Block D SAU opgaver Flashcards
The DNA of a specific bacterial cell contains 13% adenine. What are the percentages of the other nucleotides?
Chargaff’s rules state that DNA contains equal amounts of A and T and of G and C.
According to this:
A = T= 13%
G = C = 37%
The two strands of a DNA double helix can be separated by heating. lf you raised the temperature of a solution containing the following three DNA molecules, in what order do you suppose the strands would dissociate (“melt”)? Explain your answer.
a.
5’-GCGGGCCAGCCCGAGTGGGTAGCCCAGG-3’
3’-CGCCCGGTCGGGCTCACCCATCGGGTCC-5’
b.
5’-ATTATAAAATATTTAGATACTATATTTACAA-3’
3’-TAATATTTTATAAATCTATGATATAAATGTT-5’
c.
5’-AGAGCTAGATCGAT-3’
3’-TCTCGATCTAGCTA-5’
See answer in ECB 4th edition question 5-8.
- Make a schematic drawing of an interphase chromosome.
a. Place the localization of replication origins, telomeres and centromere. What is the function of replication origins, telomeres and centromere?
b. What is the difference between a chromosome and a chromatid?
a. Figure 5-15 in ECB 4th edition.
Replication origin: Nucleotide sequence at which DNA replication is initiated.
Telomeres: Repetitive nucleotide sequence (TAAGGG) that caps the ends of linear chromosomes
(approximately 3000 repeats per chromosome in human cells).
Centromere: Specialized DNA sequence that allows duplicated chromosomes to be separated during M phase;
can be seen as the constricted region of a mitotic chromosome.
b. A chromatid is either of the two strands (sister chromatids) joined together by a single centromere, formed
from the duplication of the chromosome during the early stages of cell division (the typical “X” seen in figure
5-16). The sister chromatids separate to become individual chromosome during the late stages of cell division.
- The length of the DNA double helix of the human (haploid) genome is 1 meter. The haploid genome contains 3x109 base pairs.
a. What is the “thickness” (height) of each base pair (the distance between the base pairs) in the human genome ?
b. Discuss the chromatin structure, with specific focus on the nucleosome structure, in the human genome (see figure 5-23 in ECB, 5th edition).
c. The 30-nm chromatin fiber contains about 20 nucleosomes (200 bp/nucleosome) per 50 nm of length. Calculate the degree of compaction of DNA associated with this level of chromatin structure.
d. What fraction of 10,000-fold condensation that occurs at mitosis does this level of DNA packing represent?
The histones making up the nucleosome are chemically modified.
e. Describe the chemical modifications of the histones and the functional implications of these modifications.
f. H3 tails in euchromatin are usually heavily acetylated at lysine 9 and 14. Explain what consequences acetylation of lysine would have for histone-nucleosome interaction.
g. What is the function of chromatin-remodeling complexes?
a. The (haploid) genome contains 3 x 109 base pairs. Therefore the “thickness” (height) of each base pair is
1m/3 x 109 = 0.333 nm (= 3.33 Ångstrøm (Å)).
b. See Figure 5-21 and Figure 5-24 in ECB 4th edition. Focus on the structure of the nucleosomes and the 30 nm
fiber.
c. (20 nucleosomes x 200 bp/nucleosome x 0.34 nm/bp)/50 nm = 27-fold.
d. This corresponds to 0.27% of the 10.000-fold condensation of the human mitotic chromosomes.
e. The chromatin structure can be altered by reversible chemical modifications of the histones. The tails of all
four of the core histones are particularly subjected to these covalent modifications. Acetyl, phosphate and
methyl groups can be added to or removed from the tails by enzymes. For example acetylation of lysine
residues can reduce the affinity of the tails for adjacent nucleosomes thereby loosening chromatin structure
and allowing access to particular nuclear proteins. Most importantly these modifications can serve as docking
sites on the histone tails for regulatory proteins. Different patterns of modifications (“the histone code”)
attract different proteins some of which promote chromatin condensation whereas others decondensate
chromatin facilitating access to the DNA.
