transfer RNA

A
  • coiled or folded version of RNA
  • amino acid site at one end
  • anticodon at the other end
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26
Q

explain relationship between codons and anticodons

A

anticodons are complementary to codons so they can bind together.

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27
Q

describe how translation is terminated

A

Stop codon signals end of amino acid sequence
ribosome releases amino acid chain and separates into 2 pieces

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28
Q

explain why product of translation is a polypetptide chain, not a functional protein

A

Translation produces a polypeptide chain which is a linear sequence of amino acids. TO become a functional protein, it needs to undergo further folding into its specific 3D shape.

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29
Q

polymerisation

A

the process of bonding monomer molecules in a chemical reaction to form polymer chains

30
Q

The genetic code is described as:

A

degenerate/redundant
- each amino acid can be coded for by multiple codons
- acts as a buffer against mutations
- mutations may result in same amino acid

universal
- same codon codes for same amino acid across all organims

31
Q

Genome vs proteome

A

Genome: all the genetic material in a cell/organism (about 20000-25000 genes)

Proteome: all the proteins that (or can be) expressed by a cell, tissue or organism
- much larger and more complex than genome (over 1 000 000 proteins)

32
Q

What makes the proteome more complex and larger than the genome?

A
  • alternative splicing
  • occurs during splicing (RNA processing)
  • some exons removed/rearranged
  • different arrangement of exons leads ro differenent mRNA sequences, amino acid order, polypeptides and proteins
  • post-translational modifications
33
Q

Directionality of mRNA:

A

read up: 3’ to 5’
write down: 5’ to 3’
mRNA is ANTIPARALLEL to template strand

34
Q

what is the 3’ end?

A

the hydroxyl end of the sugar, associated with the 3rd carbon of the sugar

35
Q

how are the nitrogenous bases bonded?

A

A and T bond with 2 hydrogen
bonds
C and G bond with 3 hydrogen bonds

36
Q

structure of dna to chromosomes

A

The DNA double helix wraps around histones to form nucleosomes, which then condense to form chromatin and chromosomes.

36
Q

what is splicing done by?

A

spliceosomes,
recognise specific base sequences at the ends of the introns: GU at the 5′end (beginning of intron) and AG at the 3′ end (end of intron)

37
Q

Structure of gene: (general)

A

Coding region:
- contains the coded information for making a polypetide chain
Flanking region:
- region on either side of coding region
- before the start of the coding region it’s the upstream region
- after the end of the coding region it’s the downstream region

38
Q

Structure of coding region of a gene:

A
  • segement of DNA that includes DNA template strand (always starts with TAC- codes for AUG/methionine) and some distance away there is (ACT, ATT, ATC)
  • made up of exons (contain the instrcutions for protein synthesis)
  • exons separated by introns
39
Q

Promoter and upstream region: what it is, what can bind, what would happen if altered by mutations

A
  • region of DNA on the template strand located in upstream region (check which strand!!!)
  • TATA box (rich in As and Ts)
  • location where RNA polymerase and transcription factors bind to initiate transcription
  • site for proteins called transcription factors can bind and regulate gene expression
  • upstream sequences are found in all organisms
  • if upstream reagion is altered by mutations, activity of the coding region may reduce or become inactive, can affect final protein produced (e.g. thalassaemia)
  • upstream region has segments of DNA to which regulators (such as proteins and hormones) can attach to regulate rate of transcription
40
Q

Operator region

A
  • in an operon, between promoter and the gene being transcribed in prokaryotes
  • binding site for repressor proteins
  • when a repressor binds to the operator, prevents RNA polymerase binding to the promoter
  • transcription cannot be initiated
41
Q

Distinguish between introns and exons

A

Introns are interfering sections of DNA that are transcribed
but not translated, and are thus non-coding. Exons are
sections of DNA that are coding and are translated into the
final polypeptide product in the ribosome.

  • introns are thought to have regulatory functions
42
Q

What would happen to a protein if introns were not removed?

A
  • polypeptide formed would be
    different
  • The addition of extra amino acids will change the protein
    produced.
  • inclusion of extra nucleotides may
    change the reading frame within the exons, affecting the amino acids that are translated.
43
Q

regulatory vs structural gene

A
  • regulatory gene codes for proteins that control the action of other genes.
  • whether they are active, inactive, and rate at which their products are made)
    (e.g. DNA binding proteins bind directly to DNA, signalling molecules bind to receptors on the cell surface)

Structural genes code for proteins that contribute to the structure or functioning of the organism

44
Q

Explain how the expression of a single gene can lead to the production of different proteins.

