Outcome 2 Flashcards

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1
Q

Describe genome strucutre in a prokaryote

A

Prokaryortic cells are haploid (one copy of every gene)
The prokaryotic genome is composed of a circular chromosome and often contiains a plasmid. The chromosome contains all of the genes required for growth and reproduction of the bacterium while the plasmid carries genes that provide a survival advantage i.e. resistance to an antibiotic. The chromosome does not contain much non-coding DNA and genes are located close to each other. Genes with related functions tend to be located close to each other on the chromosome and are under the control of a single promoter in arrangements called operons.

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2
Q

Descrime the organisation of the eukaryotic genome

A

In Eukaryotes, genes are divided between a number of separate, linear chromosomes. Each structural gene has its own set of response elements to which transcription factors bind. The chromosome consists of telomeres, centromemeres and DNA wrapped around histone proteins into nucleosomes. The chromatin can be loosely or tightly packed. There is an excess of junk DNA interspersed amongst the genes.

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3
Q

What is an operon

A

A set of adjacent structural genes which are functionally related as well as a shared promoter and regulator genes.

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4
Q

How is gene expression regulated in prokaryotes

A

At the transcriptional level by regulatory proteinds which bind to the sequences which control the transcription of structural genes.

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5
Q

Describe expression regulation in the Lac operon

A

The lac operon is an inducible operon. When lacose is absent the constitutively produced repressor protein binds to the operator preventing RNA polymerase from binding to the promoter. When lactose or IPTG are present they bind to the repressor protein, preventing the repressor protein from binding to the operator and allowing RNA polymerase to bind to the promoter.

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6
Q

Give an example of a repressible operon

A

The trp operon is a classic example of a repressible operon. When tryptophan accumulates, tryptophan binds to a repressor, which then binds to the operator, preventing further transcription.

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7
Q

At what three levels is gene expression regulated in eukaryotes

A

Transcpitional, post-transcriptional and post-translational

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8
Q

How are eukaryotic gene expression regulated at transcriptional level

A

By chromatin packing or by regulatory sequences. Regulatory sequences are dsDNA sequences that control transcription by bonding to regulator proteins.Contain promoter, enhancer, silencer operator and 5’utr,

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9
Q

What are introns and exons

A

In eukaryotes after transcription a primary mRNA transcript is formed in the nucleus that contains introns and exons.
Introns are intervening non-protein coding sequences
Exons code for protein expression.

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10
Q

What is a spliceosome

A

A large strucuture which contains proteins which are responsible for splicing.

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11
Q

How does a spliceosome know where to splice

A

It recognises splice sites - GU and AU which are cut.

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12
Q

What is alternative splicing

A

Not all exons in primary mRNA are necessarily incorporated into the final mRNA transcript. By altering which exons are spliced together, its possible to make different proteins. So alternative splicing allows for the production of different proteins from one gene.

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13
Q

Describe the poly(A) tail and its purpose

A
The poly(A) tail is a large sequence of adenosine around 200 bases long added to the 3' end of the primary mRNA transcript in the nucleus.
It protects the mRNA from degradation in the cytoplasm and allows mRNA to enter ribosomes.
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14
Q

Describe the 5’caps purpose

A

Protects the mRNA from degradation and allows entry into the ribosomes for translation

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15
Q

What does post-translational regualtion consist of

A

The control of levels of active protein in the cell proteolysis and post-translational modifications are type of post-translational regualtion.

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16
Q

3 examples of post-translational modification

A

Glycosylation, phosphorylation and acetylation- adding sugar groups, phosphate groups and acetyl groups respectively

17
Q

How can gene expression in a cell be measured

A

By analysing the type of and number of mRNA transcripts per cell.

18
Q

Describe the process of extracting mRNA from cells

A

Cells are lysed into buffer containing agents that disrupt cell structure, disassociate proteins and inhibit RNases.
Cell lysate is loaded onto an affinity column containing immobilised oligo(dT). The column is washed through with a high salt buffer. The mRNA is able to bind to the oligo (dT). The cell components which are not bound to the oligo(dT) are washed out and removed.
The mRNA is eluted from the column by lowering the salt concentration of the buffer disassociating the mRNA from the oligo(dT)
The mRNA can be analysed by agarose gel electrophoresis.

