M2M LO Unit1

Question 1

Distinguish purines and pyrimidine bases, ribose and deoxyribose, ribo- and deoxyribo nucleosides, nucleotides, nucleoside di and triphosphates.

Answer 1

Purines and pyrimidines make up the two groups of nitrogenous bases,including the two groups of nucleotide bases.Two of the four deoxyribonucleotides and two of the four ribonucleotides,the respective building blocks of DNAand RNA,are purines (G/A).
Purines: Fifty percent (50%) of the bases in nucleic acids,adenineand guanine,are purines. In DNA,these bases form hydrogen bondswith their complementarypyrimidinesthymineand cytosine,respectively. This is called complementary base pairing.
Pyrimidine: Three nucleobases found in nucleic acids,cytosine(C), thymine(T), and uracil(U), are pyrimidine derivatives.
Ribose is a five-carbon sugar that’s the primary building block of ribonucleic acids. Deoxyribose has been de-hydroxylated at the 2’ position.
A ribonucleoside is a type of nucleoside including ribose as a component.
A deoxyribonucleoside is a type of nucleoside including deoxyribose as a component.
Nucleosides are the central ribose sugar and a base attached to it at the 1’ position.
A nucleoside with one phosphate group attached to the 5’ position of the ribose is called either a nucleotide or a nucleoside monophosphate.
If there’s a chain of two phosphates tagged on at the 5’ position, it’s a nucleoside diphosphate
If the chain is three phosphates long, it’s a nucleoside triphosphate.

Question 2

Discuss the relative solubility of the components of nucleotides and the diseases related to their insolubility.

Answer 2

Phosphates are highly hydrophilic
Purine is less soluble than pyrimidine.
Base is less soluble than nucleoside.
Nucleoside is less soluble than nucleotide.
Phosphates>Nucleotide>Nucleoside>Base
Diseases related to purine insolubility, leading to build up of derivatives in tissues and kidney issues:
Gout- build up of uric acid in joints as a precipitant of purines
Lesch-Nyhan Disease- Causes severe neurologic symptoms

Question 3

Identify the chemistry in the phosphodiester linkage of DNA and RNA polynucleotide strands and discuss the reason for 5’ -3’ polarity.

Answer 3

Each nucleotide has an open hydroxyl group at the 3’ ribose position and a phosphate at the 5’ position (as well as a base stuck onto the 1’ carbon). It’s the binding of one nucleotide’s phosphate group to another’s OH group that makes the “chain” of DNA/RNA. Thus, at the end of each DNA or RNA chain, there’s either a ‘spare,’ unlinked phosphate group (what’s called the 5’ end) or a ‘spare,’ unlinked hydroxyl group (called the 3’ end). The phosphodiester linkage is the link between phosphate and hydroxyl.

Question 4

Discuss the important experiments that helped to establish DNA as the genetic material.

Answer 4

Avery, McCloud and McCarty: established DNA as the genetic material with their Pneumococcus experiments (Smooth strain killed mice, Rough strain did not. DNA from heat-killed S cultured with R then killed mice).
Franklin and Wilkins: x-ray diffraction suggesting a helical structure.
Watson and Crick discover definitive double-helical structure.

Question 5

Describe the implications of Chargaff’s rules.

Answer 5

The molar ratios of total purines and total pyrimidines are roughly equal (G+A=C+T)
The molar ratios of adenine to thymine, and guanine to cytosine, are roughly equal. (G=C, A=T)
G+C / A+T ratio is different for different organisms

Question 6

Describe the Watson-Crick three-dimensional model for DNA structure and recognize the major and minor grooves, the phosphodiester backbone, and the base pairs.

Answer 6

DNA is a right handed double helix. The sugar phosphate backbone is hydrophilic and is located on the outside of the molecule. The bases are paired and stacked on the inside due to their relative hydrophobicity. The major groove is the larger groove seen on the double helix and constitutes the ‘gap’ between the curves on the same single strand of DNA. The minor groove is the ‘gap’ between the backbones of complementary DNA strands. The relative distance crossed between AT bonds and GC bonds is pretty much the same. There are about 10 base pairs per turn of the helix.

Question 7

Describe the chemical basis for the stability of the double helix DNA in solution.

Answer 7

Electrostatic repulsions between phosphate groups (a) are mostly neutralized by positively charged species in the cell(magnesium); (b) the base pair linkages give the helix a lot of stability; and (c) adjacent base pairs “stack” on top of each other, providing additional delocalization options for the electrons and giving, often, more stability than comes from the base pairing itself. The stacking interactions are stronger between G/C than A/T.
Increased salt concentrations will increase the stability of the DNA molecule (thus increase its melting temperature Tm)
Extremes of pH alter ionization of bases which form H-bonds and thus decreases stability
Increase in DNA length will increase stability
The higher the GC content the more stable the DNA (more H-bonding/delocalization potential).

Question 8

Distinguish between linear and circular and relaxed and supercoiled forms of DNA.

Answer 8

Linear vs circular: Linear DNA is large and the segments can not distribute twisting so they can become supercoiled. Circular DNA is shorter and bound to itself (in prokaryotes) so can’t really become supercoiled.
Relaxed vs supercoiled: Relaxed DNA is essentially a straight ribbon of DNA with appropriate twisting while supercoiled DNA has an additional level of twists and coils on the DNA that influences its shape… important for DNA packaging in eukaryotes.

Question 9

Describe the chemical modifications of bases in DNA including different forms of DNA damage (methylation, deamination, depurination, UV cross linking) and their significance to disease.

Answer 9

Methylation: DNA methylationin vertebrates typically occurs at CpG sites(cytosine-phosphate-guanine sites, that is, where a cytosineis directly followed by a guaninein the DNA sequence). This methylation results in the conversion of the cytosine to 5­methylcytosine.The formation of Me-CpG is catalyzed by the enzyme DNA methyltransferase.Human DNA has about 80%­90% of CpG sites methylated, but there are certain areas, known as CpG islands, that are GC-rich (made up of about 65% CG residues), wherein none are methylated. These are associated with the promoters of 56% of mammalian genes, including all ubiquitously expressed genes. One to two percent of the human genome are CpG clusters, and there is an inverse relationship between CpG methylation and transcriptional activity.
Deamination: C can lose an amine to become U, which results in mutated DNA. Hopefully a repair enzyme would recognize this, but not always. Additionally, deamination of guanine can cause C-G to be turned to A-T, and vice versa for deamination of Adenine.
Depurination: Depurination is an alteration of DNAin which the purinebase (adenineor guanine)is removed from the deoxyribosesugar by hydrolysisof the beta-N-glycosidic link between them. After depurination, the sugar phosphate backbone remains and the sugar ring has a hydroxyl(-OH) group in the place of the purine. It’s usually caught by DNA repair enzymes.The problem is that it significantly weakens the phosphodiester backbone at the depurination site—so if you have a couple of these nearby, it can break the backbone.
Ultraviolet light: can covalently bond thymines, which distorts the DNA helix and can block replication enzymes. Generally repaired by nucleotide excision repair
Alkylating agents: Nucleophilic attack of bases on nucleotides (mustard gas, cisplatin)

Question 10

Explain the chemistry of DNA polymerization and how nucleoside analogues are used as drugs.

Answer 10

DNA polymerization is covered in the next section. Essentially, you can use molecules that are very similar to nucleosides to block replication of virally infected cells by having the replicating DNA (which recruits free-floating nucleosides as it replicates) incorporate the analogues into the growing chain. The analogues are different enough from actual nucleosides to ensure that the resulting DNA chains are nonfunctional.
Notice you can also use differences in the viral reverse transcriptase pathways to design nucleoside analogues preferentially incorporated by reverse transcriptase pathways.
More-specific nucleoside analogues usually used against retroviruses. Less specific nucleoside analogues usually used as chemotherapy against cancer.
Attacking DNA metabolism can occur through 4 methods
Blocking synthesis of precursors
Intercalation (getting in the middle)
Covalently binding bps
Attack topoisomerases which relieve supercoiling

Question 11

Describe how DNA melting temperature/annealing to complementary sequence is negatively affected if there is a mismatch and how this can be taken advantage of in diagnostic techniques to distinguish the presence of a particular unique sequence using a “probe” that is completely complementary to that sequence.

Answer 11

Using base pairing as a diagnostic tool: because perfectly complementary single strand DNA molecules pair with higher stability, a “probe” DNA single strand can be used to test whether a sample has the perfect complementary DNA sequence. The probe is tested for “hybridization” to an unknown DNA sample. Even a single mismatch can be distinguished. Variations on this phenomenon have been used for many diagnostic purposes in identifying genetic abnormalities that can arise from as little as one base pair change compared to unaffected individuals.

Question 12

Define the major similarities and differences between DNA and RNA.

Answer 12

DNA: has no hydroxyl group at the 2’ position of the ribose, which makes it more stable and less prone to hydrolyzation by nucleophilic attack at the 2’ location. DNA binds cytosine, guanine, adenine, and thymine as its bases.
RNA: hydroxylated at its 2’ ribose position; binds uracil instead of thymine. RNA is usually single-stranded, although it can form double-stranded loops (often called ‘hairpin loops’) with itself.

Question 13

Define 3 classes of RNA in a human cell.

Answer 13

Structural
rRNA (ribosomal RNA)- make up ribosomes and translate tRNA (transfer RNA)- move RNA around
snRNA (small nuclear RNA) and snoRNA (small nucleolar RNA) for a variety of in-cell modifications such as splicing.
Regulatory
miRNA (mirco RNA) and siRNA (small interfering RNA) to downregulate gene expression.
Information-containing
mRNA (messenger RNA) to be translated into proteins.

Question 14

Explain how puromycin mimics amino-acyl tRNA to terminate translation.

Answer 14

Puromycin is an antibiotic that mimics the “acceptor” 3’ end of a tRNA that is charged with an amino acid. Puromycin can bind in the ribosome as it is translating and covalently attach to a growing polypeptide chain, preventing the completion of translation.

Question 15

Discuss the meaning of these terms in the context of DNA Replication: “semi-conservative”, “bidirectional”, “Okazaki fragments”, “origin”, “fork”.