f. Acetylation of lysine residues can reduce the affinity of the tails for adjacent nucleosomes thereby loosening
chromatin structure and allowing access to particular nuclear proteins. In general, interactions influenced by
histone modifications can either be intra-nucleosomal (histone-DNA interactions) or inter-nucleosomal
(histone-histone interactions). What is important here is that lysine residues, which has a positive charge at pH
7, can interact with DNA and/or acidic amino acids and these interactions will be weakened after acetylation.
g. Chromatin–remodeling complexes are protein machines that use the energy of ATP hydrolysis to change the
position of the DNA wrapped around nucleosomes. The complexes which attach to both the histone octamer
and the DNA wrapped around it, can locally alter the arrangement of nucleosomes on the DNA, making the
DNA either more accessible or less accessible to other proteins in the cell including those involve in
transcription.
- In the experiment presented in the figure, agarose gel electrophoresis was used to demonstrate the organization of DNA in nucleosomes. First, nuclei were obtained from cells by cell rupture followed by separation of nuclei by centrifugation. Next, a nuclease (Micrococcus nuclease - (MNase)) was added to the nuclei. MNase digests any accessible DNA, irrespective of its sequence, into nucleotides. However, if the DNA is bound to proteins, it is protected from the action of the nuclease. After incubation of the nuclei with different concentrations of MNase, the DNA was isolated from the samples and analysed by gel-electrophoresis. For comparison, samples of de-proteinized DNA (naked DNA) were digested with MNase and analysed in parallel.
a. Describe how gel-electrophoresis of DNA is performed and how the DNA is visualized.
b. How will you describe the DNA in the untreated samples (lanes 1 and 5)?
c. The naked DNA treated with MNase has a “smeary” appearance. Why is that? Upon treatment of the naked DNA with a high concentration of MNase (lane 4), the DNA has seemingly disappeared. What has happened?
d. The DNA isolated from MNase-treated chromatin forms a “ladder” (lane 6 and 7). What are the steps in the ladder? These steps (“bands”) are fuzzy in appearance compared to the bands in the marker. Why?
e. Treatment of chromatin with a high concentration of MNase results in a single band (lane 8). What does this single band represent?
f. How can the length of the nucleosomal repeat (nucleosome plus linker) be determined from the gel?
Amazingly, the nuclear DNA is degraded in a similar fashion when cells undergo a form of programmed cell death known as apoptosis ( discussed in Sau E). In fact, staining kits have been developed to detect apoptotic cells in tissue sections based on labelling of the DNA ends left by the endogenous DNase.
a. The principle in gel electrophoresis is described on the pages 327-329 in ECB 4th edition.
b. High molecular weight DNA.
c. The naked DNA has been cleaved by the MNase at random into many fragments of variable size. At the
highest MNase concentration the DNA has been cleaved into small fragments that have migrated out of the
gel.
d. The first “step” of the ladder is the DNA found in the nucleosome core particle. The next is DNA from a dinucleosome
particle consisting of two nucleosomes and the connecting linker and so on. These DNA bands are
not as sharp as the bands in the marker lane (M) because the linker is of variable size and because the
“trimming” of the ends by the MNase may vary.
e. The single band represents DNA found in the nucleosome core particle.
f. Size of second “step” – size of the first “step” of the ladder (i.e. di-nucleosome minus mononucleosome).
In the lab-exercise II a similar experiment is performed in order to determine the nucleosome repeat length in
DNA isolated from pig small intestine.
a. Draw the newly synthesized ”leading” and ”lagging” strands with orientation in the figure.
b. Indicate on the figure where the replication of the newly synthesized DNA strands is initiated.
a. See the orientation of the leading and lagging strand on the figure.
b. The replication starts in the middle of the replication origin (bobble) (see mark on the figure).