A
  • alternative splicing
  • post transcriptional modifications (addition of 5’ methly cap, poly-A tail, splicing)
  • different nucleotide sequences/mRNA sequences code for a
    different protein (e.g. mutations)
  • post-translational changes to the protein, such as alternative folding (even with same amino acid sequence, different folding can effect function of protein)
45
Q

Tryptophan:

A
  • amino acid that E.coli ingests from its enviroment or makes itself using enzymes
  • these enymes are encoded by 5 genes
  • since making tryptophan requires energy, the enzymes are only produced when tryptophan is not present
  • genes that synthesise them are only active when required and controlled by the trp operon
46
Q

trp operon:

A
  • contains 5 structural genes that encode for the enzymes that synthesis Tryptophan
    genes: (trpA, trpB, trpC, trpD and trpE.)
  • operon regulates transcription when tryptophan is in low abundance
  • operator partially overlaps with promoter
  • leader (contains trpL gene which encodes a leader peptide and attenuator section to regulate transcription)
  • regulatory gene (encodes a repressor which, when active, binds to DNA and decreases rate of transcription)
47
Q

operon:

A

group of linked structural genes with a common promoter and operator that is transcribed as a single unit (only prokaryotes)
- expression of operons is controlled by regulator genes that produce repressor proteins.

48
Q

what happens when tryptophan is present and absent? (regulation through repression)

A

Present:
- tryptophan binds to the repressor protein, causing a change in shape, allowing it to be active
- allows repressor to bind at the operator
- RNA polymerase unable to bind to promotor
- operon is off

Absence:
- repressor unable to bind to operator
- RNA polymerase can bind to promoter and start transcription of structural genes

49
Q

Attentuation vs repression:
When does regulation occur?
what is involved?
When is the operon turned off?

A

repression:
- prevents initiation of transcription
- repressor protein binds to operator

Attenutation:
- prevents transcription being completed
- occurs during transcription
- When tryptophan is high, and the leader peptide allows for the production of the termination hairpin loop, stopping transcription.
- When tryptophan is low, and translation of the leader peptide is slowed, so the termination hairpin loop cannot form.

50
Q

How is the structure of an operon different to eukaryotic genes?

A
  • operon contains a common promoter and operator
  • eukaryotic genes, no operator, instead a longer upstream region where regulation can occur.
51
Q

Transcription factors what is it?

A

proteins that directly influence the process of transcription by binding to specific regions of DNA to regulate gene expression

52
Q

In what two ways is the process of gene regulation in eukaryotes different from gene regulation in prokaryotes?

A

Use comparative language! (whereas, while)

In eukaryotes, genes go through transcription, RNA processing and translation, while in prokaryotes, they only have transcription and translation.

Transcription occurs simulatenously in prokaryotes while it occurs separately in eukaryotes.

Transcription occurs in the nucleus in eukaryotes, whereas it occurs in the cytosol in prokaryotes.

Perhaps not (related to structure not process)? ask!
- In prokaryotes: repressor proteins bind to operator
- In Eukaryotes: no operator, so they bind to upstream region ???

53
Q

what does condensation polymerisation mean?

A

condensation- water is released
polymerisation- monomers being bonded together to form a polymer

54
Q

forces that maintain the tertiary structure of proteins are:

A
  • hydrogen bonds (Weak attractions between a hydrogen atom and an electronegative atom.)
  • ionic attractions being charged R groups
    (Attractions between oppositely charged R group)
  • interactions between hydrophobic R groups in the protein interior
    (Nonpolar R groups, tend to avoid water and cluster together in the protein’s interior.)
  • covalent disulfide cross links. (strongest bond, between the sulfur atoms of two cysteine amino acids)
55
Q

What is the basic formula of an amino acid molecule?

A

H2NCHRCOOH

56
Q

How is a peptide bond formed? (and what type of bond)

A
  • covalent bond between carboxyl group of one amino acid and amino group of adjacent amino acid
  • occurs through condensation polymerisation
  • ATP is added and water is released
57
Q

How does the R group of an amino acid determine the secondary and tertiary structure of a protein?

A
  • The R group is the variable group in amino acids that determines the property of the amino acid.
  • Some amino acids are positively charged, negatively charged, hydrophobic, hydrophilic, others contain sulfur
    (cysteine).
  • interactions between different
    R groups determine how folding in the protein occurs.
  • e.g.
  • 2 cysteine residues (amino acids after being incorporated into a popypeptide chain) will form disulphide
    bridges.
  • Ionic bonds form between positively and negatively charged groups
  • Hydrophobic groups tend to fold to be close to each other, as do hydrophilic groups.
58
Q

Disulfide bridges can only form between cysteine amino acids.
a. Suggest a reason that this may be the case.
b. Explain what you would expect if a protein did not contain any cysteine.
c. Methionine also contains sulfur but cannot form disulfide bridges. Why?

