pro euk ll

Question 1

what are the aims of regulating gene expression

Answer 1

1. Cellular differentiation: by regulation of genes, different proteins are produced expressed, so that in a multicellular organism, all somatic cells carry identical genes but cells show a wide variation in structure and function
2. Adapt to changes: the proteins that are produced by an organism vary according to circumstances and demand
3. Conserve resources (transcriptional level control predominates as it is the most efficient mechanism with minimal wastage, especially in prokaryotes)
4. More varied proteome despite limited genome size, achieved through RNA splicing

Question 2

control at transcriptional level in prokaryotes by operons

Answer 2

Genes with related functions are clustered in an operon
 are controlled by a single promoter and operator region and
 transcribed into a single polycistronic mRNA and are expressed together

Question 3

control at transcriptional level in prokaryotes by repressors

Answer 3

negative gene regulation
Repressor binds to the operator, preventing RNA polymerase from binding to the promoter so transcription frequency decreases

Question 4

control at transcriptional level in prokaryotes by activators

Answer 4

Activator (in lac operon) is the activated Catabolite Activator Protein (CAP) which binds to the CAP binding site at the promoter of the lac operon and increases the affinity of RNA polymerase to the promoter, increasing transcription frequency (postive gene regulation)

(CAP is activated by binding to cAMP)

Question 5

control at translational level in prokaryotes by mRNA stability

Answer 5

prokaryotic mRNA has a short half life, resulting in rapid degradation

to control its half life, an anti sense mRNA complementary to the mRNA binds to it, and blocks its translation as well as target the RNA for degradation

Question 6

how is short half life beneficial to prokaryotes

Answer 6

enables prokaryotes to synthesise different proteins in response to changes in external environment

Question 7

result of formation of heterochromatin in eukaryotes

Answer 7

heterochromatin is highly compacted DNA where DNA is winded more tightly around histones
it limits access of RNA polymerase and general transcription factors to po

Question 8

result of formation of euchromatin in eukaryotes

Question 9

regulation at genomic level in eukaryotes by histone acetylation

Answer 9

1. addition of acetyl group to certain amino acids (e.g. lysine) in the histone by histone acetyltransferase/acetylase
2. removes +ve charge on histones
3. decreases electrostatic attraction between –vely charged DNA and histones
4. promoter region is more accessible to RNA polymerase and general transcription factors
5. promotes transcription as it promotes assembly of transcription initiation
   complex

Question 10

regulation at genomic level in eukaryotes by histone deacetylation

Answer 10

1. removal of acetyl groups from histones by histone deacetylase
2. restores +ve charge on histones
3. restores tighter interaction between DNA & histones
4. promoter region is less accessible to RNA polymerase and general transcription factors
5. inhibits transcription as it prevents assembly of transcription initiation complex

Question 11

regulation at genomic level in eukaryotes by chromatin remodelling complex

Answer 11

Chromatin remodeling complex are protein complexes which alter structure of nucleosomes temporarily
1) (-) can cause DNA to be more tightly coiled around histones, inhibiting transcription as it prevents assembly of transcription initiation complex
2) (+) can cause DNA to be less tightly coiled around histones, promoting transcription as it promotes assembly of transcription initiation complex

Question 12

control at transcriptional level in eukaryotes by repressors

Answer 12

Repressors bind to silencers, allowing bending of spacer DNA and allowing repressor bind to general transcription factors hence preventing activators from
binding to general transcription factors
this prevents assembly of transcription initiation complex and transcription frequency decreases

(Bound repressor may recruit histone deacetylase and repressive chromatin
remodeling complex to decrease accessibility of promoter to general transcription
factors and RNA polymerase)

Question 13

control in transcriptional level in eukaryotes by activators

Answer 13

Activators bind to enhancers,
resulting in bending of spacer DNA to allow binding of activators with RNA polymerase and/or GTFs at the promoter, promoting assembly of transcription initiation complex and transcription frequency increases

(Bound activator may recruit histone acetyltransferase and chromatin remodeling
complex to increase accessibility of promoter to general transcription factors and RNA
polymerase)

Question 14

how can repressors interfere with action of activators

Answer 14

1. competitive DNA binding: enhancer region overlaps with silencer region so binding of reporessor prevents binding of activator
2. making activation surface: repressor binds to activator to prevent it from interfering with general transcription factors
3. direct interaction with general transcription factors: repressor interacts with general transcription factors to prevent assembly of transcription initiation complex