- Describe DNA replication with the help of the figure and the following terms:
Replication origin Replication fork DNA denaturation Helicase
Topoisomerase Okazaki fragment SSB proteins Lagging strand
DNA ligase DNA polymerase Proofreading Leading strand
DNA primase RNAse 3’-5’ exonuclease Clamp
Clamp loader 5’ – 3’ for template strands Primer
See figure 6-19 on page 208 in ECB 4th edition.
The speed of a eukaryotic DNA polymerase is about 100 nucleotides per second and the estimated
number of origins of replication in a human cell is 10.000. Recall that a human cell contains two copies
of the human genome, one inherited from the mother and one inherited from the father, each
consisting of 3 x 109 nucleotide pairs (bp).
a. How fast could the human genome be replicated if all origins of replication were in use at the same time?
b. What is the speed of an E. coli DNA polymerase if the genome (4.6 x 106 bp) is replicated once every 40
minute?
a.
Nucleotide polymerisation from 10.000 origins of replication per minute:
4 polymerases x 100 nucleotides/second x 10.000 x 60 seconds/min = 0.24 x 109 nucleotides/min
The human genome is 6x109 bp = 12x109 nucleotides
The time required to replicate the human genome:
12x109 nucleotides / 0.24 x 109 bp/min = 50 min
The replication of DNA in the nucleus of a typical proliferating human cell last for approximately 8
hours. This is because not all origins of replication are in use, not all origins of replications are
initiated at the same time, the distance between the origins of replication varies and the DNA polymerase stalls when DNA repair is required. In addition, numbers of origins of replication and
the speed of the DNA polymerase are estimates.
b. 4.6 x 106 bp x 2 nucleotides/bp / (40 min x 4 polymerases x 60 sec/min) = 958 nucleotides/sec.
Explain why telomeres and telomerase are needed for replication of eukaryotic chromosomes but not
for replication of a circular bacterial chromosome.
See answer to 6-14 and figure 6-22 and 6-23, EBC 5th edition.
- Below are five different types of DNA damage (A-E):
A
5’-AATCGTACCGCTTAGCTGACTGA-3’
3’-TTAGCATGGUGAATCGACTGACT-5’
a. What has happened here?
b. What is the cause of the damage and could it have been avoided?
c. How is the damage repaired (name the steps in the repair process)?
d. What is the name of this type of DNA repair?
B
5’-AATCGTACCGCTTAGCTGACTGA-3’
3’-TTAGCATGGCGAATCG-CTGACT-5’
e. What has happened here?
f. What is the cause of the damage and could it have been avoided?
g. How is the damage repaired (name the steps in the repair process)?
h. What is the name of this type of DNA repair?
C
5’-AATCGTACCGC AGCTGACTGA-3’
3’-TTAGCATGGCGAATCGACTGACT-5’
i. What has happened here?
j. What is the cause of the damage and could it have been avoided?
k. How is the damage repaired (name the steps in the repair process)?
l. What is the name of this type of DNA repair?
D
5’-AATCGTACCGCTTAGCTGACTGA-3’
3’-TTAGCGTGGCGAATCGACTGACT-5’
m. What has happened here?
n. What is the cause of the damage and could it have been avoided?
o. What is the requirement for the cell to repair this type of DNA damages?
p. How is the damage repaired (name the steps in the repair process)?
q. What is the name of this type of DNA repair?
E
5’-AATCGTACCG AGCTGACTGA-3’
3’-TTAGCGTGGC TCGACTGACT-5’
r. What has happened here?
s. What is the cause of the damage and could it have been avoided?
t. How is the damage repaired (name the steps in the repair process)?
u. What is the name of this type of DNA repair?
A. a. Spontaneous loss of an amino group from a cytosine produces the base uracil.
b. Spontaneous hydrolysis and it cannot be avoided.
c. Excision of the damage, re-synthesis of the original sequence and ligation.
d. Excision repair.