Answer 15

Semi-conservative: Each DNA strand is preserved as one half of a new double DNA strand; thus each new DNA strand has one-half original material and one-half newly synthesized material.
Bidirectional: Means that when the replication machinery attaches to the DNA double helix, replication proceeds in both directions along that helix at once.
Okazaki fragments: Small stretches of DNA synthesized during replication in the 5’ to 3’ direction (on the lagging strand). Since the synthesized strand is running in the 3’ to 5’ direction, and since new dNTPs can only be added at the 3’ hydroxyl group, the DNA synthesis process takes a kind of “leapfrogging” approach whereby small segments on that strand are copied 5’ to 3’ and then melded together later.
Origin of replication, aka replication origin: Specific sequences for recognition by binding proteins. Usually contain multiple short repeats, as well as an A-T rich streak. There are hundreds per chromosome, 1 in a prokaryote.
Replication fork: where the DNA helicases have unwound the double helix. Effectively the H-bonds of the base pairs have been split apart and the two strands are peeled away from each other, thus forming a “fork” in which the replication machinery sits and synthesizes complementary strands.

Question 16

Describe the functions of the following proteins during DNA replication: Origin binding proteins, Helicases, Single-strand binding proteins, Primase, DNA polymerase I, DNA polymerase III, DNA ligase, Sliding clamp, Topoisomerase/gyrase, Telomerase/Reverse transcriptase.

Answer 16

Origin binding proteins: Proteins bind to the origin and become part of the complex, also recruits Pol III.
Helicases: enzymes that catalyze the breaking of H-bonds between base pairs and the subsequent ‘unwinding’ of the helix.
Single-strand binding proteins: bind to the melted strands of original DNA to prevent them from re-annealing or getting messed up. More important for Okazaki fragments as they spend more time single stranded (you really don’t want single stranded stuff hanging around).
Primase: Enzyme that catalyzes the addition of the RNA primer to begin replication.
Pol I: DNA polymerase I, “distributive.” Versitile enzyme that replaces the RNA primers using 3 different functions: DNA polymerase, 3-5’ exonuclease activity and 5-3’ exonuclease activity. Does not have the sliding clamp so it is slow and distributive.
Common to both Pol III and Pol I: “proofreading” activity, or 3’ to 5’ exonuclease activity. If the wrong dNTP is added during DNA synthesis, the synthesis stops and ‘backs up’ slightly (in the 3’ to 5’ direction) and chops off the last nucleotide added.
Pol III: DNA polymerase III, “processive.” Synthesizes DNA strand from its complement on both leading and lagging strand. High processivity due to a sliding clamp mechanism that holds the polymerase tightly to the DNA. No 5’ to 3’ exonuclease activity (thus can’t be used to remove RNA primers).
Processivity clamp: the aforementioned ‘sliding clamp’ mechanism. Present in Pol III.
DNA ligase: enzyme responsible for sealing Okazaki fragments together once the RNA primers have been replaced by Pol I.
Telomere: sequence at the ends of chromosomes, consisting of a large number of repeating segments. Gets consistently shorter every time the chromosome is replicated, since the RNA primer on the very last O. fragment can’t be replaced by Pol I (Pol I needs to have a nearby 3’ OH from the next fragment to bind and replace the RNA primer). After a certain point, the telomeres get short enough that the cell becomes unstable and is destroyed.
Telomerase: Enzyme responsible for ensuring that the telomeres of chromosomes in certain immortal structures, such as germ cells, never shorten. Effectively, they act as reverse transcriptases, binding to the ends of DNA sequences and adding on some extra dNTPs. The reason this works is that the telomeres have more or less a uniform repetitious sequence.
Topoisomerase/gyrase: Enzyme responsible for relieving torsional strain in the DNA helix in the region ahead of the replication fork. It does this by clipping the phosphodiester backbone in selected places and then putting it back together without strain. Topogyrase is specific to prokaryotes.
Reverse transcriptase: Enzyme responsible for copying a base sequence INTO DNA (as opposed to out from it), usually from RNA. This can be endogenous (ie. telomerases) or exogenous (ie. retroviruses).

Question 17

Describe how DNA polymerase creates the phosphodiester bond during addition of deoxyriboncleotides (dNTPs).

Answer 17

It breaks off a diphosphate group from the dNTP and uses the energy liberated from that reaction to bind the remaining phosphate group to the hydroxyl group of the previous nucleotide on the chain.

Question 18

Recognize that DNA polymerase requires an RNA primer, that DNA synthesis only occurs in the 5’-to-3’ direction, and that errors during replication are corrected by the 3’-to-5’ exonuclease proofreading activity of the DNA polymerase.

Answer 18

none

Question 19

Describe the order of events that occur during, the differences between, and coordination of, DNA synthesis on the leading strand and the lagging strand.

Answer 19

Leading strand: pretty simple, relatively speaking:
origin binding proteins bind to origin.
DNA melted apart locally by helicases.
Topoisomerases relieve tension ahead of the replication fork.
Pol III elongates DNA complementary to leading strand.
The two strands, one new, one old, are annealed.
Lagging strand: similar but a little more complicated since DNA synthesis can only occur in the 5’ to 3’ direction. Share first 3 steps with leading strand, then:
Primase attaches RNA primer to lagging strand segment
Pol III elongates DNA from RNA primer back a short ways, forming an Okazaki fragment.
RNA primer is removed and replaced with DNA by Pol I.
Fragments are sealed together with DNA ligase.
The two strands, one new and one old, are annealed.

Question 20

Describe the “end replication problem” and the activity of telomerase.

Answer 20

The “end replication problem” refers to the fact that the leading stand can be synthesized to the very end, but the lagging strand cannot during DNA replication.
This is because you need an RNA primer to begin synthesis of each piece of the lagging strand DNA, but at the end of the DNA there is nothing for this piece to attach to thus the last section of the lagging strand cannot be synthesized. The result is that the telomeres (i.e. the end of chromosomes) get shorter and shorter as a cell replicates its genomes and divides, until they are so short that they signal for cell death – a normal healthy process in ageing cells.
Telomerase is an RNA-dependent DNA polymerase that maintains chromosomal ends by copying the telomeric repeat sequence from an RNA template. Telomerase activity is repressed in normal somatic cells.
In cancer cells, the telomerase enzyme is de-repressed, restoring the ends of the chromosomes to their full length and therefore blocking the normal cell death of old cells, promoting tumor growth. As such, telomerase is a potential target for anti-cancer drugs.

Question 21

Describe the relationship between mutations, DNA repair and cancer. Give examples of heritable human diseases that are caused by defective DNA repair pathways.

Answer 21

A mutation or mismatch is needed in a gene associated with cell proliferation
The cell must not notice the change or be unable to fix it to lead to proliferation
Self destruction pathways must be repressed or inactivated
These three criteria lead to a cell being cancerous
Examples of heritable human disease from defective DNA repair: xeroderma pigmentosum and cockayne syndrome.

Question 22

Describe the sources and nature of damage to DNA, the type of machinery used to repair the damage, and the molecular consequences of failure in DNA repair, e.g. thymine dimers, uracil mis-incorporations, bulky chemical adducts, and double-strand breaks.

Answer 22

Thymine dimers: Good candidate for nucleotide excision repair. Usually caused by UV radiation causing linkage between adjacent thymine residues, causing a bulging deformation of the helical structure. If unrepaired, can cause problems with normal processing due to malformed helix. Also will cause Pol III to fall off and Pol II take over, leading lots of errors
Uracils in DNA: Good candidate for base excision repair. Usually caused by the deamination of a cytosine residue to produce uracil in the DNA. If uncorrected, can cause problems with both the process of replicating/transcribing this portion of the DNA and also with recognition sites of transcription enzymes.
Specifically, if this portion of the DNA is transcribed or replicated as is, it will pair with an adenine residue, not a guanine; thus will have effectively swapped a C for a T.
Bulky chemical adducts: like thymine dimers, except usually caused by chemotoxic binding of large molecules to bases in a DNA helix. (nucleotide excision repair)
Double-strand breaks: Good candidate for either homologous recombination repair or non-homologous end joining (good to know: non-homologous end joining is the major form of double-strand break repair). Caused by a double break of the phosphodiester backbone (not sure what underlying causes are). If unrepaired, since a chromosome is one long DNA sequence, can lose up to half of the chromosome (very bad).

Question 23

Explain the basic steps of mismatch repair, describing the type of damage repaired by this pathway, and understand the marking of the old strand of DNA by methylation in E. coli.

Answer 23

Mismatched base recognized soon after synthesis (before methylation) on new strand (not old strand; old strand recognized by methylation in E. coli, mechanism of recognition unknown in humans). A stretch of DNA behind and in front of the mismatch is clipped by endonucleases, excised by helicase and exonucleases, and replaced with the correct sequence by DNA Pol III (and sealed with ligases).

Question 24

Describe the basic mechanisms of base excision repair, nucleotide excision repair, double-strand break repair by homologous recombination and NHEJ, and the types of lesions corrected by these DNA repair pathways.

Answer 24

Nucleotide-excision repair (NER): tends to repair more overt modifications that alter the helical pattern of the affected DNA. Process: recognition, clipping the backbone by endonucleases, excision of the affected part, replacing by Pol I, resealing by DNA ligase.
Notice that the recognition pathways here need a transcription factor, TFIIH, to work properly. (needs the helicase to melt the DNA)
Notice also that there’s two kinds of NER:
Transcription coupled NER: the distortion is within a gene being actively transcribed
Global Genome NER: the distortion isn’t within a gene being actively transcribed, goes back over the whole thing.
Base-excision repair (BER): tends to repair subtler modifications, like a mismatched base pair not caught by either proofreading or mismatch repair. Process: recognition, clipping off the inappropriate base by glycosylases, clipping the backbone by endonucleases, chewing off by exonucleases of the affected part, replacing by Pol I, resealing by DNA ligase.
Recombinatorial repair : also called homologous recombination. Repairs double-stranded breaks or crosslinks in DNA. Process: partially degrades sides of the break to create primers for DNA synthesis. Copies intact, homologous sequence from other chromosome that aligns with it. Each strand aligns itself with a strand on homologue and fills in its gap from that strand.
NHEJ [Non-Homologous-End-Joining]: A form of double-stranded break repair that doesn’t involve the homologous chromosomes. Essentially you unwind the two ends with helicases, pair up a few matching bases, and reseal the phosphodiester backbone. Note that this can be inaccurate, as you often lose a few bases off the unpaired strands during the resealing.