19
Q

How is RNA purity and concentration determined

A

By spectrophotometry - nucleic acid absorbs light strongly between 259nm and 271nm. By dissolving RNA in water and measuring absorbancy at 260nm

20
Q

Why is RNA difficult to work with

A

RNA is unstable and degrades easily. RNases are found in all cel types, in the air and on surfaces in the lab. RNases absorb strongly to glass

21
Q

What steps should be taken when working with RNA

A

Gloves worn at all times, DIsposable plastics used. All RNA solutions should be treated with diethylpyrocarbonate (DEPC). Work carried out at 4C if possible.

22
Q

What is DEPC

A

A RNase inhibitor that can be added to solutions and left shaken for 2 hours. It can then be removed by breaking down at 70C for an hour.

23
Q

How should RNA be stored

A

At -70C if a highly pure RNA sample

At -20C if in ethanol solution

24
Q

What is cDNA

A

Complementary DNA - DNA that is originally synthesised from mRNA as a single strand of DNA complementary to the mRNA sequence.

25
Q

WHat is the process of cDNA synthesis called and what enzyme is used?

A

Reverse transcription

Reverse transcriptase

26
Q

What is reverse transcriptase and what does it requiere

A

A type of DNA polymerase that is able to catalyse the polymerisation of a DNA strand using RNA as a template. Originally found in RNA viruses.
Requires a primer with a free 3’ OH group

27
Q

Describe first strand cDNA synthesis

A

A mixture of poly(A)mRNA, reverse transcriptase, dNTPs, oligo dT primers is prepared.
The mixture is incubated in a buffer containing Mg2+ at 42C. The oligo dT primer binds to the poly A sequence of the mRNA, the reverse transcriptase is able to use the oligo dT primer to catalyse polymerisation of the first cDNA strand. Creating a RNA/cDNA duplex

28
Q

Describe the synthesis of the second cDNA strand

A

The RNA/cDNA duplex is exposed to RNaseH which is an RNase capable of creatingg single stranded nicks and gaps in the mRNA by breaking phosphodiester bonds. The remaining short sequences of RNA can acts as a primer. DNA polymerase is added and synthesises the second strand. DNA ligase is then added to ligate the remaining gaps in the second cDNA and completing synthesis of the cDNA molecule

29
Q

Why are alternative primers used

A

Reverse transcriptase can “fall off” the mRNA chain. Because oligo dT primers starts at the 3’ end of the mRNA chain it can make producing a full length cDNA strand difficult.

30
Q

What are specific primers

A

Primers designed when a full or partial nucleotide sequence of the specific mRNA is known.

31
Q

What are random primers

A

A mixture of random hexanomers which contain all possible hexanucleotides.

32
Q

What is the overview of making a cDNA library

A
  • Isolate and purify the mRNA from the cells of interest
  • Synthesise cDNA from the mRNA
  • Add linkers to cDNA to convert blunt ends into sticky ends
  • Ligate cDNA into vector and transform with e.coli
  • Screen the transformants for clones of interest
33
Q

Why are DNA linkers used

A

The formation of the second cDNA strand will result in blunt ends being formed. Linkers are small pieces of DNA used to convert blunt ends into sticky ends.

34
Q

Describe inserting oligo linkers into cDNA

A

Short oligonucleotides that have been designed to contain recognition sites are synthesised. Since restriction sites are palindromic the oligonucleotides will form dimers. They initially contain OH groups which will not ligate, but these are replaced with phosphate groups by t4 kinase. The oligos are blunt end ligated to the cDNA. After ligation the cDNA can be cut with restriction enzymes creating stick ends.

35
Q

Describe inserting homopolymer linkers into cDNA

A

The enzyme terminal trasnferase will randomly add nucleotides to the 3’ends of the cDNA indepently of template sequence. If the terminal transferase has access to only one dNTP (eg dCTP) it will add a poly C tail to the cDNA. If a vector was treated with a terminal transferase that only had dGTP then it would have a poly G tail.

36
Q

What problem does directional cloning solve

A

If both ends of a cDNA have the same linker it is possible for the cDNA to insert into the vector in either orientation but only one orientation is correct for expression. Directional cloning produces cDNA that can only insert into a vector in one orientation

37
Q

How is directional cloning acheived

A

By using oligo dT primers with added nucleotides to their 5’ end which contain a restriction recogntion site. After the RNA is removed terminal transferase and one dNTP can be added to created a different linker at the other end of the strand
When DNA polymerase synthesises the second cDNA strand it will form two different recognition sequences. at each end of the cDNA.

38
Q

What are the three methods for screening a cDNA library

A

DNA hybridisation
PCR screening
Antibody screening

39
Q

How are antibodies used to screen cDNA libraries

A

WHen a cDNA is used to generate an expression library andtibodies can be introduced to screen the expression library by recognising specifc protein sequences.