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Year 3 - Biotechnology > 1 - Intro > Flashcards

1 - Intro Flashcards

1
Q

Describe chromosomes

A
  • Composed of a centromere from which four arms protrude, and a telomere at each end
  • Telomere helps to confer stability to the ends of the chromosome
  • Composed of chromatin which is compactly folded mixtures of DNA and basic DNA-binding proteins
  • Chromatin is composed of individually packaged units called nucleosomes
  • Chromosome -> centromere -> chromatin -> nucleosomes
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2
Q

Describe nucleosomes

A
  • Appear as beads on a thin string by electron microscopy
  • Each nucleosome consists of octamer of basic DNA binding proteins called histones w/ a constant length of DNA wound around it
  • Each octamer consists of 2 copies of histones – H2A, H2B, H3, and H4
  • Each nucleosome is separated from its neighbours by a length of DNA that is bound to histone H1
  • Nucleosomes pack DNA but also help to regulate gene expression
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3
Q

Describe the major differences in a prokaryotic gene structure compared to eukaryotic

A
  • No splice sites, so no splicing
  • Promoters are much less complex
    • Consensus sequence for binding RNA pol
  • No poly A signal so no poly A tail
    • Half-life of RNA controlled by other mechanisms
  • No 5’ methyl-guanosine cap
    • Ribosomes bind to an RNA sequence called the Shine-Dalgarno sequence
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4
Q

DNA replication

A
  • Synthesis is 5’ to 3’
  • Incoming nucleotide triphosphate is added to 3’ OH of existing DNA strand
  • Diphosphate (PPi) is a good leaving group; along w/ hydrolysis of diphosphate –> phosphate drives the reaction forward
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5
Q

DNA synthesis, mutation, and editing

A
  • DNA polymerase has “proof-reading” activity, which means it can replace improperly paired bases w/ correct ones
  • DNA polymerases also have exonuclease activity, so can remove the mismatched base and add the right one
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6
Q

Replication fork

A
  • Single-strand binding proteins bind to single-stranded DNA to prevent re-forming double-stranded DNA during replication
  • Topoisomerase I cuts one strand of DNA, allowing it to twist around by one turn and then re-ligate
    • Does this repeatedly to relieve the torsional strain in the DNA created by the helicase
  • Leading strand is synthesized continuously in the 5’ to 3’ direction
    • Lagging strand also synthesized in 5’ to 3’ direction, but in small discontinuous stretches (Okazaki fragments)
    • Fragments are joined by ligase
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7
Q

Describe the lac operon of a prokaryote (what it uses as a carbon source)

A
  • Glucose is a preferred carbon sources, but if not present and lactose is, lactose can be digested into galactose and glucose for carbon
  • Genes for digesting and taking up lactose = beta-galactosidase, permease, and transacetylase = lacZ, lacY, and lacA respectively; lie in tandem on the E. coli chromosome
  • If lactose not present or if sufficient glucose is, there is no need to make genes for digesting lactose
  • Therefore, expression of genes lac Z, Y, and A are controlled by what is known as the lac operon
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8
Q

Describe Lac 1, RNA polymerase, and CAP

A
  • Lac I produces the lac repressor which binds to the operator in the absence of lactose (repressor means no lac operon produced b/c no lactose needed to be digested)
    • If lactose is present, it binds to the repressor and prevents it from binding to the operator (so repressor can’t stop lac operon from being produced, therefore it is produced to digest lactose)
  • RNA polymerase binds to the promoter
    • If repressor is bound to the operator, RNA polymerase can’t transcribe lac Z, Y, and A (so no lac operon)
  • cAMP receptor protein binds cAMP and only then will it bind to the CAP site, which helps RNA polymerase bind to promoter
    • [cAMP] increases as [glucose] decreases
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9
Q

Which factors will be bound to lac gene based on whether glucose and lactose are present?

A
  • [cAMP] low, glucose and lactose present = very low expression of lac gene (b/c can just use glucose, doesn’t need lactose)
  • [cAMP] high, glucose absent, lactose present = high expression (RNA polymerase and CAP bound b/c lactose needs to be used)
  • [cAMP] high, glucose and lactose present = no expression (CAP, RNA polymerase, and repressor bound – repressor = most important; don’t need lactose)
  • [cAMP] low, glucose present, lactose absent = no expression (RNA polymerase and repressor bound – CAP only bound when cAMP present)
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10
Q

Gene regulation in eukaryotes (describe RNA polymerase, cis acting element, and chromatin structure)

A
  • More complicated than in bacteria
  • RNA polymerase doesn’t bind to the promoter by itself, group of proteins must also bind to the promoter
    • Transcription factors, co-activators, and co-repressors
  • Depending on the proteins that are bound, expression can be higher or lower
  • *Cis acting element is a DNA portion that a DNA binding protein (trans acting element) will bind to
  • DNA sequences involved in regulating expression are far more extensive
  • Chromatin structure can also regulate access of the transcriptional machinery to the DNA regulating expression
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11
Q