Question 25

Describe the mechanism that enables replication to continue in the face of DNA lesions that other repair pathways fail to remove, and know the unfortunate consequence of this process for the cell.

Answer 25

The mechanism is called lesion bypass polymerization and usually occurs when the cell doesn't have enough resources to fix all the thymine dimers occasioned by UV exposure. Big damage stops DNA Polymerase III, if NER cannot occur then two bypass polymerases bind to the complex. This creates a conformational change in the complex placing the bypass polymerases across from the damage where they add nucleotides without proof reading. After the damage the complex changes conformation and returns to DNAP III doing its business with proof reading. The error rate is 2 to 4 orders of magnitude higher than normal replication, thus frequently results in cancers, etc.

Question 26

Explain the concept of DNA damage checkpoint and its role in maintaining genome stability.

Answer 26

The cell cycle is paused if damage is sensed so that defects aren’t passed along to future generations of cells. During this arrest, repair machinery is up-regulated. 2 pairs of protein kinases are central to this mechanism (ATR for stalled forks and ATM for double strand breaks). They amplify the signal and phosphorylate Chk1 and Chk2 DNA for repair, arrest, or apoptosis. Chk1 and Chk2 phosphorylate p53 which causes cell cycle arrest or activate cell death program. Chk1 and Chk2 kinases are important to ensure that problems have been fixed. Mutations in any of these checkpoint DNAs lead to genome instability because the cell is now unable to regulate itself and may proliferate into a cancer. The checkpoint is the first step in guarding against cancer.

Question 27

Describe the chemical reaction catalyzed by RNA polymerase and why it is unidirectional.

Answer 27

RNA polymerase catalyzes phosphodiester bond formation between ribonucleasides in 5’ to 3’ direction, based on template of 3’ to 5’ Need triphosphate nucleoside to make hydrolysis spontaneous Catalysis means this reaction is done 1 way (unidirectional)

Question 28

Distinguish five steps in the transcription cycle common to bacterial and eukaryotic RNA polymerases.

Answer 28

1: RNA polymerase binds to promoter sequence on the helical DNA in a "closed complex." 2: Polymerase melts DNA strands apart near transcription start site, forming an "open complex” aka the "transcription bubble." 3: Polymerase catalyzes phosphodiester linkage of two initial rNTPs. 4: Polymerase advances 3' to 5' down template strand, melting DNA and linking rNTPs. 5: At transcription stop site, polymerase releases completed RNA and dissociates from DNA. Note that steps 1-3 are called initiation, step 4 is elongation, and step 5 is called termination.

Question 29

Name the four cellular RNA polymerases and their main functions.

Answer 29

RNA pol1: makes ribosomal RNA(rRNA) RNA pol2: makes messenger RNA (mRNA), small nuclear RNA (snRNA), micro RNA (miRNA) RNA pol3: makes transfer RNA (tRNA), lncRNA (long non-coding RNA for regulatory functions) Mitochondrial RNApol: makes RNA for the mitochondria

Question 30

Define a promoter and name sequence elements characteristic of promoters in human genes.

Answer 30

A promoter is a sequence of DNA upstream of the transcription start site that positively affects the expression of the gene. (where RNA Polymerase binds) Consensus elements: -30 (TATA box), initiator, promote The TATA box is a frequently conserved TATA sequence about 30 bases upstream from the start site. Mutations in the TATA box often result in reduced expression of the gene (ie beta-thalassemia with B-hemoglobin), because the TATA box binding protein (TBP) helps in assembly of the pre-initiation complex of general transcription factors at the promoter Initiator is +1 in some but not all eukaryotic gene Promoter proximal elements are promoter DNA sequences between 30-1000 bp upstream of the start site Enhancer elements are promoter DNA sequences much farther upstream (10,000-50,000 bp). This acts through DNA looping.

Question 31

Describe how α-amanitin and rifampicin block transcription.

Answer 31

alpha-amanitin: extremely toxic substance found in death cap mushrooms. Acts by inhibiting the movement of RNA Pol II, binding its bridge substructure so that translocation of the polymerase down the DNA chain can't happen. rifampicin: broad-spectrum antibiotic. Acts by binding the beta subunit of bacterial RNA polymerase, plugging up the exit chamber where assembled RNA exits the transcriptional complex. Thus elongation is prevented from going farther than a few base pairs due to having nowhere to go.

Question 32

Name four components of the RNA polymerase II pre-initiation complex.

Answer 32

RNA pol2, general transcription factors (TF2a,b,d,e,f,h), promoter DNA, mediator

Question 33

Describe the clinical syndromes caused by mutations in TFIIH subunits.

Answer 33

TF2H is a transcription factor that also serves to repair damages to DNA Problems with nucleotide excision repair: Cockayne's syndrome, Trichothiodystrophy Xeroderma pigmentosum - light sensitivity, abnormal pigmentation, cancer susceptibility, neuorological abnormalities, unscheduled DNA synthesis).

Question 34

Describe the three major ways in which most pre-mRNA's are processed

Answer 34

Capping: replacement of the 5' triphosphate (from the first rNTP to be added) with a backwards, 7-methylguanosine (no phosphate group). Splicing: excision of introns and desegmentation of exons. Cleavage/Polyadenylation: cleavage of RNA at 3' end past the consensus sequence and polyadenylation (> 200 A's) of cleaved site. Note that these processing steps take place while the RNA is still being made, not after it's finished and released. For example, the cleavage/polyadenylation step is partly responsible for Pol II being released from the DNA.

Question 35

Compare and contrast a pre-mRNA with a mature mRNA.

Answer 35

Pre-mRNA-Has its 3’ phosphate still intact, has large caps of non-coding sequences between coding exons called introns, the 3’ end is un-modified Mature mRNA-5’ phosphate has been capped by 7-methyl guanisine, introns have been spliced out, poly A tail has been added to the 3’ end

Question 36

List the functions of the 5' cap of the mRNA.

Answer 36

It makes the 5' end resistant to exonucleases (which target the "lone ends" of single D/RNA strands). It helps with splicing and processing through a cap-binding complex that recognizes the cap. (primes for splicing, 3’ tail, translation) Translation factor eIF4E (eukaryotic initiation factor 4E) recognizes the cap for transport to the ribosomes. When the cap is eventually removed, it signals for the mRNA to be degraded.

Question 37

List the three reactions required to add a 5' cap to pre-mRNA.

Answer 37

(1) Cut off the last PO4 from the triphosphate group at the 5' end of the mRNA. (2) Add guanosine triphosphate (GTP) backwards via guanylyl transferase. It loses two PO4 groups of its own (making GMP) and forms a 5'-to-5' triphosphate bond (2 PO4 from mRNA, 1 PO4 from GMP) to the end of the mRNA. (3) Methylate the 7-position of the guanosine cap via S-adenosyl methionine (SAM; nearly universal methyl donor in cell; donates methyl to form homocysteine) to form 7-methyl-guanosine at cap.

Question 38

Recall the conserved sequences at the 5' and 3' ends of most introns and the consensus sequence at the polyA site.

Answer 38

Splice site at 5' end of intron: GU. Splice site at 3' end of intron: AG. Consensus sequence at poly-A site: AAUAAA.

Question 39

Describe how alternative splicing permits multiple proteins to be produced by splicing defects.

Answer 39

(normal) Retain or remove exons, mutually exclusive exons, exon truncation.extension at 5’ end, exon truncation/extension at 3’ end, intron retained or removed. Since the 5' splice site (the one identified by snRNA that recruits the spliceosome) at any given intron is part of the DNA sequence, it's vulnerable to being corrupted by mutation and being unrecognizable as an intron, in which case the finished mRNA has an extra sequence in it and the final protein winds up significantly different

Question 40

Give examples of genetic disorders caused by splicing defects.

Answer 40

Marfan's syndrome: caused by mutations that disrupt splicing of the fibrillin gene transcript (fibrillin is a connective tissue protein that is important for the integrity of the walls of the heart and blood vessels). They are tall and prone to aneurysms. Abnormal splicing of CD44 (cell-surface glycoprotein) is a predictor of tumor metastasis. Used as diagnostic and prognostic marker.

Question 41

Describe the function of U1 and U2 snRNA's in splicing.

Answer 41

U1snRNA binds to the GU 5' splice site of introns. U2snRNA binds to the branch point (A) on the pre-mRNA sequence between the 5' and 3' splice sites U2AFsnRNA binds to the 3’ splice site of intron (AG) Lariat splicing mechanism: the U1snRNP brings the 5' splice site into proximity to the branch point. The U2snRNP activates the 2' hydroxyl group at the branch point, which attacks the phosphodiester bond just past the GU 5' splice site. Now there's a free 3' OH on the end of the 5' exon, which attacks the phosphodiester bond at the AG 3' splice site-- linking the two exons and excising the snRNP/intron complex to be degraded.

Question 42

Identify on a diagram of a gene the following:

Answer 42

transcription start site: Look for either the +1 position or the little bent arrow coming out of the sequence at this point. introns: Should be between the GU (5') and AG (3') splice sites. 5' splice sites: Look for GU. 3' splice sites: Look for AG. branch points: should be in the introns, between the GU (5') and AG (3'). exons: should be between the introns. Mark the introns with GU at 5' and AG at 3', then look between them for the exons. 5' UTR: look for a region between position +1 (start of transcript) and the start codon (see below) on the processed mRNA strand. 3' UTR: look for a region after the stop codon (see below) until the end of the transcript, including the consensus sequence and the poly-A tail. initiation codon: also called start codons. Encodes methionine. Look for 5' AUG. termination codon: also called stop codons. Look for 3' UAG, UAA, or UGA. poly A site: look for consensus sequence AAUAAA

Question 43

Describe the two reactions that make the mature 3' end of mRNA's.