B. e. Depurination – loss of a purine base; A or G.
f. Spontaneous hydrolysis and it cannot be avoided.
g. Excision of the damage, re-synthesis of the original sequence and ligation.
h. Excision repair.
C. i. Generation of a pyrimidine dimer (T-T, T-C or C-C).
j. Ultraviolet radiation promotes covalent linkage between two adjacent pyrimidine bases. This could
be avoided by no exposure to sunlight.
k. Excision of the damage, resynthesis of the original sequence and ligation.
l. Excision repair.
D. m. A wrong base pair (T-G) is found in the DNA sequence.
n. The DNA polymerase has made a replication error and it cannot be avoided (1 mistake/107
nucleotides copied).
o. The repair machinery must be able to recognize which of the two mismatched bases is the newly
synthesized in order to differentiate between the correct and the incorrect base.
p. Using a DNA repair system which is able to distinguish between the new and old DNA strand.
q. DNA mismatch repair system.
E. r. A double strand break is found in the DNA sequence.
s. Radiation or chemical assaults can fracture the backbone of DNA and it is difficult to avoid.
t. Cleaning of the broken ends and ligation (non-homologous end joining /quick and dirty
mechanism) or homologous recombination
u. Non-homologous end joining or homologous recombination (fig. 6-29 page 216 in ECB 4th edition).
- Many hereditary diseases are caused by mutations arisen from spontaneously hydrolytic deamination of
5-methylcytosin. Repair of such mutations is very difficult and often not performed at all. Why?
Hint: make a drawing of 5-methylcytosin and the base produced after deamination.
Methylation of cytosine produces 5-methylcytosin. If 5-methylcytosin undergoes spontaneous deamination
the base thymine (T) is produced (examine panel 6-2 page 76 in ECB 4th edition).
It is very difficult for a DNA repair system to distinguish between a naturally occurring T and a T which has
been produced by deamination of 5-methylcytosin and therefore the damage is often left unrepaired.
- Polymerase chain reaction (PCR) is a method in which it is possible to amplify a specific DNA
sequence. A commonly used variant of PCR is RT-PCR (reverse transcription PCR) in which RNA
is first converted to cDNA by the used of the enzyme reverse transcriptase (see figure below).
a. Discuss the principle in RT- PCR by the use of the following terms and the figure:
Denaturation Hybridization (Annealing) Extension
Upstream primer Downstream primer Taq polymerase
Reverse transcriptase dNTPs
a. PCR and RT PCR are described on pages 335-339 in ECB 4th edition. The methods are also
described in the theory section in lab-excersize I.
b. Downstream primer 1: 5’ – CAGCCCGTAGTTCTTG – 3’
Upstream primer 2: 5’- GCCTGCCCTGGCTTCA – 3’
- Make a pie chart of the human transcriptome divided into
rRNA, mRNA, tRNA and other small RNAs.
rRNA: 80%, mRNA: 1-2%, tRNA and other small RNAs: 5-10%. The remainder are mostly various types of ncRNA.
- A stretch of DNA that is transcribed is referred to as a transcription unit (TU). This is a useful term because methods exist to precisely map the ends of a transcription unit in contrast to the borders of a gene. Thus, the term “gene” is used in a loose sense in moleclular biology. The figure introduces the basic terminology associated with transcription units. Fill in the terms in the dashed boxes.
- A gel electrophoretic analysis of whole cell RNA from eukaryotic cells is shown in the figure below. The RNAs were separated on an agarose gel and stained using an RNA-binding dye.
a. Locate the RNA main types discussed in problem 1 on the gel.
