Lecture 16

Question 1

What is meant by the term isoform

Answer 1

An isoform is a protein variant that differs based on posttranscriptional modifications

Question 2

What process is used to create different protein isoforms

Answer 2

Alternative splicing

Question 3

What regions of the immature mRNA transcript are removed during splicing

Answer 3

Introns

Question 4

Splicing means that different proteins can be created from the same gene, T or F

Answer 4

T

Question 5

All eukaryotic genes contain introns and exons, T or F

Answer 5

F – yeast do not contain introns

Question 6

What components of genes allow for alternative splicing to occur

Answer 6

Optional introns and exons, mutually exclusive exons and internal splice sites

Question 7

40% of the Drosophila genome is alternatively spliced. What percentage of human genes are also spliced

Answer 7

0.75

Question 8

Other than splice donor and acceptor sites within gene transcripts, what other features of the mRNA allows for alternative splicing

Answer 8

Other sequences contained within the mRNA and the secondary structure also affects the choice of splice sites

Question 9

How is splicing regulated

Answer 9

By RNA binding proteins

Question 10

Give an example of a gene that undergoes alternative splicing

Answer 10

Dscam in Drosophila is a very large gene that produces a massive mRNA transcript containing 100 exons. The final mature mRNA will contain one exon from 12 A exons, 48 B exons, 33 C exons and 2 D exons, creating 38,000 splice variants of the dscam gene product. This is indicative of its role in the Drosophila nervous system

Question 11

What are the three key genes in determining sex in Drosophila and what are their roles

Answer 11

Sex lethal (sxl), a RNA binding protein and splicing repressor, transformer (tra), a RNA binding protein that acts as a splicing activator and, doublesex (dsx) a transcription factor.

Question 12

How are male Drosophila determined using alternative splicing and the interactions between the three sex determining genes

Answer 12

Male Drosophila have one X chromosome and this acts as the default pathway for sex determination in fruit flies. The transcripts for sxl and tra are spliced to give rise to inactive proteins. The dsx transcript is also spliced but this gives rise to a male specific transcription factor that acts as a transcriptional repressor of female-specific genes.

Question 13

How are female Drosophila determined using alternative splicing and the interactions between the three sex determining genes

Answer 13

Female Drosophila have two X chromosomes and a sex chromosome to autosome ratio of 1. The presence of two X chromosomes results in the transient activation of an alternative sxl promoter sequence which leads to the production of the sxl transcript which is then spliced and translated to form a splicing repressor. The sxl protein produced binds to other sxl transcripts and represses splicing by blocking binding of U2AF. This feeds back to result in more production of functional sxl transcripts. The sxl protein also binds to the tra transcripts causing an alternative splice that produces a functional tra protein after translation. The functional tra protein is a splicing activator and causes splicing of the dsx transcript. Splicing of the dsx transcript produces the female dsx transcript which is translated to the female dsx isoform. The female dsx protein is a transcriptional repressor of male specific genes.

Question 14

Give an example of how polyadenylation can act as a regulation of gene expression

Answer 14

The site of polyadenylation within the mRNA can be regulated. B lymphocytes for example can produce two different isoforms of an antibody. The antibody gene for a specific antigen has two possible positions for cleavage and polyadenylation. This determines whether the antibody is to be secreted or to remain membrane-bound. To produce the membrane-bound antibody, the cell produces the long transcript of the antibody. In this case the first stop codon within the antibody mRNA transcript is spliced out. This results in the translation of the transmembrane domain. Once the protein is secreted it remains tethered to the membrane via the transmembrane domain. For an antibody to be secreted the short transcript is produced which results in the loss of a splice acceptor site. Thus, no splicing of the transcript occurs and the first stop codon isn’t lost. This results in a termination of translation at the first stop codon, prior to the transmembrane domain region. This means that when the antibody is secreted it isn’t tethered to the membrane by a transmembrane domain.

Question 15

What is meant by the term leaky scanning

Answer 15

Sometimes the first AUG codon can be missed by the ribosome

Question 16

Sequences around the start codon help to initiate translation, T or F

Answer 16

T

Question 17

What is meant by the Kozak sequence

Answer 17

The Kozak sequence is the optimal translation initiation sequence that contains the start codon and ideal bases adjacent.

Question 18

Recall the Kozak sequence

Answer 18

ACCAUGG

Question 19

How can leaky scanning lead to the production of different protein products

Answer 19

If the sequence is less than optimal the ribosome can miss the first start codon and begin at the second or third AUG. These proteins will all be produced in the same reading frame, differing only by the sequence in the N-terminus

Question 20

High levels of what translation cofactor increase the probability that the first start codon will be recognised

Answer 20

eIF-4F

Question 21

How does the HIV virus make use of regulated nuclear transport to influence gene expression

Answer 21

After integration of the HIV genome into the host cell, the whole genome is transcribed as one piece of mRNA. Alternative splicing allows for the many different protein products to be made. Although the full-length mRNA is needed to make new virions the unspliced mRNA containing the entire virus genome cannot leave the nucleus. One of the proteins encoded by the virus genome is the rev protein. During the early stages of infection when only the alternatively spliced viral mRNA can leave the nucleus, the rev transcript moves through the nuclear pore and is translated. The rev protein then interacts with the nuclear pore in late stage infection to allow the exit from the nucleus of the unspliced mRNA.