Answer 43

1) recognition of consensus sequence at pre-mRNA's 3' end (AAUAAA) and cleavage of the mRNA soon after this sequence. (2) polyadenylation of free hydroxyl at 3' end. (not coded for)

Question 44

Describe the relationship between 3' end processing of the pre-mRNA and termination of transcription at the end of a gene.

Answer 44

The cleavage and polyadenylation occur while the RNA polymerase continues, it could soon fall off or it could continue to make another mRNA. The poly-A tail seems to be necessary for the RNA polymerase complex to detach from the DNA (termination).

Question 45

Describe the two major functions of the mRNA's poly A tail.

Answer 45

protection from degradation | export of mRNA from nucleus

Question 46

Give an example of how alternative poly A sites can be used to make more than one protein from a single gene.

Answer 46

two different forms of immunoglobulin M (IgM), membrane-bound and secreted, are formed by alternative poly-A sites in their common gene. This can occur because transcription continues past poly-A site depending on which one is used it will make 2 variants (heavy and light chains).

Question 47

Discuss the basics of eukaryotic transcription (DNA control elements, transcription factors) and the role they play in human disease and list the different eukaryotic DNA control elements.

Answer 47

DNA control elements: DNA elements that act locally. Binding of transcription factors to these elements controls expression of the gene that the element is associated with. TATA box/Initiator sequence- This element is generally 25-35 bps upstream of the transcription start site. It determines the site of transcription initiation and directs binding of RNA polymerase II. This is the site at which general transcription factors bind. Promoter proximal element Enhancers Transcription factors: Proteins encoded by one gene that act on other genes to regulate their transcription. Can therefore diffuse around the nucleus and affect transcription of numerous genes. Can either activate or repress transcription.

Question 48

Define promoter proximal element and enhancer.

Answer 48

promoter proximal element: usually within 200 bp upstream of start site, about 20 bp long. Bound by transcription factors to regulate transcription. enhancers: usually much farther upstream, or downstream, than promoters, although still fairly short in themselves (8-20 bp). Can be upstream of the start site, downstream of the last exon, or within introns in the gene itself. Similar function to promoters.

Question 49

Describe a disease that arises from a mutation in a DNA control element, and how the mutation leads to the disease state.

Answer 49

Beta-thalassemia (from thalassa, Greek for sea): mild inherited anemia (low hemoglobin count). Caused, here, by a mutation in the promoter of the b-globin gene, resulting in lowered rate of production of b-globin protein. (less promotion = less transcription) Gamma-delta-beta thalassemia: more serious anemia caused by a deletion in the locus control region for the transcription of all globin genes, resulting in the loss of globin transcription. Hemophilia B Leyden: X-linked disease (usually males) that affects clotting. Again, a problem in the promoter region of a clotting protein gene. Tends to get partially better at puberty. Fragile-X syndrome: Again, usually a disease of men. Results in mental retardation and atypical development of the face with enlarged testicles (macroorchidism). Caused by an expansion in the CGG count upstream of a particular gene (the FMR1 gene), which results in an abnormally high rate of methylation in that region and transcriptional silencing of the gene.

Question 50

Describe the role of transcriptional activators and repressors.

Answer 50

They bind to either the DNA control elements (ie, to the DNA itself) or to other factors bound to control elements (ie, to other proteins that are bound to the DNA). They increase or decrease the rate of transcription of the gene's protein(s).

Question 51

List the two classes of activators and repressors.

Answer 51

Those that bind to the control elements in the DNA are called sequence-specific DNA binding proteins or SSDBPs. They usually bind to short (6-8 bp) sequences by inserting their alpha-helices into the major groove of the sequence in question. Those that bind to SSDBPs are called co-factors. These can both increase or decrease efficacy of transcription factors.

Question 52

Describe the domains of a sequence specific DNA binding protein.

Answer 52

Generally two domains. Notice that these domains are modular-- that is, if you pull one domain off of one protein and attach it to another, the second protein will have the function of the domain you just attached. First domain: DNA binding domain. Very highly structured and evolutionarily conserved. Binds to the DNA target. Second domain: activation (or repression) domain. Fairly unstructured and less conserved; they recruit other proteins (either co-factors or general transcription factors) to bind and affect transcription. Note the difference between co-factors and general transcription factors: co-factors influence the rate of transcription, while GTFs provide the pre-initiation complex needed to begin transcription.

Question 53

List the four major families of sequence specific DNA binding proteins and describe the means for categorizing the proteins into these families.

Answer 53

These are categorized based on tertiary structure differences: Homeodomain proteins have a helix-turn-helix structure. They tend to be regulators of development and affect many genes at once. Zinc-finger proteins have a "finger" made up of two antiparallel beta sheets and an alpha helix, held together by a zinc ion. This finger is what binds with the DNA. The largest family of SSDBPs. Include androgen and estrogen (nuclear hormone) receptors. Basic leucine zipper proteins (bZIP): (the "basic" here refers to the fact that they have a high-pH region that binds to the DNA) Chop sticks. Hydrophobic residue every 7 amino residues, (like leucine). Dimerizes to bind DNA. (c-fos and c-jun). Basic helix-loop-helix proteins (bHLH): also has basic region for DNA binding. Muscle group (MyoD, myogenin).

Question 54

Describe a particular human disorder that arises from a mutation in a sequence specific DNA binding protein, explaining how the mutation leads to the disorder.

Answer 54

Craniosynostosis: premature closure of the skull sutures in infants. Arises from a mutation in the homeodomain protein that causes the protein to bind more strongly, creating a "hypermorphic allele" that activates genes more strongly than it should. (upregulates proteins that close sutures via homeodomain mutation) Androgen insensitivity syndrome: Feminization or undermasculinization. Indifference of androgen receptors to androgen hormones. Caused by a mutation in the zinc-finger androgen receptor binding domain or ligand binding domain. This downregulates the transcription of genes controlled by male androgens. Waardenburg syndrome: deafness, pigmentation defects. Associated with mutations in the MITF ( phthalmia-associated transcription factor) gene (which codes for a bHLH binding protein that regulates melanocyte development).

Question 55

Describe combinatorial control as a mechanism for controlling gene expression.

Answer 55

Combinatorial control refers to the fact that SSDBPs can dimerize, making possible a wide variety of possible DNA binding sequences. Note that this is always within classes of SSDBPs (e.g. zinc-finger with zinc-finger) For example: SSDBP can dimerize into homodimers and heterdimers, the combination of these proteins leads to many different transcription factors. If you have 4 monomers that can mix and match into dimers you have 2^4 transcription factors. Larger complexes, trimers and tetramers, are also possible. With 1,000s of TF genes you can have 10,000s of TFs (Not 2^1000 - not all TFs can form heterodimers with each other).

Question 56

Describe how chromatin structure affects transcriptional control and list the two classes of chromatin remodeling factors and briefly describe how they work.

Answer 56

DNA-dependent ATPases: these disrupt the histone octamers, opening up the chromatin and exposing DNA for binding via hydrolysis. (SWI/SNF complex) Factors that reversibly modify histones through acetylation of the N-termini: Acetylators, co-activors (histone acetyltransferases, or HATs De-acetylators, co-repressors (histone deacetylatases, or HDACs) Acetylation is associated with increased transcription of the DNA on the affected histone; deacetylation is associated with the reverse. Current theory is that HATs' pattern of histone acetylation recruits co-factors to effect increased transcription rather than directly affecting it themselves by opening up the histone complexes.

Question 57

Define HATs and HDACs and describe how their activity influences transcription.

Answer 57

see card 56 (past card)

Question 58

Give an example of a disease in which histone acetylation is altered, and describe the defect that leads to altered histone acetylation.

Answer 58

Leukemia: haematopoietic (formation of blood cell components) disease involving chromosomal translocations over-activating fusion proteins that alter the activity of HATs or HDACs. Rubinstein-Taybi syndrome : growth and mental retardation, broad thumbs and toes, craniofacial dysmorphism. Results from mutations in one copy of CREB binding protein gene (haploinsufficiency); CBP is a widespread HAT important in development and its insufficiency has particularly drastic effects.

Question 59

Describe how activators/repressors modulate transcription via their interaction with general transcriptional machinery vs. with chromatin.

Answer 59

Activators/Repressors can either: bind to general TFs or the RNA Pol II complex to influence initiation or elongation of the primary transcript interact with chromatin to regulate the accessibility of the DNA to the Pol II transcription apparatus via acetylation, phsophorylation, methylation, or ubiquitination.

Question 60

Discuss the basic principles of transcriptional regulation including how specificity is achieved and how protein-DNA interactions contribute to transcriptional control.

Answer 60

(1) Specificity depends on binding of transcriptional activators/repressors to DNA control elements. (binds to specific section of DNA) (2) Regulation depends on DNA-protein and protein-protein interactions. (activation, repression) (3) The interactions affect the conformation of DNA, modification of chromatin structure, and formation of the transcription initiation complex. (change access to gene) (4) Control elements are combinatorial, which allows for thousands of transcriptional activators/repressors to alter the expression of various genes in response to varying stimuli. (mix and match monomers)

Question 61

Describe how sequence specific DNA binding proteins are themselves regulated, and how their dysregulation may lead to human disease and list at least four mechanisms by which sequence specific DNA binding proteins are regulated.

Answer 61

1) Alter the conformation of the TF with ligand binding (2) Regulate entry into the nucleus (regulate access to DNA) (3) Regulate amount of TF in the cell (4) Regulate DNA binding action of protein (5) Phosphorylation/dephosphorylation of TF (affecting degradation, recruitment of of co-activators, DNA binding)

Question 62

Describe how the activity of nuclear hormone receptors is controlled, and how tamoxifen acts in breast cancer therapy.

Answer 62

- Steroid hormone enters cell and binds to the nuclear hormone receptor, leading to conformational change, recruitment of coactivators/repressors, and entry into the nucleus. They have zinc-finger DNA binding motif - Tamoxifen acts as an antagonist to estrogen, binding to the estrogen receptors as a ligand without providing dimerization-- thus effectively preventing estrogen from binding to its site of action and preventing the transcriptional effect of estrogen receptors.

Question 63

Give an example of a sequence specific DNA binding protein regulated by nuclear entry and describe the mechanism by which its entry is controlled.