(hint: rRNA comprises the Small Subunit rRNA (SSU; also known as 18S), the Large Subunit rRNA (LSU; also known as 28S), and 5S and 5.8S rRNA.
b. Would RNA extracted from different tissues be similar or different in this type of analysis?
c. The migration of the RNA measured from the top is related to the size (chain length) of the RNA. The intensity of the signal is related to the amount of the RNA. Is the gel analysis consistent with your pie chart?
d. Why is the upper band more intense than the middle band?
e. Why is the lowest band fuzzy?
a. The upper band is the Large Subunit ribosomal RNA (LSU rRNA), the middle band is the Small Subunit ribosomal RNA (SSU rRNA), and the lower band a pile-up of various small RNAs including 5.8S and 5S ribosomal RNAs, tRNAs and many others. The underlying smear (not noticeable on this gel) is mostly mRNAs of varying sizes.
b. The RNA would look essentially the same irrespective of the tissue source.
c. At least it is consistent with the correct answer to the pie chart shown above.
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d. The LSU and SSU rRNAs are stoichiometrically 1:1. However the sizes are 5035nt:1871nt (human). Thus, there is 2-3 times more RNA in the upper band. If the ratio is less than expected this might be due to unequal degradation of LSU and SSU.
e. The lower band is fuzzy because it includes RNAs of different size (see explanation for 3a).
- A gel electrophoretic analysis of whole cell RNA from human liver (1), brain (2), kidney (3), spleen (4), pancreas (5) and intestine (6) is shown in the drawing below to the left. A northern blot analysis was performed with probes directed against mRNA for insulin (lower band) and actin (upper band) to analyze insulin expression (drawing below to the right).
a. Briefly describe how a northern blotting analysis is performed.
b. Insulin mRNA is only detected in lane 5. Why?
c. Beta-actin mRNA bands are detected in all lanes. Why?
d. Why it is necessary to visualize beta-actin mRNA in this experiment?
a. A northern blot analysis is used for detection of specific RNA sequences. A mixture of RNAs is separated according to length by gel electrophoresis where after a sheet of nitrocellulose paper is laid over the gel and the RNAs are transferred to the sheet by blotting (simple migration or forced migration by applying a voltage gradient). The sheet containing the bound RNAs is exposed to a radioactive, single-stranded DNA probe (usually) specific for the RNA sequence of interest under conditions that favour hybridization. The sheet is washed thoroughly, so that only probe molecules that have hybridized to the RNA on the sheet will show up as a band on the autoradiograph (other detection methods are available).
b. Because in the adult insulin is almost exclusively expressed in the pancreas (although low amounts of insulin has been detected in several other tissue types).
c. Because all tissues (present) expresses actin.
d. Detection of actin can be used as loading (and blotting) control because most tissues expresses similar amounts of actin mRNA.
- Compare DNA replication and transcription based on the figure. List three similarities and three differences between the two processes
Similarities: 1) template-dependent synthesis according to the Watson-Crick base pairing principle 2) the building blocks are nucleoside triphosphates that are linked by 3’to 5’ phosphodiester bonds 3) the direction of synthesis is 5’ to 3’.
Differences: 1) replication comprises the entire template; transcription only selected parts (the transcription units) 2) replication requires an RNA primer *3) replication has proof-reading whereas no proof-reading is associated with transcription (although back-tracking occurs). This can be related to the fundamentally different roles of the two processes. *Notice that the primer in the figure consists of deoxyribonucleic acid and the figure therefore neither represent replication nor transcription.
The transcription process can be visualized by electron microscopy (“Miller spreads”). The figure shows transcription of a ribosomal RNA gene by RNA polymerase I. These ribosomal genes are among the most abundantly transcribed genes in the human genome. Many protein encoding genes (RNA polymerase II genes) have only a single polymerase transcribing the gene at any time.
a. Are the RNA polymerases moving from right to left or from left to right?