Initiation of transcription – promoter region of eukaryotic gene (TATA box, GC box, CCAAT box)

A
  • TATA box located ~ 25-30 bp upstream of the start point of transcription
    • Its consensus sequence = TATAAAA
  • TFIID (transcription factor IID) composed of subunits called TATA-binding protein (TBP) and TBP-associated factors (TAFs)
    • TBP binds in the narrow (minor) groove of DNA at the TATA box and bends DNA
    • Forces helix to open slightly, probably allowing better access to RNA polymerase
  • GC box = common element in eukaryotic promoters
    • Its consensus sequence = GGGCGG
    • May be present in 1 or more copies
    • Located between 40-100 bp upstream of the start point of transcription
    • Transcriptional activator Sp1 binds to this region
  • CCAAT box often found between 40-100 bp upstream of the transcription start points
    • Consensus sequence = GGCCAATCT
    • Transcriptional activator C/EBP (CCAAT box/enhancer binding protein) binds to this sequence and recruits co-activators and even more transcription factors
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12
Q

Enhancers

A
  • Bind activators and repressors
  • Located at some distance from the gene
  • Enhancer sequence can be relative to the promoter or flipped in opposite orientation
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13
Q

Signal integration at enhancers

A
  • Activators bound at an enhancer can work together to switch on a gene
  • Human interferon-beta gene
  • Virus infects a cell => triggers 3 different activators (NF-kB, IRF, and JUN/AP1)
    • Only bind if all 3 are present (cooperative binding)
  • Enhance ability to bind DNA
  • HMG-1 binds w/in the enhancer, bending the DNA to facilitate activators binding and interaction
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14
Q

What are the 5’-methyl-guanosine cap and poly A tail involved in?

A

RNA processing

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15
Q

RNA splicing – what is it and what are the steps?

A
  • The process of removing introns from pre-mRNA to form mRNA
  • A two-step process of sequential trans-esterification reactions
    1) The cleavage of the 5’ intron/exon junction occurs simultaneously w/ the formation of a new phosphodiester bond between the phosphate at the 5’ end of the intron and an adenine residue just w/in the 3’ end of the intron
    2) Another trans-esterification reaction occurs, in which the phosphodiester bond at the 3’ intron/exon junction is cleaved and the 2 cleaved exons are joined
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16
Q

What are some major differences between translation in bacteria and eukaryotes?

A
  • Polycistronic mRNA
    • mRNA encode more than 1 protein
    • Info for each is arranged in tandem
  • Shine-Dalgarno box in mRNA
    • Specific ribosome-binding sequence
    • Adjacent to the start site for each protein
  • Ribosomes
  • Post-translational modifications
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17
Q

Examples of protein modification after synthesis

A
  • 3D structure – protein folding
  • Membrane proteins – fatty acid side chains
  • N-linked glycosylation
  • O-linked glycosylation
  • Phosphorylation
  • Acetylation
  • Alkylation
  • Gamma-carboxylation
18
Q

What is needed to make protein drugs?

A
  • Must add the gene that codes for the intended protein to the microorganisms or cultured mammalian cells
  • Need:
    • The gene (not normally synthesized chemically)
    • A vector to carry the gene (usually plasmid)
    • Way to get the gene into the vector
    • Way to get the vector w/ the gene into the microorganisms or cultured mammalian cells
    • Way to select for the microorganisms or cultured mammalian cells that have the vector w/ the gene
    • Way to control the expression of the gene once it is in the microorganisms or cultured mammalian cells
    • Way to lyse the cells and purify the protein drug away from other contaminating proteins made by the microorganisms or cultured mammalian cells
19
Q

How is a gene obtained through cloning?

A
  • Get the gene from the human cell by making what is known as cDNA
  • Amplify the gene from the cDNA using polymerase chain reaction (PCR)
20
Q

Describe a plasmid as a vector for transferring genes

A
  • Vectors can be used to transfer genes to mammalian, yeast, insect, and bacterial cells
  • Each organism requires different sequences for proper translation and transcription and in some cases, replication of plasmid
  • Plasmids are put into bacteria and yeast by a process known as transformation, in insect and mammalian cells it’s known as transfection
  • Antibiotic selection – this allows to select for cells that have the plasmid (ex: if we are using bacterial cells we will use an antibiotic resistance gene such as beta-lactamase; if we grow the bacteria in the presence of penicillin only, the bacteria that have the plasmid and hence express the gene can grow)
  • Affinity tag (not always present)
    • Expressed as part of the protein of interest
    • Allows us to capture the protein we expressed to purify it
  • Protease cleavage site (not always present) – a specific proteolytic cleavage site w/in the expressed protein usually 4-5 amino acids long to cut off the affinity tag from a protein
  • We can use expression control mechanisms such as lac operon to control when the gene is expressed once it is in the bacteria
21
Q