Answer 63

NF-κB: normally bound to IκB hiding the NLS, which holds it in the cytoplasm (and away from genetic material). Under certain conditions, IκB is phosphorylated, which targets it for degradation. Degrading IκB shows the NLS of NF-κB to migrate into the nucleus and affect transcription. (Asprin blocks phosphorylation of IκB and acts as an anti-inflammative and one of the reasons low-dose aspirin is given to prevent buildup of atherosclerotic plaque).

Question 64

Describe how the amount of an activator/repressor can be regulated within the cell.

Answer 64

Specific genes can target activators or repressors for degradation (like APC targets beta-catenin, a cell proliferation activator protein). Lack of Wnt signaling leads to phosphorylation and thus degradation of beta-catenin via the ubiquitin pathway. Clinical fact: ~50% of colon polyps observed in the clinic are caused by mutations in the APC gene, resulting in insufficient degradation of (and thus proliferation of) B-catenin. Not enough APC to phosphorylate the beta-catenin and degrade it to prevent it from getting into the nucleus.

Question 65

Describe a mechanism by which the DNA binding activity of a sequence specific DNA binding protein can be inhibited.

Answer 65

Id proteins have a helix-loop-helix (HLH) domain, but no basic domain-- recall that the basic domain is what allows HLH proteins to bind to DNA, while the HLH domain allows dimerization of these proteins. Recall also that bHLH proteins usually dimerize to be able to bind to DNA. When an Id protein heterodimerizes to a good one then it doesn’t bind well to the DNA and thus decreasing the effectiveness of the HLH proteins present. (diluting effectiveness). Also faster and more energy efficiency.

Question 66

List a protein modification that can alter the activity of a sequence specific DNA binding protein, and explain the mechanism by which the activity is altered.

Answer 66

Ligand binds G-protein, phosphorylates CREB protein, recruits CBP (CREB binding protein – a HAT) which recruit RNA Pol II leading to transcription of the gene. No phosphorylation, no CREB mediated transcription.

Question 67

Aside from transcriptional regulation, list at least 3 additional mechanisms to control levels of gene expression.

Answer 67

Control of mRNA export from nucleus (cap and poly-A tail) Control of mRNA degradation (ie small interfering RNA action) Control of efficiency of translation (ie next lecture IRE/IRP at 5' UTR) Control of protein degradation (ie ubiquitination)

Question 68

Name and describe the roles of the parts of the machinery that drives translation, specifically: ribosomes, mRNA, tRNA, aminoacyl tRNA synthetases, initiation factors, elongation factors, and release factors.

Answer 68

Ribosomes: site for mRNA translation into amino acids mRNA: transcribed and spliced copy of DNA that provides the 3 base codons that code for an amino acid’s addition to a pep.de tRNA: small piece of RNA with an amino acid added to its 3’ end aminoacyl tRNA synthetase: the enzyme responsible for adding the correct amino acid to the tRNA with the correct anticodon by hydrolysis of ATP. Initiation factors: proteins needed to initiate ribosome assembly or mRNA translation (IF1, IF2, and IF3 are the initiation factors in bacterial ribosome assembly) Elongation factors: factors associated with the elongation of a newly synthesized protein. One elongation factor is PTC (peptidyl transferase center) which is a ribozyme associated with catalyzing pep.de bonds by using ATP coupled to tRNA. EF2 (elongation factor 2) by way of GTP move the newly made pep.de bond over in the 3’ direction exactly one codon. Release factors: proteins that bind to the stop codon can terminate polypep.de synthesis.

Question 69

Be able to explain the nature of the genetic code, how it is read, and the effects of mutations to the mRNA. Name the start codon and what amino acid it encodes.

Answer 69

The genetic code from DNA, is transcribed to messenger RNA where it is modified to be read by ribosomes. A codon is a 3 base sequence and each codon on the mRNA triggers a different tRNA to come in to the A site of the ribosome and deliver the corresponding amino acid. As they are read from the mRNA the amino acids are added and elongated at the P site of he ribosome. A mutation in the DNA will affect the codon to be read off the mRNA from the ribosome which can lead to the wrong amino acid being added, early termination of the protein assembly, late termination of the protein assembly, or faulty initiation. Any of these can be the result of a mutation and can lead to an inac.ve or defec.ve protein ultimately leading to disease. The start codon AUG codes for methionine which is found at the beginning of each protein built. Additionally, there can be missense, silent, and frameshift mutations. Missense mutations lead to a different amino acid to be added to the pep.de. Silent mutations are a mutation in the codon but not in the amino acid that codon corresponds with. A frameshift alters the reading frame of the mRNA which can be more catastrophic.

Question 70

Describe the four phases of translation: initiation, elongation, termination, and ribosome recycling, and describe the basic elongation cycle.

Answer 70

Initiation: The 2 subunits of the ribosome are brought together around a piece of mRNA to begin translation Elongation: The subsequent codons are being read and the corresponding tRNA is brought in to the A site of the large ribosomal subunit. The tRNA is then showed to the P site and a peptide bond is made with the peptidyl transferase center (PTC). The new amino acid on the chain is then moved by elongation factors and GTP to the E site of the ribosome where the uncharged tRNA is released. Termination: At the end of the mRNA, release factors bound to the stop codon terminate protein synthesis Ribosome Recycling: Following termination of translation, the ribosome is dismantled into its subunits and available to translate another mRNA

Question 71

Identify the important differences between bacterial and eukaryotic translation, especially in regard to initiation.

Answer 71

Prokaryotes: alignment of the proper AUG codon to indicate ‘start’ is achieved by recognition of the Shine-Delgarno sequence upstream of the DNA. Initiation factors IF1 and IF3 bind the 30S subunit. The mRNA binds to the 30s subunits by way of the Shine-Delgarno sequence placing the start codon (AUG) in the P site of the small subunit of the ribosome. IF2 delivers a special methionine (formylmethionine) to the ribosome to be the first amino acid. Hydrolysis of GTP leads to a release of the remaining initiation factors and binding of the 50s subunit while the next codon is placed in the P site for translation and addition to the growing amino acid. Eukaryotes have Kozak sequences [gccRccAUGG-capital lesers indicate the most heavily conserved regions] in introns that help in the differentiation of a start codon (AUG) for a simple peptide AUG. Prokaryotic initiation of translation: dependent on a specific RNA sequence called the Shine-Dalgarno sequence (8-13 bp) that is upstream from the start codon (AUG). S-D sequence binds to 30S subunit of the prokaryotic ribosome. There can be multiple S-D sequences per mRNA transcript, allowing one transcript to code for multiple proteins (ie polycistronic). Eukaryotic initiation is complex to allow for regulation. It relies on multiple initiation factors assembling a translation complex at the Kozak sequence to position the ribosome at the start codon (eIF = eukaryotic initiation factor): eIF-4E (cap-binding protein) binds to 5' cap. eIF-4G (scaffolding protein) is recruited by eIF-4E. eIF-3, which binds to the 40S ribosome eIF-2, which binds to a tRNA charged with methionine initiates protein synthesis eIF-4A, a helicase to melt any RNA structure encountered After the pre-initiation complex (40S + eIFs) becomes the initiation complex (adds cap-binding complex and attaches to RNA), it "scans" down the mRNA until it finds the start codon. Then the eIFs break off and the 60S joins to form the full ribosome (80S). Notice that the 5' UTR can influence the rate of this process or can stop it entirely, depending on how convoluted it is (remember that the ribosome complex has to scan through the 5' UTR to get to the AUG start codon). This is the cap/scanning model. Notice that the start codon, AUG, means that the first amino acid on the translation is Met.

Question 72

Explain the significance and effects of the following processes: cap-independent initiation, interferon stimulation, mRNA editing, rapamycin treatment, eIF2-alpha phosphorylation. Be able to describe when these might be important.

Answer 72

cap-independent initiation: viruses can bypass the need for a cap to translate proteins. Some viruses produce a protease that cleaves eIF4G (eukaryotic initiation factor 4 gamma) to shut down cap-dependent translation. The virus can translate its genes using the IRES (internal ribosome entry site) on the RNA interferon stimulation: When a virus invades a cell and the cell cannot protect itself, it releases interferon into the extracellular space to come into contact with neighboring cells that recognize the interferon and translate antiviral proteins to protect themselves from the threat. It also produces 2-5 synthase mRNA editing: The RNA code can be altered after its made. In one example apolipoprotein is made but translated differently in different parts of the body due to the gene’s RNA editing resulting in one gene being truncated to from a stop codon rapamycin treatment: Rapamycin is a drug that affects m-TOR (mammalian target of rapamycin; a protein associated with regula.ng global protein translation). It prevents m-TOR from being translated and can thus inhibit protein translation eIF2-alpha phosphorylation: When interferon is recognized by a cell, it phosphorylates eif-2a to effectively turn it off. This enzyme can no longer bring in tRNA for translation

Question 73

Identify antibiotics that operate by affecting translation.

Answer 73

Some of the big ones: aminoglycosides (streptomycin), tetracyclines (tetracycline), macroclides (erthromycin), other category (chloramphenicol) Antibiotics inhibit translation by interfering with the ribosome (both 30S and 50S subunits) through tRNA binding, elongation and peptidyl transferase. The cancer treatment Rapamycin phosphorylates 4E-BP so that the initiation complex cannot be formed, downregulating translation.

Question 74

Describe how intracellular levels of iron can be regulated by translation, using this as an example of how protein-mRNA interactions regulate translation.