Prior to transcription, the RNA polymerase is recruited to the transcription start site. This is dependent on the presence of a promoter (a stretch of DNA with regulatory function) to which the polymerase binds as well as other DNA stretches that bind factors that stimulate or inhibit polymerase recruitment. A stringent definition of a promoter is the stretch of DNA to which the polymerase and the basal (general) transcription factors (the “TFII’s” – i.e. transcription factors for RNA polymerase II) bind. A larger DNA region encompassing the promoter and DNA stretches that bind stimulatory or inhibitory factors (the specific transcription factors) is then referred to as the promoter region. Sometimes the term promoter is used in a broader sense equivalent to “promoter region”. More distantly located enhancers (and silencers) are furthermore critical in regulation of transcription (discussed further in sau 19).
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b. Why are the transcripts so much shorter than the DNA that encodes them?
c. Calculate the number of base pair (bp) in the transcription unit using the given length (2.5μm) for the transcription unit and information on the geometry of the DNA helix (0.34nm/bp)
d. Estimate the number of transcribing polymerases and calculate the density of polymerases.
e. “Miller spreads” are popularly referred to as “X-mas trees”. What could the decoration on the X-mas tree be (the electron dense structures towards the end of the nascent transcripts)?
f. All three RNA polymerases are required to make a functional ribosome. Explain why this is.
g. The transcription rate is for RNA Polymerase I is 20-30 nt/sec but it takes about 1 hr to make a ribosome. What is the additional time used for?
h. The human genome has 200 ribosomal RNA genes. What is the output in terms of number of ribosomes per hour?
a. Left to right.
b. Because they fold co-transcriptionally into shorter length compared to the extended DNA helix.
c. If the length is of the transcription unit is 2.5 μm (xenopus oocyte), the calculation is: 2500 nm x 10 bp/ 3.4 nm = 7353 bp. The length of the human precursor RNA varies because of considerable variation in ITS1 and ITS2 (internal transcribed spacers). The length of the coding part is 7062 bp.
d. There are roughly 150 RNA polymerases per transcription unit in the figure (but this is hard to count). Using the number from above, this gives a density of one RNA polymerase per 49 bp. Using the transcription unit length of 13.3 kb, the result is one RNA polymerase per 89 bp. In any case, this gene is really crowded!
e. The decoration is the processosome (a protein/RNA complex that process pre-rRNA into mature rRNA) assembling on the pre-rRNA. (The explanation from figure 7.8, ECB 4th edition is “ribosomal proteins” but these small proteins only become visible very late in the ribosome biogenesis).
f. The ribosome is composed of LSU, SSU, and 5.8S rRNA (made by RNAPI), ribosomal proteins (translated from mRNA transcribed by RNAPII), and 5S rRNA (transcribed by RNAPIII).
g. Processing, folding, modification and assembly (with proteins) of the ribosome precursor. Most of this takes place in the nucleus with the final step occurring in the cytoplasm.
h. If the density of RNAPI on the gene is one polymerase per 90 bp and the transcription rate is 20-30 nt/ sec, the output of pre-rRNA should be one every four seconds corresponding to 900 per hour. If this is multiplied by the number of genes (say 300), the number of ribosomes synthesized per cell per hour is 270.000! (provided that all pre-rRNA are processed into ribosomes). There are approximately 107 ribosomes per cell. With the calculated rate of synthesis, the ribosome population in a cell will double in 40 hrs. Proliferation is discussed in blok E. Duration of a cell cycle varies greatly from one cell type to another but many cells in culture divides about once a day.
- Make a sketch of a typical eukaryotic mRNA.
a. What processing reactions were involved in producing mature mRNAs?
b. mRNA was originally known as “unstable RNA”. To which other RNAs is the comparison ? How are the mRNAs degraded ( “turned over”)?
a. Capping, splicing and polyadenylation (in addition, some mRNAs are edited). In contrast, the prokaryotic mRNA is not processed (i.e. the primary transcript is being translated) – homework at Absalon. Many prokaryotic mRNAs are polycistronic (contain multiple ORFs). This is described at pp. 254 in Essential Cell Biology 5th edition.
b. The “stable RNAs” are the rRNA and tRNA. Turn-over of mRNA is a major topic and a key process in regulation of gene expression. Prokaryotic mRNAs generally have shorter half-life (few minutes) and are degraded if not translated. Eukaryotic mRNAs have highly variable half-lifes (from a few minutes to days). Lifetimes of eukaryotic mRNAs are in part controlled by nucleotide sequences, most often in the 3’UTR region.