A

a.
- Only cysteine and methionine contain sulfur.
- only cysteine can form disulfide bridge due to the location of sulfur atom in the thiol group (-SH) .
- the thiol group (-SH) in cysteine is reactive,
attached hydrogen can be removed to form a bond between sulfur atoms and form a disulfide bridge.(-S-S-)

b. no disulfide bridges
Disulfide bonds are extremely strong covalent bonds so proteins without these bonds would likely be more easily
denatured in heat

c. To form a disulfide bridge the highly reactive thiol group is required which is in cysteine not in methionine.
- In methionine, sulfur is in the middle of the R group, unable to form disulfide
bridges.

59
Q

Define genome and proteome

A

Genome: all the genes or genetic material in a cell or organism

Proteome: all the protein that are or can be expressed by a cell, tissue or organism.

60
Q

Why is the proteome more complex than the genome?

A
  • alternative splicing
  • mRNA editing
  • post-translational modifications
61
Q

Protein secretion pathway

A

POLYPEPTIDE synthesied on ribosome LOCATED on rough endoplasmic reticulum

Rough endoplasmic reticulum:
- folded membranes studded with ribosomes
- proteins inserted in lumen through pores in membrane
- lumen: fluid-filled space between membranes where modifications occur

Transport vesicle:
- forms by budding off rough ER
- PROTEINS are packaged to protect them

Golgi apparatus:
- complex of folded membranes
- transport vesicles fuse with golgi membrane
- post-translational modifications (include removing sugar monomers and substituting them with other chemical groups like carbohydrates)

Secretory vesicle:
- buds off golgi

exocytosis:
secretory vesicle migrates towards plasma membrane, fuses with it, releases PROTEIN into extracellular environment VIA exocytosis

62
Q

triplet vs codon

A

triplet is a sequence of three bases in DNA that codes for one amino acid

codon is a sequence of 3 bases in mRNA (complement of triplet)

anitcodon is sequence of 3 bases in tRNA

63
Q

what does a gene encode?

A

Encodes the instructions for making a specific protein and forms of RNA, containing a nucleotide sequence with triplets that code for an amino acid which make up a protein.

64
Q

leader region in trp operon sections

A
  • contains sections 1,2,3,4
  • codes for leader peptide (trp L) (region 1) (transcribed and translated)
  • followed by 3 attenuator regions (2,3,4) (not translated)
  • attenuator (after domains?) contain many As and Ts/Us, weak hydrogen bonds make it easier for RNA polymerase to detach
65
Q

trp L polypeptide

A

-polypeptide coded for by trpL contains 14 amino acids
- 2 adjacent amino acids are tryptophan

66
Q

Hairpin loops in trp operon (and what they are)

A
  • hairpin loops: (mrna) contain a section in which nucleotides are complementary and can bind
  • terminator hairpin occurs between regions 3 and 4 (RNA polymerase detaches from DNA)
67
Q

Attenuation with high tryptophan

A
  • ribosome translates trp L quickly, and quickly detaches from mRNA
  • hairpin loop forms between mRNA in sections 1 and 2
  • terminator hairpin loop forms in regions 3 and 4 of mRNA
  • terminator hairpin loop causes RNA polymerase to detach from DNA and transcription is stopped
68
Q

Attenuation with low tryptophan:

A
  • trp L is translated slowly. As ribosome translates gene, ribosome pauses at trp codon to wait for enough tryptophan to be available
  • ribosome stays attached to region 1 so no hairpin loop can form between sections 1 and 2
  • hairpin loop forms between 2 and 3 (antiterminator loop)
  • does not cause RNA polymerase to detach, prevents terminator hairpin loop from forming
  • transcription occurs
69
Q

how many amino acids in a polypeptide?

A

two or more

70
Q

Transformed bacterial cells can be used to produce large quantities of human insulin. In the process of producing human insulin, scientists
A. extract insulin from a human.
B. insert the human insulin gene into the bacterial chromosome.
C. use DNA ligase to isolate the insulin gene from human DNA.
D. enable a transgenic bacterium to reproduce millions of identical transgenic bacterial cells.

A

D

A. Extract insulin from a human → False

Scientists do not extract insulin from human tissue. Instead, they use recombinant DNA technology.

B. Insert the human insulin gene into the bacterial chromosome → False

The human insulin gene is usually inserted into a plasmid.

C. Use DNA ligase to isolate the insulin gene from human DNA → False

DNA ligase is used to join DNA fragments, not to isolate them. Scientists typically use restriction enzymes to cut out the insulin gene or synthesize it directly.

71
Q

Scientists wanted to check whether their results were reproducible. To do this, they repeated the experiment and changed the
A. type of investigation.
B. variable being measured.
C. method of the experiment.
D. person recording the observations.

A

D

To check if results are reproducible, scientists need to ensure that another person can follow the same method and obtain the same results. Reproducibility means that different researchers can conduct the same experiment independently and achieve similar outcomes.