Answer 74

``` Iron regulation is critical to proper biological functioning, high levels are toxic. There are two mechanisms to control iron levels, both working through Iron Response Biding Proteins 1 & 2. Iron response element (IRE) is an RNA stem-loop structure found in mRNAs that bind IRPs. IRPs (Iron response binding proteins) bind iron and regulate the expression of ferritin and TFR. Low iron means little will bind to IRE-BP allowing it to bind to the Iron Response Element. This stops Ferritin production and stabilizes Transferrin receptor mRNA. High iron environments mean iron will bind to IRE-BP preventing it from binding to the IRE, thus ferritin mRNA is translated and TFR mRNA is degraded. Ferritin sequesters iron (thus lowering the level of free iron in the cell) Transferrin receptor (TFR) transports iron into the cell. When cellular iron levels are low, the translation of transferrin receptor is up-regulated and the ferritin levels go down. When cellular iron levels are high, the transferrin receptor is down-regulated and the ferritin levels go up. ```

Question 75

Identify α carbon, the NH2, COOH, and side chains (R groups) of an amino acid.

Answer 75

alpha carbon sits between the NH2 and COOH. The alpha carbon also sits between a H and the R group/side chain.

Question 76

Distinguish between amino acids with hydrophobic, polar, acidic and basic side chains.

Answer 76

Nonpolar/Aliphatic R groups: MAGLIV (great lecturers value my awesome intellect) (methionine, alanine, glycine, leucine, isoleucine, valine) These are not very reactive, only hydrocarbon R groups (+methionine). Aromatic R groups: FYW (phenylalanine, tyrosine, tryptophan) These are mostly hydrophobic and have a ring or two. Polar uncharged groups: SCTPQN (serine, cysteine, threonine, proline, asparagine, glutamine) These have hydroxyl, carboxamide and sulfur groups where electronegativities are unbalanced (+proline) Polar positively charged groups: HRK (histidine, arginine, lysine) These are basic with charged amines (+histidine). Polar negatively charged groups: DE (aspartate, glutamate) These are acidic with resonant oxygens.

Question 77

Describe the function of disulphide bonds within proteins.

Answer 77

S-S bonds (formed from two proximal cysteine residues) stabilize both intra-protein (tertiary) and inter-protein (quaternary) structure. Examples are insulin (A and B), keratin in hair and ribonuclease.

Question 78

Describe the major post-translation covalent modifications of amino acid side chains in proteins and identify the post-translational modification targeted by the disease process or medicine discussed in class.

Answer 78

Hydroxylation of proline residues in collagen stabilizes the structures, mediated by vitamin C. Scurvy is caused by lack of vitamin C and thus underhydroxylation of collagen and is thus weak. Ergo Limey. Carboxylation of glutamate residues on prothrombin, mediated by vitamin K, is required for effective blood clotting. Vitamin K insufficiency leads to improper clotting. Warfarin prevents carboxylation of glutamate residues and acts as an anticoagulant. Glycosylation of asparagine residues of proteins on cell membranes and that are secreted increases hydrophilicity. Congenital Disorder of Glycosylation (CDG) has malfunctions in this mechanism. Acetylation and de-acetylation of histones in gene regulation (HDACs and HATS) Cancer treatment can involve blocking HDACs Reversible phosphorylation via kinases (adds P) and phosphatases (removes P) affects signal transduction Gleevec (tyrosine kinase inhibitor) is a cancer treatment that competitively inhibits bcr-abl kinase such that the substrate is not phosphorylated and active and the tumor cell cannot proliferate via an abberant bcr-able gene. Ubiquitin added to proteins signals them to be sent to the proteosome for destruction Velcade inhibits a proteosome that degrades good proteins resulting in multiple myeloma.

Question 79

Distinguish three covalent bonds including the peptide bonds that make the backbone of a polypeptide chain.

Answer 79

Peptide bond (C1-N): formed from dehydration reaction of COO- and NH3+. Stabilized by partial double bond character from proximal carboxyl group. This also makes it a rigid bond that cannot rotate. Bond from alpha carbon to carbon of peptide bond/carbonyl (C2-C1). Free rotation. Bond from amide nitrogen to alpha carbon (N-C3). Free rotation.

Question 80

Distinguish that the amino acid sequence determines the function; mutations in amino acid sequence can cause genetic disease.

Answer 80

Two important models relate protein structure to function: lock and key and induced fit. Lock and Key refers to the idea that the protein has a specific structure that will fit only very specific substrates Induced Fit refers to the idea that when a protein and substrate come together there is a conformational change More generally amino acid sequence determines the structure of the protein. Proteins with similar structures have similar functions. Mutations in even a single amino acid codon can lead to proteins with a different structure, and thus will not perform the same function as well. However, 20-30% of proteins are polymorphic with differences in non-essential amino acids and thus structures that are similar enough to perform the same function. Depending on the specific assortment and order of side chain from the peptide backbone, the pattern of hydrogen- and sulfur-bonds in a protein will be very different. Collagen is a good example, as without the proline residues available for hydroxylation, the strong H-bonds holding it together couldn't happen. Keratin is another example-- without being cysteine-rich, the multiple disulfide bonds that give it its tensile strength would be absent. Re mutations: if the ratio of glycine residues is messed with in collagen (thus deforming the tight helix structure), it results in a fatal condition in infants

Question 81

Recognize that proteases and the specific breaking of peptide bonds can have important functions.

Answer 81

Proteases break down peptide bonds through hydrolysis. These can be general proteases such as trypsin, chymotripsin and pepsin. These proteases help in breaking down protein in food during digestion. There are more specific proteases that often help to activate precursor proteins to their active form. In the clotting cascade , blood clotting factors are proteases that upon trauma lead to a cascade of activated factors resulting in clots. Angiotensinogen is cleaved by Renin to form Angiotensin I which is cleaved by ACE to Angiotensin II which is the active hormone regulating blood pressure.

Question 82

Describe hydrogen bonds and their role in secondary structure formation.

Answer 82

Weak, but numerous, bonds between a hydrogen atom and an electronegative atom such as oxygen. The H is the "donor" of electron density; the electronegative atom is the "acceptor" of electron density. "H bonds between the protein backbone amide nitrogen and carbonyl oxygen are the driving force to form secondary structures." On alpha-helices , the H-bonds between every n and n+4 amino acid within the same polypeptide chain, every 8th amino acid is right on top of the 8th before it. These bonds force the chain into a right-handed helix. All side chains point out of the helix. For beta-sheets the hydrogen bonds are between two polypeptide chains (or one chain with a turn). The chains can be parallel or anti-parallel.

Question 83

Describe two major types of protein secondary structures.

Answer 83

alpha-helix: ~30% of all protein structures (also see above) All biological alpha-helices are right-handed screws All side chains point towards the outside of helix Every five residues there's a H-bond between the carbonyl O and amine H. Every 8th amino acid is aligned with the one 8 before Hemoglobin is a good example. Thalassemia is a genetic disease where mutations destabilize hemoglobin by disrupting alpha helices. beta-sheet: ~30% of all protein structures Polypeptide chains are held together by H-bonds Strands can be parallel or anti-parallel, anti-parallel is more common. Immunoglobulin or antibody is an example. Notice that the primary structure of the protein determines its secondary structure-- different side groups influence the formation of different patterns of H-bonding. There's also turns and loops to compact proteins (mainly glycine, because it's small and flexible, and proline, because it's slightly kinked). Notice there's a triple helix structure in collagen: different conformation from either alpha-helices or beta-sheets. Has a lot of glycine and proline, like turns or loops, but the proline residues are often hydroxylated (via vitamin C), resulting in massive H-bonding between strands to form the characteristic triple helix. Every three residues there's a glycine residue (the smallest AA, in order to compact the helix).

Question 84

Explain tertiary and quaternary structures.

Answer 84

Tertiary structure is the spatial arrangement of the polypeptide chain. The two major classifications of tertiary structure is globular (most of the proteins in the body, lipid or water soluble, diverse functions) versus fibrous (long and composed of alpha helices OR beta sheets, insoluble and have structural or protective role). Separate domains having independent secondary structure can come together to form the tertiary structure. Features which control the tertiary structure of proteins: N-terminus of protein is synthesized first, meaning that the AA's at that end have an opportunity to fold first (affects overall folding and refolding after denaturation). Interactions between adjacent AA's in secondary structures Hydrophobic AA's tend to cluster in the cores of proteins (vs. hydrophilic AA's). Various proteins assist with "chaperone" activity in folding. (about 30% of all eukaryotic proteins are folded by chaperones.) Quaternary structure is how multiple polypeptide chains come together to form a single functional protein. Hemoglobin is made up of four subunits.

Question 85

Explain the role of loops in protein structure and function.

Answer 85

Enable polypeptide chain to form particular structures instead of staying as extended linear chains Can interact with other proteins such as with variable loops of in immunoglobulin molecule Loops and turns are mostly Glycine and Proline amino acids

Question 86

Explain how to use Kd to represent binding strength.

Answer 86

Kd is the dissociation constant | Kd= [ligand] when 50% of ligands are bound

Question 87

Explain how binding specificity can be achieved.

Answer 87

Lock and Key complementary model: the ligand and protein have high complementarity where the ligand fits very nicely onto a site of the protein. There is complementarity of size, shape, charge, hydrophobicity Induced fit model: There is not preformed complementarity (as in lock and key model) but instead both can undergo conformational changes upon binding of the ligand. For enzymes the ligand binds into its transformational state reducing activating energy and speeding up the reaction.

Question 88

Explain how heme enables myoglobin to bind oxygen.

Answer 88

Free or exposed Fe would be oxidized irreversibly, and the protein itself can’t bind O2. A heme group isolated within the protein works perfectly (without producing free radicals). Myoglobin is the main oxygen storage protein in mammals

Question 89

Explain the molecular basis of carbon monoxide poisoning.

Answer 89

Carbon monoxide has a similar size and shape to O2 so it also can fit to the same binding site CO binds over 200,000 time stronger than O2 (though the structure of myoglobin and hemoglobin reduce that to ~200:1) so it will out compete O2 every time, blocking the functions of myoglobin, hemoglobin and mitochondrial cytochromes involved in oxidative phosphorylation

Question 90

Explain why hemoglobin is a good oxygen transporter.

