Lecture 16 & 17

Question 1

How big is the human genome in diploid cells

Answer 1

6 x 10^9 bp
2 m/cell

Question 2

How is DNA protected

Answer 2

DNA is organized into histone octamers
Four histone proteins (H2A, H2B, H3, and H4) were found in equimolar amounts and H1 in a half molar amounts

Question 3

describe the nucleosome structure

Answer 3

Beads on a string
histone octamer is composed of a tetramer of H3-H4 and two dimers of H2A-H2B
DNA (200bp) wrapped around the histone octamer composed of the core histones (H2A, H2B, H3, and H4)
Nucleosomes are linked together by DNA with one bound histone H1

Question 4

Structure of the core histones

Answer 4

The core histones have long N-terminal histone tails that protrude from the histone octamer

Question 5

Why are histone tails important

Answer 5

Important for forming higher-order, more compact, forms of chromatin.
Histone tails are subject to many post-translational modifications that affect the compaction level and chromatin which affects its accessibility to enzymes involved in replication and transcription.
Histone tails facilitate inter-nucleosome interactions that result in higher-order packaging/compaction of the DNA

Question 6

What does the histone H1 do

Answer 6

Binds the nucleosome and organizes chromatin into a zig-zag structure. When H1binds it protect and additional 20-22 bp of DNA and now the DNA enters and exits the nucleosome on the same side

Question 7

Describe what happens without H1.

Answer 7

Without H1 the nucleosome wraps
147 bp of DNA and the DNA enters and exits on different faces

Question 8

Histone H1 allows assembly of chromatin into ____?

Answer 8

Histone H1 allows assembly of chromatin into a 30 nm fibre that increases the compaction afforded by the nucleosome by the factor of 35-40 times

Question 9

what are the two things required for 30 nm fibre formation

Answer 9

H1 histone and N-terminal tails of the core histones

Question 10

What are the two alternate arrangements of the 30nm fibre

Answer 10

Solenoid model (one-start helix)
Zigzag model (two-start model)

Question 11

What are SMC complexes

Answer 11

30 nm fibres are arranged in 40-100 kbp loops within chromosome by SMC complexes that affords a much greater degree of compaction than the 30 nm fibre on its own

increases compaction of the fibre

Question 12

What is gene expression

Answer 12

Series of events where the information in a DNA sequence is converted into an RNA product then to a protein product which performs biological function

Question 13

Includes 5 things

What is a gene

Answer 13

It is the entire DNA sequence that is necessary for the synthesis of a protein or RNA molecule

This includes: the promoter, coding region, untranslated regions, introns, transcription termination signal

Question 14

What is the amount of protein in a cell determined by

Answer 14

Abundance of mRNA
Efficiency of translation
Processing and stability of the protein

Question 15

What is the abundance of mRNA determined by

Answer 15

Rate of its synthesis (transcription of the gene)
The rate of mRNA degradation (mRNA stability)
The availability of the RNA molecule (mRNA sequestration)

Question 16

What is the purpose of prokaryotic gene control

Answer 16

Primarily to allow cells to optimize growth and division in response to changing environment

Question 17

What is the purpose of eukaryotic gene control

Answer 17

Primarily to regulate a genetic program that underlies embryonic development and tissue differentiation, but also to respond to changes in the environment

Question 18

What is a chromosomal scaffold

Answer 18

a proteinaceous scaffold that retains the overall structure of the chromosome after histones have been extracted

Question 19

Pol, mRNA processing, Compatmentation, regulation, chromatin, intron

Prokaryotic vs eukaryotic gene transcription (6 things)

Answer 19

3 types of RNA polymerases (1 in prokaryotes)

mRNA undergoes significant processing (mRNA not processed in prokaryotes)

Compartmentation of transcription and translation; gene expression is slower! (no compartmentation in prokaryotes)

Regulation is more complex (regulation is much simpler in prokaryotes – polycistronic mRNAs – more than one protein can be translated from a single mRNA)

Presence of introns (no introns in prokaryotes)

DNA is packaged into chromatin (no histones in prokaryotes)

Question 20

Which RNA polymerases drive the expressions of what classes of genes?

Answer 20

Pol I = pre-rRNA
Pol III = tRNAs and 5S rRNA
Pol II= everything else

Question 21

What is the most abundant product of transcription

Answer 21

pre-rRNA (80% of all transcription in eukaryotic cells)

Question 22

What is the pre-rRNA produced by RNAP I processed into

Answer 22

mature 5.8S, 18S and 28S rRNAs

Question 23

Where the RNAP I transcription occur

Answer 23

In the subcompartment of the nucleus called the nucleolus

Question 24

What are the two elements of the Pol I promoter

Answer 24

Core sequence and an upstream control element (UCE) located 100-150 bp upstream from the transcription start site

Question 25

What kind of transcription levels can be maintained using RNAP I

Answer 25

Low basal levels of transcription can occur from the core sequence using RNAP I and selectivity factor 1 (SL1 = TBP + 3 TAFs). More robust transcription requires binding of the upstream binding factor (UBF) to the upstream control element (UCE).

Question 26

Common features of Pol II core promoters

Answer 26

Pol II promoters share common features of
1) a TATA-box near -30
2) a TFIIB recognition element (BRE) upstream of TATA
3) an initiator sequence near the +1position and
4) downstream promoter element around the +30 position of the gene

Most pol II promoters also require other regulatory sequences called enhancers

Question 27

What is the most complex RNAP

Answer 27

Pol III

Question 28

two main types of Pol III promoters

Answer 28

1) for tRNA genes
2) 5S RNA genes

Question 29

Pol III promoter elements include _____

Answer 29

box A-C elements that reside downstream of the transcription start site (+1) that recruit transcription factors which then recruit TFIIIB and TBP to the promoter elements upstream of +1.
These then recruit RNAP III to the promoter

Question 30

Bacterial vs Eukaryotic RNAP II

Answer 30

Eukaryotic RNAP II must make mRNA
and miRNAs from a much larger genome than found in bacteria.

Eukaryotic genome is highly packaged in the form of chromatin, so is harder to access.

RNAP II is more complex than the bacterial RNAP.

Nonetheless, RNAP II, overall, is very similar in overall structure and function to the bacterial counterpart.

RBP1 & 2 are structurally similar to b’ and b, while RBP11 & RBP3 are similar to a1 and a2

Question 31

5 steps of transcription at Pol II promoters

Answer 31

1) The pre-initiation complexis formed and recruits RNAP II to the promoter.

2) A transcription bubble is formed

3) the C-terminal domain (CTD) of RNAP II is phosphorylated and some pre-initiation complex factors are released

4) elongation proceeds as in bacteria

5) transcription is terminated, the RNAP II CTD is dephosphorylated and the cycle is ready to restart

Question 32

Pol II transcription resembles bacterial transcription

Answer 32

True

Question 33

What does Pol II transcription require

Answer 33

Requires a large set of transcription factors

Mediator complex

Question 34

How does the mediator complex function

Answer 34

The mediator complex functions as an intermediary between the general transcription factors bound at the core promoter and more specific transcription factors
Mediator has ~20 subunits in yeast arranged into head, middle and tail domains -/+ a kinase complex that represses expression from a subset of genes. Mediator can facilitate the rate or efficiency of pre-initiation complex formation at the core promoter

Question 35

How is TBP recruited to genes

Answer 35

Binds TATA-box
only 25% of genes of TATA-box sequences
Other genes must be recruited to TATA-less promoters by TBP-associated factors (TAFs)