Question 22

Signals within which regions of mRNA transcripts target them to particular parts of the cell

Answer 22

3’ and 5’ untranslated regions

Question 23

How are 3’UTRs recognised by cellular proteins which lead to the sequestration of RNA in one part of the cell

Answer 23

Intermolecular base pairing within the 3’UTRs form stem loops which are recognised by cellular proteins

Question 24

What is the role of ferritin in the regulation iron availability

Answer 24

Ferritin is a protein that stores iron inside the cell leading to a decrease in Fe availability

Question 25

What is the role of transferrin in the regulation iron availability

Answer 25

Transferrin is a receptor that imports iron into the cell leading to an increase in available Fe

Question 26

How does aconitase interact with ferritin and transferring mRNA

Answer 26

Aconitase binds to stem loops in the 5’UTR of ferritin mRNA and in the 3’UTR of transferrin mRNA

Question 27

Aconitase can also bind to iron itself, T or F

Answer 27

T

Question 28

Explain the role of aconitase in increasing Fe availability when iron levels are low

Answer 28

Aconitase binds to stem loops in the 5’UTR of the ferritin mRNA and blocks translation by physically blocking the ribosome from moving along the transcript. Aconitase also binds to stem loops in the 3’ UTR of transferrin mRNA and blocks its degradation. Binding of aconitase stabilises the transferrin mRNA thus increasing transferrin synthesis. Decrease ferritin translation and increase transferrin stability and synthesis results in an increase in intracellular [Fe]

Question 29

Explain the role of aconitase in decreasing Fe availability when iron levels are high

Answer 29

Aconitase binds to iron itself in the cytoplasm which causes a change in its conformation. A change in aconistase comformation causes it to dissociate from the stem loops in the 5’UTR of the ferritin mRNA and the 3’UTR of the transferrin mRNA. The now unstable transferrin mRNA transcript is then degraded quickly and the ribosome is now free to move along and translate the ferritin mRNA. This leads to a decrease in Fe availability

Question 30

What co-factor of the translation machinery is required for all mRNA translation

Answer 30

eIF-2

Question 31

When a cell is in the G0/quiescent phase of the cell cycle, globally, translation is turned down. How is this achieved

Answer 31

Phosphorylation of eIF-2 causes its tight binding to eIF-2B. eIF-2B is usually required as a guanine nucleotide exchange factor but tight binding of eIF-2B to the phosphorylated eIF-2 prevents it’s recycling by GTP displacing the bound GDP. By preventing guanine nucleotide exchange, eIF-2 is prevented from initiating translation.

Question 32

eIF-2 with its bound GTP binds to Met-tRNA to start ribosome scanning, T or F

Answer 32

T

Question 33

What are IRES sequences and how can they influence gene expression

Answer 33

Internal ribosome entry sequences are stem loops contained within RNA that can initiation formation of the ribosome independent of the Cap/PolyA initiation complex by mimicking it.

Question 34

IRES translation initiation is an effective way of encoding a second reading frame within RNA, T or F

Answer 34

F – this is not a very effective method with less than 10% of second ORF translation

Question 35

Which translation initiation cofactor is required to bind to the IRES stem loop and initiate translation independent of the 5’-Cap and PolyA tail

Answer 35

eIF-4G

Question 36

Where are IRES sequences commonly found

Answer 36

IRES sequences are often found in viral transcripts

Question 37

How can viruses use IRES sequences to promote translation of their transcripts

Answer 37

Viruses favour translation of their transcripts by cleaving eIF-4G in a way that prevents it from binding to eIF-4E but can still bind to IRES stem loops

Question 38

What is the significance of eIF-4G cleavage in apoptosis

Answer 38

eIF-4G is cleaved during apoptosis to prevent its binding to eIF-4E and hence prevent any more translation

Question 39

RNA stability is defined by the half-life of the different mRNAs and varies greatly, T or F

Answer 39

T

Question 40

How does the polyA tail change over time and act as a clock to determine mRNA age

Answer 40

The polyA tail usually starts at around 200 residues in length but as time goes by this is gradually degraded by exonucleases. Once it reaches 30 nucleotides in length it is de-capped and degraded. Hence the number of adenines in the PolyA tail can determine the age of the mRNA transcript

Question 41

How can mRNA half-lives be extended

Answer 41

Re-adenylation of the polyA tail

Question 42

Often factors that promote translation also promote mRNA degradation to prevent overexpression, T or F

Answer 42

F – de-adenylating nuclease (DAN) is responsible for de-adenylation of polyA tails and competes with eIF-4E binding to the mRNA cap.