Answer 90

Myoglobin binds O2 too strongly, so it would not be released in the tissues Hemoglobin has four subunits each with heme groups that interact with each other in positive cooperativity (thus the sigmoidal binding curve). Hemoglobin has a Tense, lower binding affinity state and a Relaxed, higher binding affinity state triggered by initial O2 binding. Thus in the lungs with lots of O2 it will easily bind in the Tense state creating the Relaxed state bringing O2 to the tissues, where the pH is lower and O2 does not bind as well causing it to release and go back to the Tense state. The pH difference is called the Bohr effect. Tense: low affinity state (in lungs) Relaxed: high-affinity state (leaving lungs) High pH = high O2 binding (in lungs) Low pH = low O2 binding (in tissues)

Question 91

Identify the factors that cause protein denaturation.

Answer 91

Heat- think of cooking egg whites pH- yeah acids and bases are never good Chemicals- organic solvents, urea, detergent

Question 92

Describe the most fundamental conclusion drawn from the Ribonuclease refolding experiment.

Answer 92

Used reducing agent to denature Ribonuclease A, then after dialyzing out the urea it slowly refolded and restored almost 100% of activity All the information needed to fold the protein correctly is embedded in the primary amino acid sequence The environment provided by the inside of the cell is not always required in order for proteins to fold correctly The protein does not explore all possible structures while folding, there is instead a pathway it follows. (Levinthal’s paradox)

Question 93

Describe the two classes of chaperones and the general function of chaperones.

Answer 93

Heat Shock Proteins (Hsp70): induced at elevated temperatures and binds to hydrophobic region of unfolded proteins to prevent aggregation, can also help transport some proteins across membranes in unfolded states, works with other heat shock proteins Chaperonin: consists of a cap and two 7-subunit rings. The hydrophobic region of the unfolded protein binds to the hydrophobic region of the chaperonin then with some ATP and a conformational change of the chaperonin the protein is folded at least partially so that it can only continue to the final native shape. Ex: GroEL/GroES complex in E. coli.

Question 94

Explain why sometimes protein disulfide isomerase or protein prolyl isomerases are required for protein folding.

Answer 94

Disulfide isomerase: incorrect disulfide bond formation between free cysteines means this enzyme needs to come in, reduce the improper ones and reform them correctly Protein prolyl isomerase: This protein reforms proline from the typical trans formation to the cis formation (6% of proline in mammals) for proper folding and functioning

Question 95

Recognize that protein mis-folding is the major cause of prion disease, Alzheimer's disease, Parkinson's disease, and amyloidosis.

Answer 95

Prion disease - the prion protein (PrP) simply misfolds, which then causes other normal prion proteins to misfold. These are altered from alpha helices to beta sheets causing them to aggregate within amyloid plaques leading to neuron loss and gliosis Alzheimer’s disease: the normal AB-40 folds correctly, however the slightly different AB-42 misfolds and aggregates into amyloid plaques. This leads to B-amyloid plaque and tau tangles. Parkinson’s: Beta-synuclein misfolds into Lewy Bodies. Amyloidosis: Generalized protein misfolding in the rest of the body leading to a variety of disease (from Type II diabetes to Cardiac amyloidosis).

Question 96

Identify the secondary structure changes and the infectious agent in prion disease.

Answer 96

The normal prion protein is misfolded into an infectious form. It then converts other normal prion proteins into the bad form. These are altered from alpha helices to beta sheets causing them to aggregate within amyloid plaques leading to neuron loss and gliosis.

Question 97

Describe the major approaches for purifying a protein.

Answer 97

According to size by gel filtration chromatography: Have proteins run through a filtration column; large proteins flows through quickly, small proteins more slowly. According to charge by ion exchange chromatography: Have proteins run through a cation exchange column filled with negatively charged beads; positively charged proteins stick and can be eluted off with salts (more slowly), negatively charged proteins flow directly through. According to ligand binding properties by affinity chromatography: Attach a given ligand to beads of column (eg glucose)-- run protein solution through column and only the ligand-binding proteins should stick. Then elute the proteins off the beads into a separate container Ligands: enzyme substrates, DNA, metal ions, carbohydrates, peptides, etc. Extra info for proteins: Electrophoresis: (SDS-PAGE) determine size of protein. Apply a detergent to denature proteins and coat them in an even negative charge. Use a polymerized acrylamide gel to separate based on size. Electric field causes denatured, negatively charged proteins to run down gel at a rate dependent on size. Also need to run a "marker" of known molecular weights to give a "size ladder" to determine the approximate size of the protein under investigation. Mass spectrometry: determine sequence of (unknown) protein by molecular mass. Edman Degradation: Label and remove the N-terminal amino acids one at a time and identify. Western Blots: use immunology to identify proteins on an acrylamide gel. Transfer proteins onto a membrane, react with primary antibody for a particular protein, wash off anything unbound, react with fluorescent secondary antibody to detect. One use is to identify HIV infection: use patient's serum as primary antibody (checking for HIV antibodies in serum) against HIV proteins.

Question 98

Write out three palindromic double-stranded DNA sequences that are likely to be cut by restriction endonucleases (a known restriction enzyme does not have to exist for your sequences).

Answer 98

Palindromic usually means they read the same forwards and backwards (ie "A man, a plan, a canal, Panama"). Here it seems to mean that the same sequence, read 5' to 3' on the two complementary strands (NOT on the same strand) will read the same. Here's our sequence: 5' TACGTA 3'. This implies its complementary sequence: 3' ATGCAT 5'. If you read the 5' to 3' sequence on each, it's the same. Note that TACGTA is not a classical palindrome (it doesn't, by itself, read the same backwards) so don't get confused. Some other sequence examples: TTTAAA TCCGGA CTATAG

Question 99

Explain the principle of electrophoretic separation of DNA to a pre-med student.

Answer 99

You want a method to separate DNA strands by size. You do it with a polarized electrical field in a gel that will exert a uniform electromotive force on all the DNA in that gel to move down its gradient. The gel is thick but porous, meaning that smaller strands of DNA will move through it faster than larger strands. It's made of polymerized acrylamide (or other stuff) Because DNA has a uniform negative charge per unit length (from the phosphate backbone) only size of the strand affects its speed. The DNA samples are placed in wells in the gel, and a current is applied - the DNA will migrate toward the positive pole. A "ladder" mixture containing known sizes of DNA is run alongside the samples for comparison. Remember: Smaller strands migrate farther down the gel than larger ones A substance that intercalates into the DNA and fluoresces, such as ethidium bromide, is used to visualize the results. A Southern Blot is performed by denaturing the DNA then blotting the gel (after electrophoresis) onto a membrane that binds DNA. The membrane is then washed with short, labeled "probe" sequences of DNA or RNA, and visualized. This technique is used when examining a large quantity of many types of DNA for particular sequences, e.g. for diagnosis of genetic diseases.

Question 100

Give an example of a disease that can be diagnosed using a restriction fragment length polymorphism (RFLP) and a use of DNA fingerprinting. Describe at least three experimental stages required in each of these procedures.

Answer 100

RFLP: Example of looking for the HbS mutation of sickle cell anemia. Sickle Cell anemia results from a mutation that, coincidentally, lies in a restriction site. Digest patients' DNA with diagnostic restriction enzymes (MstII) Electrophorese against a normal genome Southern blot with P32 labeled B-globin gene- if 1.1kb and 0.2kb then normal, sickle cell is both together at 1.3kb DNA fingerprinting: Example: paternity testing PCR with primers that surround variable number tandem repeat (VNTRs) sequences Electrophoresis/detection of altered size of DNA fragment patterns. Compare sample against target(s) for similarity

Question 101

List the names given to the transfer of DNA, RNA and protein respectively from an electrophoresis gel to a membrane.

Answer 101

Transfer of DNA: Southern blot (actually named after somebody) Transfer of RNA: Northern blot (to follow Southern) Transfer of proteins: Western blot (to follow Northern)

Question 102

Describe three characteristics of a hybridization probe that you will use to detect a specific DNA sequence on a membrane.

Answer 102

Specific length: you need to know what temperature to anneal the probe to the DNA with, and that's dependent on the primer's length Specific sequence: so it can bind to the DNA you're interested in. Radioactivity (or some other visualization characteristic): so you can detect it later. Quantity: add it in sufficient quantities to outcompete the other strand of DNA in annealing to its target DNA.

Question 103

Recall the classes of enzymes that are used in recombinant technology to: a) copy a DNA sequence into a DNA sequence, b) copy an RNA sequence into a DNA sequence, and c) join DNA fragments.

Answer 103

copy a DNA sequence into a DNA sequence: DNA polymerases. copy an RNA sequence into a DNA sequence: reverse transcriptases. join DNA fragments: DNA ligases.

Question 104

Describe the three main stages that are repeated during PCR amplification. State the approximate temperature of each step, and relate this temperature to the state of the DNA molecules in the PCR reaction.

Answer 104

Step 1: add thermal-stable DNA polymerase (usually Taq polymerase) and dNTPs (deoxyribonucleoside triphosphates) and desired primers. Step 2: Heat to 95 degrees C (denaturing DNA). Step 3: Cool to 55 degrees C (allows primers to hybridize). Step 4: Warm to 72 C at which time the polymerase copies the DNA.

Question 105

Describe at least 1 distinct use for PCR amplification in the diagnosis of a genetic condition in your patients.

Answer 105

Prepare a primer that hybridizes with a mutant copy of a gene but not the normal copy. Then hybridize and replicate as per normal PCR with your patient's chromosomes. PCR will generate a detectable amplification of genetic material if the mutant gene is present. Thus can use PCR as a diagnostic test for discovering mutant genes that are markers for cystic fibrosis, beta-thalassemia, etc.

Question 106

Compare and contrast the molecular details of the processes of DNA sequencing and PCR amplification in a short paragraph.

Answer 106