Protein + affinity tag

A
  • Once the protein is expressed it can be purified w/ affinity resin
  • Affinity resin – N tag - Protein
22
Q

Restriction enzymes

A
  • Have codes for names – indicate name of bacteria where they were derived from and number (roman numerals)
  • Often make staggered cuts in DNA leaving sticky ends that can be exploited to link DNA fragments cut w/ the same restriction enzymes
23
Q

Multiple cloning region

A
  • Short sequence w/ multiple restriction sites
  • Restriction sites: sequences that are specifically cleaved by restriction enzymes
    • Restriction sites in the multiple cloning region only appear once in the whole plasmid
  • This is the place we insert our gene of interest
    • This is done by exploiting restriction enzymes to cut the plasmid and our gene using compatible sticky ends to hold the gene in the plasmid temporarily
    • Then use ligase to fuse the plasmid (vector) and gene together
24
Q

Subcloning

A
  • The process of putting the gene of interest into a plasmid

- Can be done w/ restriction enzymes

25
Q

Describe advantages and disadvantages of the yeast expression system

A
  • Good alternative when bacterial expression system isn’t adequate
  • Advantage = some post-translations modifications, faster than mammalian, some recombinant proteins that are insoluble when expressed in bacteria will be soluble in yeast (due to complex protein processing)
  • Disadvantage = handling may be difficult, both cells and medium may contain compounds that affect binding of affinity-tags to the resin (making it difficult to isolate proteins)
26
Q

Describe advantages and disadvantages of the mammalian cell expression system

A
  • The best expression system for generating recombinant eukaryotic proteins
  • Advantage = post-translational modifications, uses same synthesis and processing signals found in mammalian cels
  • Disadvantage = slow, more complex, expensive, low expression levels
27
Q

Describe protein folding for bacteria vs. yeast vs. mammalian cells

A
  • Bacteria = refolding usually required
  • Yeast = refolding may be required
  • Mammalian cells = proper folding
28
Q

Describe N-linked glycosylation for bacteria vs. yeast vs. mammalian cells

A
  • Bacteria = none
  • Yeast = high mannose
  • Mammalian cells = complex
29
Q

Describe O-linked glycosylation for bacteria vs. yeast vs. mammalian cells

A
  • Bacteria = no

- Yeast and mammalian cells = yes

30
Q

Describe phosphorylation for bacteria vs. yeast vs. mammalian cells

A
  • Bacteria = no

- Yeast and mammalian cells = yes

31
Q

Describe acetylation for bacteria vs. yeast vs. mammalian cells

A
  • Bacteria = no

- Yeast and mammalian cells = yes

32
Q

Describe alkylation for bacteria vs. yeast vs. mammalian cells

A
  • Bacteria = no

- Yeast and mammalian cells = yes

33
Q

Describe gamma-carboxylation for bacteria vs. yeast vs. mammalian cells

A
  • Bacteria and yeast = no

- Mammalian cells = yes

34
Q

What makes the small RNA primer? What is its function? What happens to it later?

A
  • Made by RNA primase to initiate synthesis

- Later degraded by RNase H followed by filling in DNA by DNA polymerase

35
Q

Function of DNA helicase

A
  • Helicase unwinds DNA so DNA polymerase can gain access to the DNA
    • Unwinding causes torsional strain on DNA
36
Q

Where does DNA replication start?

A

Starts at a specific sequence called the origin of replication sometimes abbreviated ori

37
Q

Describe the structure and function of the 5’ methyl-guanosine cap

A
  • Composed of a methyl-guanosine nucleotide covalently attached to the 5’ phosphate of the first nucleotide of the mRNA
  • Functions = recognition signal for ribosome, signal to export out of nucleus, and increases half-life of mRNA
38
Q

Describe the function of the Poly A tail for RNA processing

A
  • Specific mRNA sequence AAUAAA is recognized by the enzyme polyadenylate polymerase
  • Cleaves the primary transcript 11-30 bases at 3’ end
  • Add a stretch of 200-250 adenosines – increases mRNA half-life and nuclear export
39
Q

What are the RNA splicing consensus site sequences?

A

GGU - AG

40
Q

What is the purpose of having a long sequence of circular double stranded DNA for transferring genes?

A
  • Place to put gene
  • Selection mechanism
  • Method of self-replication
  • Promoter (place to initiate translation)
41
Q

Describe advantages and disadvantages to the bacteria expression system

A
  • Advantage = easily transformed and selected, fast, can produce large quantity of protein rapidly; bacterial genome relatively well-characterized
  • Disadvantage = no post-translational modifications, protein might not fold properly, and many recombinant proteins become insoluble and are trapped in inclusion bodies

Year 3 - Biotechnology (11 decks)

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