The following is a crude generalization of what is known (and more detailed than described in Essential Cell Biology 5th). Prokaryotic mRNAs: there are several degradation pathways, but a major one starts with removal of the two outer phosphates at the 5’ end followed by consecutive cleavages by the 5’ end dependent endonuclease RNaseE. Eukaryotic mRNAs: many are degraded if not translated, but some are stored. There are two major pathways. 90% or so are degraded by de-adenylation and de-capping followed by degradation from the ends by exonucleases. The remaining 10% are first cleaved in the 3’ UTR by an endonuclease and subsequently degraded bi-directionally by exonucleases.
- Sequence exercise/ part A [the sequences are found on the next page, and as a separate txt file]
a. Use the cDNA sequence to identify and mark (e.g. by underlining) the exonic parts of the genomic sequence (use word search tool!). How many exons did you find?
b. Compare the splice site sequence ( denotes the actual site of splicing) to the consensus sequences:
Consensus 5’ splice site: 5’- AGGTPuG Intron 1: Intron 2:
Consensus 3’ splice site: 5’-PyAGG Intron 1: Intron 2:
Introns are usually pyrimidine-rich at the 3’ end and the branch site (consensus 5’ PyPuNAPy with the branch nucleotide underlined) located immediately upstream. Is this also the case for the -globin introns?
c. Locate the TATA-box (consensus 5’ PyATAA) and mark the transcription start site with a broken arrow.
d. Find the initiation codon. Starting with this codon mark the entire open reading frame (i.e. the translated region) and divide it into triplet codons. What is the sequence of the stop codon? What is the number of amino acids in -globin? Are the introns inserted between codons or within codons?
e. Locate the polyadenylation signal (consensus 5’ AATAAA). What is the distance from the polyadenylation signal to the site of polyadenylation?
Regarding branch site:
Potential branch sites (bold and underlined sequence).
Intron 1 has the sequence 5’ tctat 18 nt upstream of the 3’ splice site as a potential branch site. The last 18 nt have 14 pyrimidines.
Intron 2 has the sequence 5’ ttcat 26 nt upstream of the 3’ splice site as a potential branch site. The last 26 nt have 20 pyrimidines.
c. The TATAA-box in this case is CATAA and is in bold in the above sequence. The transcription start site is at the beginning of the first exon (sequence in upper case letters).
d. The start (ATG) and stop codons (TAA) are underlined. The open reading frame encodes 147 amino acids. The first intron is inserted within a codon, the second between codons.
e. The polyadenylation signal is in bold in the above sequence at a distance from the polyadenylation site of 20 nt counting from the 3’ end of the signal. However, it is not clear if the following two A’s are template or untemplated in the mRNA (and the distance therefore 22 nt) because it is impossible to discriminate the two situations from a sequence comparison.
A comparison between key numbers for the β-globin transcription unit and an average transcription unit.
Feature
β-globin
Average
mRNA size (nt)
626
2200
Number of exons
3
7
Exon size (nt)
142, 223, 261
200
5’ UTR size (nt)
50
150
3’ UTR size (nt)
132
520
Protein size (aa)
147
476
Remember that the β-globin gene is an unusually highly expressed gene (in red blood cells). The short UTRs suggest that there is less regulation at the post-transcriptional level than found with the average gene.
