Bentzen 6 - RNA molecules and mRNA processing (part 1)

Question 1

What is Crick’s 1958 proposition of colinearity between genes and protein? Was it correct?

Answer 1

That there are continuous sequences of nucleotides that encode continous sequences of amino acids.

The number of nucleotides in the gene is proportional to the number of amino acids in the protein. (eg. 3 nucleotides to 1 amino acid)

More or less correct for prokaryotes.

Question 2

What four discoveries led to the discovery of split genes?

Answer 2

1. mRNA is shorter than its DNA template
2. mRNA is not encoded as an equal colinear segment of DNA
3. mRNA is derived from segments of DNA called exons
4. Exons are separated by blocks of noncoding sequences called introns

Question 3

Introns are spliced from _____ to produce _____

Answer 3

Introns are spliced from pre-mRNA to produce mRNA

Question 4

Are introns transcribed?

Answer 4

Yes, they are transcribed to pre-mRNA

Question 5

How many introns does the average human gene have?

Answer 5

8-9

Question 6

How many introns can there be in a gene?

Answer 6

0 to 0

Question 7

What is the average length of a intron?

Answer 7

200nt - 5,000 nt

Question 8

Do genes have more DNA in introns or exons?

Answer 8

Most genes have more DNA in non-coding introns than exons

Question 9

Intron number and complexity in a species is loosely correlated with what?

Answer 9

Organismal complexity

Question 10

What is a prokaryotic operon? Why don’t eukaryotes have these? What is the consequence of this for prokaryotes?

Answer 10

A continuous array in DNA, with multiple protein-coding genes. Each operon has only a single transcription start site for multiple genes.

Eukaryotes have a start site for each gene, to yield a pre-mRNA that is processed into a functional mRNA encoding a single protein.

The consequence of this for prokaryotes are that they don’t have any introns (or very few) so that DNA is transcribed directly into colinear mRNA, which is then translated into protein while the mRNA is still being produced. No pre-mRNA in prokaryotes.

Question 11

What is the primary transcript in eukaryotes?

Answer 11

pre-mRNA

Question 12

Where is the ribosome binding site in mRNA?

Answer 12

About 7nt upstream of the start codon.

Question 13

What is the Shine-Dalgarno sequence?

Answer 13

A ribosome binding site in prokaryotes only.

Question 14

What are three modifications done to pre-mRNA to convert it to mRNA?

Answer 14

1. Capping of the 5’ end
2. Polyadenylation of the 3’ end
3. Splicing of introns

Question 15

What is 5’ capping of the pre-mRNA? What are the three steps (in order)?

Answer 15

A methylated guanine nucleotide is added to the 5’ - end of the pre-mRNA.

1. Phosphate group removed from 5’ end of the pre-mRNA
2. GMP is added
3. Methyl groups are added to the quanine base and the 2’ sugar of the first 2 RNA nucleotides. 1st base may also be methylated

Question 16

How is the quanine nucleotide added to the pre-mRNA in 5’ capping?

Answer 16

The unusual 5’ to 5’ linkage involves 3 phosphate groups.

Question 17

Why is modification of pre-mRNA necessary?

Answer 17

For efficient translation and transport of mRNA from nucleus, and protects mRNA from degradation.

Question 18

What is 3’ polyadenylation? The steps?

Answer 18

Where 50 to 250 adenine nucleotides are added to the 3’ end of the pre-mRNA.

1. Pre-mRNA is cleaved at a position form 11 to 30 nucleotides downstream of the consensus sequence in the 3’ untranslated region.
2. The addition of adenine nucleotides (polyadenylation) takes place at the 3’ end of the pre-mRNA generating the poly(A) tail

In pre-mRNA processing, a poly(A) tail is added through cleavage and polyadenylation

Question 19

Where does splicing take place?

Answer 19

In the nucleus on a spliceosome

Question 20

What different types of splicing are there? (4)

Answer 20

Splicing
Self-splicing
Spliceosomal
Enzymatic

Question 21

What type of splicing does nuclear pre-mRNA undergo?

Answer 21

Spliceosomal

Question 22

What does splicing require? At which sites of the pre-mRNA?

Answer 22

Consensus sequences. Three of them in the pre-mRNA at:

5’ splice site
3’ splice site
branch point

Question 23

How are introns recognized and removed?

Answer 23

Consensus sequences are used by spliceosomes to recognize and remove introns

Question 24

Introns are removed in what form?

Answer 24

In the form of a lariat.

Question 25

How is a lariat formed? (eg. between what two points)

Answer 25

Phosphodiester bond is broken and a new one is formed between a nucleotide of the 5’ splice site (G) and the branch point (A)

Question 26

What is a spliceosome? Its components?

Answer 26

A ribonucleoprotein complex (about 300 proteins and 5 small nuclear RNAs (snRNA)

It contains five snRNPs (small nuclear ribonuclear proteins), which are snRNAs and protein. These five components are labelled U1, U2, U3, U5 and U6 (no U3)

Question 27

What are the steps of spliceosome assembly?

Answer 27

1. snRNA U1 binds to the consensus sequence at the 5’ splice site
2. U2 binds to the branch point site of the pre-mRNA
3. A complex of U4, U5 and U6 joins the spliceosome
4. U1 and U4 are released
5. Interactions between the mRNA and snRNAs hold the spliceosome together

Question 28

How does polymerase II couple mRNA transcription and mRNA processing?

Answer 28

It mediates the events by the ‘tail’ (C-terminal repeat domain - CTD) of the largest subunit of Pol II.

mRNA processing enzymes are recruited to the CTD of PolII during transcription

Question 29

When is the 5’ CAP added to the pre-mRNA? How?

Answer 29

Just as it emerges from the polymerase. Capping enzymes are recruited to the C-termal domain (CTD) of RNA polymerase II during the early stages of transcription

Question 30

How is transcription terminated by polyadenylation?

Answer 30

1. Polyadenylation factors are recruited to the C-terminal repeat domain (CTD) of Pol II
2. RNA is cleaved at the 3’ cleavage site
3. Degradation of the remaining RNA by Rat1 terminates transcription