Question 15

role of control at post transcriptional level in eukaryotes by addition of 5’ cap

Answer 15

• Addition of a 7-methylguanosine nucleotide is added to the 5’ end of the pre-mRNA and occurs shortly after transcription begins
• helps the cell to recognize mRNA (amongst other RNAs) so that subsequent steps such as splicing and polyadenylation can occur
• 2. signal to export mRNA out of nucleus
• 3. protects the growing pre-mRNA chain from degradation by ribonucleases
• 4. promotes initiation of translation as it is recognized by translation initiation factors that help recruit mRNA to small ribosomal subunit

Question 16

describe control at post transcriptional level in eukaryotes by intron splicing

Answer 16

1. splicing: the introns are excised and exons are joined together by spliceosomes which recognize the sequences at intron-exon boundaries so that functional proteins can be produced
2. Alternative splicing: where different exons of a single pre- mRNA can be joined together such that different mature mRNAs from 1 gene and different proteins can be produced

Question 17

role of control at post transcriptional level in eukaryotes by polyadenylation

Answer 17

• (role) -
  1. signal to export mature mRNA out of nucleus to cytoplasm through nuclear pore
• 2. enhances mRNA’s half life and protects mature mRNA from degradation by ribonucleases so that more proteins can be made
• 3. interacts with initiation factors, together with 5’cap for initiation of translation

Question 18

describe polyadenylation

Answer 18

polyadenylation: 3’end of pre-mRNA is cleaved by endonucleases to make it shorter and a poly-A polymerase recognises the polyadenylation signal (AAUAAA) and adds a long
sequence of adenine nucleotides to 3’ end of the pre-mRNA, forming a poly(A) tail
this occurs immediately after transcription

Question 19

control at the translational level in eukaryotes by mRNA stability

Answer 19

mRNA half-life is determined by the length of its poly-A tail. The longer the poly-A tail, the longer the half life of the mRNA and the longer the mRNA can be used as a template to make proteins.

degradation: The poly-A tail is removed by ribonucleases in the 3’ to 5’ direction until a critical length is reached which will triggers removal of the 5’cap and degradation of the mRNA from the 5’end too.

Question 20

control at translational level in eukaryotes by antisense RNA

Answer 20

anti sense RNAs which are complementary to the part of the mRNA to be degraded will bind to the mRNA to form a double stranded RNA, which will block translation of the mRNA and will be a target for degradation by nucleases

Question 21

control at translational level in eukaryotes by formation of translation initiation complex
(binding of translational repressors)

Answer 21

1) During translation initiation,
small ribosomal unit, eukaryotic initiation factors and initiator tRNA form a complex which binds to the 5’ cap and the 3’ poly A tail causing the mRNA to circularise
small ribosomal subunit binds to 5’cap of mRNA

This can be prevented by binding of translational repressor to (1) 5’cap, (2) 5’UTR (untranslated region), (3) 3’ UTR
which interferes with the interaction between the 3’ poly-A tail, the 5’cap, eukaryotic initiation factors, and the small ribosomal subunit which are needed for translation initiation

2) .

Question 22

control at translational level in eukaryotes by formation of translation initiation complex
(availability of initiation factors)

Answer 22

During translation initiation, translation initiation factors facilitate the binding of the small ribosomal subunit to the 5’cap

The availability of translation initiation factors is determined by whether or not they are phosphorylated. Some initiation factors are activated by phosphorylation while others
are inactivated by phosphorylation. Without activated translations initiation factors, translation cannot begin

Question 23

controls at post translational level in eukaryotes by formation of functional proteins

Answer 23

Covalent modification/cleavage (eg. attachment of prosthetic groups, glycosylation, disulfide bond formation) of polypeptides in the RER and Golgi apparatus make them functional proteins

Question 24

control at post translational level in eukaryotes by regulation of protein activity

Answer 24

Phosphorylation / dephosphorylation of translation initiation factors can activate / deactivate the protein and hence up / down regulate its activity.

Question 25

control at post translational level in eukaryotes by protein degradation

Answer 25

Protein degradation by proteasomes determines how long a protein remains in a cell.
Proteins targetted for degradation are tagged with ubiqutin (by ubiquitin ligase) and then recognised and degraded by the proteasome.