- Sequence exercise/ part B
We will now use the sequence information from problem 2 to discuss experimental analysis and the effect of some disease causing mutations.
a. Suppose that you want to analyse splicing of -globin mRNA by RT-PCR. Where would you place the primers and what would be their orientation (suggest two primers with orientation, each of a length of 20 nt).
b. It has been estimated that around half of the disease causing mutations affects splicing. Assume that a mutation in the -globin gene resulted in the skipping of the middle exon. What would be the consequence in terms of -globin synthesis and how could you use RT-PCR to verify the loss of exon 2?
c. Another mutation directly affects the 5’ splice site of the first intron (see sequence below). The mutation changes the splice site from AGGT to the non-functional AAGT. As a result, the spliceosome seeks an upstream “cryptic splice site” (i.e. a sequence that looks like a splice site but is not normally recognized as such). In this case it is the sequence TGGT located 16 nt upstream of the normal site. This is then spliced to the normal 3’ splice site. What is the consequence for -globin synthesis in this case and would the mis-splicing be detected by your RT-PCR analysis?
Figure: Partial amino acid and nucleotide sequence (with the G→A mutation) of β-globin.
3. a. The primers should be placed in the first and the last exon and be oriented towards each other. Examples (using the extreme 5’ and 3’ ends of the exons): Upstream primer (syn. forward, 5´, sense): 5’ – ACA TTT GCT TCT GAC ACA AC (the primer is divided into 3-nt for convenience) Downstream primer (syn. reverse, 3´, antisense): 5’ – GCA ATG AAA ATA AAT GTT TT (which is complementary to 5’- AA AAC ATT TAT TTT CAT TGC). Note that this is a particularly poor primer because it is highly biased in nucleotide composition and has a stretch of A’s at the 3’ end which could lead to mis-priming. A much better primer can be designed by choosing a mixed sequence further upstream. b. Skipping of exon 2 would result in a mRNA that is 223 nt shorter. This would be detected as a faster migrating PCR-product in our RT-PCR analysis. At the level of translation, the reading frame of exon 3 will be changed and extend into the 3’ UTR resulting in a protein without function. c. The mis-splicing changes the reading-frame. The new reading frame ends in a stop codon in exon 2. This stop codon is sufficiently upstream of the last exon-exon junction in the mature mRNA to be recognized by the non-sense mediated decay system (NMD). Thus, the mis-spliced mRNA is eliminated before being translated and no specific target will be amplified by RT-PCR. This is not always the case. Sometimes NMD only reduces protein translation.
-thalassemia is a form of anemia caused by mutations in the -globin gene.
In the experiment shown to the right, RNA and protein from three different patients (S1-3) were analyzed by northern and western blotting analysis, respectively, and compared to normal (N). Discuss the location of the mutations in the three patients based on the results of the blotting analyses.
Note: The distance of the signal from the top (the site of sample application)
is related to the size of the molecule and the intensity of the signal is related
to the amount of the molecule.
- This is modified from a previous exam set.
S1: The patient has a lower level of mRNA and protein than N, but at the right sizes. The low level of protein is probably a reflection of the mRNA level. The patient will have a β+ phenotype (as explained in #5). A mutation in the TATA-box is consistent with the observations.
S2: The patient has normal amounts of mRNA of the right size. The protein has the right size but is considerably less abundant. Most likely a β+ phenotype. The mutation could be one that affects translation, e.g. in the 5’-UTR of at the initiation codon. More likely, the mutation is a mis-sense mutation that results in a less stable protein.
S3: This patient has no β-globin mRNA and no β-globin protein. This could be due to mis-splicing and NMD as explained in problem 3C.
- Fill in the multiple choice form below. The questions are all about the structure and properties of the average human mRNA. It is made in the form of an old-fashion “tipskupon”.
Some of the numbers are experimentally based estimates. In these cases the order of magnitude is most likely correct but a precise number cannot be determined for technical reasons or because it doesn’t make sense (e.g. 11 and 12). Nevertheless, the estimate describes well the importance of the phenomenon in question.
- Extra.
2-1-2/ x-x-x/ 1-2-x/ 2-2-x/ x