DNA sequencing: uses a primer for a given start sequence (at the beginning of the region to be sequenced) and four differently labeled dideoxynucleic acids, as well as normal unlabeled dNTPs, in DNA synthesis. When the DNA synthesis incorporates a ddNTP, it stops synthesis with a particular final "color" or label, indicating which base it ended with. You run this a while and allow a lot of different lengths of DNA to be made. The product is then run through a column chromatograph that separates based on size. A detector at the bottom of the gel detects the different colors of labeled ddNTPs as they flow past. (In the "manual" version, labeled ddNTPs would be run in a reaction in four separate tubes, on for each kind of ddNucleotide, then run in four adjacent rows with gel electrophoresis) PCR amplification: uses a primer for a given start sequence and DNA polymerase to make more copies of a desired DNA sequence. Use varying temperatures to melt DNA, hybridized probe, and have polymerization reaction happen. The point of PCR is to amplify (make lots of copies of) a given DNA segment. The point of sequencing is to determine the sequence of an unknown DNA fragment. Similarities: Both use primer sequences to initiate replication of genes. Differences: PCR is used on double-stranded DNA, sequencing on a single-stranded DNA fragment. Sequencing used ddNTPs, PCR doesn't.

Question 107

Be aware of the different types of cloning vectors and their general features.

Answer 107

Plasmids: Vectors for amplifying DNA sequences in bacteria, max insertion of 20kb into E. coli, simple but inefficient. Bacteriophage: Used to infect E. coli and use its replication machinery to produce the recombinant vector. Insertion up to 25kb, more efficient than plasmid. Cosmids: hybrid of bacteriophage and plasmid: use plasmid replication origin; can take up to 45 kb insert into E. coli. BAC: bacterial artificial chromosome with insert up to 300kb, good for chromosome mapping and sequencing YAC: yeast artificial chromosome with insert up to 2mb, chromosome mapping and sequencing Retroviral vectors: can carry very large inserts; introduce DNA into mammalian cells, delivers gene therapy.

Question 108

Describe the use of microarrays for measuring levels of mRNA (e.g. gene expression).

Answer 108

Microarrays rely on hybridization, limited by background hybridization and signal saturation, can’t provide sequence level information, can miss mutations, modifications, and splice forms

Question 109

Describe how cells regulate their size by coordinating growth with division at the Restriction point ("R") in G1 phase.

Answer 109

In somatic cells, there's a "R point," or restriction point, at which the cell has to make a decision - based on whether or not various hormones or growth factors are there, as well as whether the cell is large enough - whether or not to undergo replication and mitosis. (checkpoint before leaving G1 and entering S) "R" point = first (thus most highly regulated) step in cell cycle pathway. In embryonic cells the R point is bypassed and continues to divide despite getting smaller every time. At all other times the R point prevents reducing the surface to volume ratio, "divide itself out of existence.”

Question 110

Recognize that the main goal of the somatic cell cycle is to ensure exact duplication of the genome in S phase followed by exact of division of the genome in M phase to produce identical daughter cells.

Answer 110

Don’t want extra chromosomes (reinitiation), etc., want them divvied up correctly

Question 111

Describe how cells prevent re-replication of their genomes by keeping the assembly and activation of replication complexes in separate cell cycle phases.

Answer 111

M- high CDK prevents building Pre-Replication Complex G1- low CDK allows building of Pre-RC S- high CDK activates replication and prevents any more Pre-RC building During G1: Orc proteins [Origin replication complexes] initiate replication by binding to the DNA origin and binding to it; other proteins (cdt1 and cdc6, helicase) complex with Orcs. In short: pre-RCs assembled (but not activated) during G1, and activated (but not assembled) during S, driven respectively by the low and high concentrations of CDK in the cell.

Question 112

Recognize that genomic instability either by chromosome re-replication in S phase or mis-segregation during mitosis produces human diseases such as cancer and birth defects (trisomy 21).

Answer 112

You never want excess or shortage of genetic material as it will be translated…

Question 113

Compare the cell cycle of somatic cells (mitosis) with that of germ line cells (meiosis), which produces haploid gametes.

Answer 113

End point of mitosis: 2 identical diploid cells from 1 diploid cell. End point of meiosis: 4 different haploid cells from 1 diploid cell. Start: chromosome A and chromosome a. 1st: replication of chromosomes (AA, aa) 2nd: homologous recombination of chromosomes to provide genetic variation 3rd: separate homologues (diploid to diploid, or meiosis I) (one cell: AA; another cell: aa). 4th: separate chromosomes of daughter cells (diploid to haploid, or meiosis II) (two cells: A, two cells: a).

Question 114

Recognize that differentiated, post-mitotic cells such as neurons are stuck at the "R" point in that they continue to grow without cycling.

Answer 114

These cells need to grow and serve specific long-term functions (all those stable synapses)

Question 115

Describe how alterations in many cell cycle regulators that are found in cancer cells are being used for patient diagnosis and prognosis.

Answer 115

Diagnosis: sequence genes as with LFS (ATM mutation), BRCA, etc. for specific, common mutations In general: CDKs [cyclin-dependant kinases] : 6 enzymes central to regulating the cell cycle that phosphorylate Ser, Thr, Tyr and are regulated by cyclin Active CDKs are produced by growth factor hormones (and cancer factor proteins) and result in cell replication and duplication. Particular CDKs tend to be found at particular levels in various types of cells. So if an elevated level of a given CDK is found in a tumor, that's an indication of active proliferation. Low levels of RB, or highly phosphorylated (inactivated) RB will indicate abnormally high replication rates (as it inhibits entrance into S) There's also a family of proteins called CDIs [CDK Inhibitors] that turn off CDKs (thus repressing the cell cycle]. Mutations in these proteins often turn off the ability to turn off a given CDK, thus result in uncontrolled proliferation. Most frequently seen in cancer cells: mutation in Ink4 (inhibitor of CDK4) CDI family that lets CDK4 push past the "R" point regardless of other growth factors. So can look for mutations in CDI genes as a cancer marker as well. mitogen: signal protein that leads to activation of CDK4. Notice that this leads to a positive feedback cycle which leads

Question 116

Recognize the importance and the mechanism of cell cycle checkpoints in maintaining genomic stability.

Answer 116

Checkpoint at G1: Is the cell large enough? (R point/Rb pathway) Checkpoint at S: Should DNA be replicated? Any mutations? (S point/p21 on cyclinE) Checkpoint at G2: Correct copies of new/old DNA? Checkpoint at M: Did the spindles form and act as they should? Mitogen produces CyclinD which inhibits Rb popping off E2F creating cyclinE which initiates S phase. p16 (Ink4) inhibits CDK4,6 preventing passage to S phase (R point).p21(Cip/Kip) inhibits CDK2 preventing activation of synthesis (S point). Failing any of these checkpoints results in a cessation of the cell cycle (cell stays in phase it's in when the checkpoint's failed) until the damage is repaired or the checkpoint is passed. Notice that you can repair DNA in any phase. If the damage can't be repaired, have apoptotic signaling. Mechanism: Sensor proteins on the DNA communicate with transducers which communicate with effectors that regulate the cell cycle via checkpoints. Summary: sensors (such as Rad 17) -> ATR ->Chk1,2 activating repair enzymes and BRCA1/p53 (respectively) OR ->ATM ->p53 -> p21 leading to arrest, apoptosis, DNA repair Notice that most cancer is not inherited unless the mutation is present in germ cells.

Question 117

Describe what an enzyme is, the characteristics of enzymes, and thermodynamically how they increase the rate of a reaction. Define the terms: catalyst, activation energy, free energy of the reaction.

Answer 117

Enzyme: molecule that increases the rate of a chemical reaction without itself being irreversibly changed in the process Have specific structures and active sites where the catalysis happens Some use specific cofactors or coenzymes that help in the catalysis Classified and named by the reaction that they catalyze Catalyst: something that increases the rate of a reaction (such as an enzyme) Activation energy: the hump of increased energy that must be overcome for products to become reactants, dictates the rate of the reaction. Reaction free energy: free energy differential between reactants and products (products – reactants) dictates the spontaneity of the reaction.

Question 118

Describe how enzymes work using the terms: binding of the transition state, induced fit, covalent chemistry, metal ion chemistry, and general acid-base chemistry.

Answer 118

What you're essentially talking about is the following equation: E + S -> ES -> EP -> E + P, where E = enzyme, S = substrate, and P = product. Current theory: Enzyme binds and stabilizes the substrate in its transitional state, where induced fit can change the conformation of the substrate and/or enzyme putting even more pressure on the substrate towards becoming the product. This decreases the amount of energy (activation energy) needed to form the product. Active site chemistry can rearrange covalent bonds Metal ion chemistry refers to the idea that some enzymes utilize cofactors that attract the transition state and help to form the substrate product via redox reactions (almost 1/3 of enzymes) General acid-base catalysis: side chains of enzymes donate or accept protons to stabilize transition states Covalent catalysis: transient covalent bond between enzyme and substrate

Question 119

Define the terms cofactor and coenzyme.

Answer 119

cofactor: Metal ions required for enzymatic activity. coenzyme: An organic ligand that binds to the enzyme and allows it to act on the substrate. Some terms associated with this: A ligand that is tightly or covalently linked to the enzyme is a prosthetic group. The complex of a cofactor or coenzyme with an enzyme is a holoenzyme The enzyme dissociated from its cofactors or coenzymes is an apoenzyme.

Question 120

Describe the significance of the terms Km and Kcat (know what it means if one enzyme has a lower or high Km than another, etc.) and estimate the value of Km from a graph of reaction velocity versus [S].

Answer 120

Km is the substrate concentration at which the 'enzyme velocity' of the reaction is 1/2 the maximum velocity of the enzyme. Generally, enzymes in biological systems tend to operate around ½ Vmax, as this is the typical substrate concentration in the cell. If an enzyme has a lower Km than another, then it needs less substrate to operate at ½ Vmax [[http://en.wikipedia.org/wiki/Lineweaver–Burk_plot|Lineweaver-Burke plot]] (Double-reciprocal plot) : a way to empirically figure out Km and Vmax. kcat : "turnover number", a rate constant for a given enzyme of how many substrate molecules are converted to product in a given unit of time by a molecule of that enzyme under saturated conditions. Kcat/Km tells you the overall efficiency of an enzyme Smaller Km = less substrate needed to work at ½ Vmax. Bigger kcat: faster action of enzyme. Thus a large Kcat/Km is indicative of a highly efficient molecule.

Question 121

Describe four different types of inhibitors and, in general, how each works (competitive, uncompetitive, mixed, and irreversible).

Answer 121

Reversible inhibitors: Binds to enzyme and decreases enzyme function until released. Competitive inhibitor: inhibitor goes into the substrate binding site. Competes for the binding site with the normal substrates. Increases Km. Uncompetitive inhibitor: inhibitor binds somewhere other than the binding site. Means it doesn't have to compete with normal substrates for binding. Only binds to the ES complex. Lowers Vmax and changes Km (note that Noncompetitive is not the same - it is a type of mixed inhibitor) Mixed inhibitor: inhibitor binds outside of active site to either E or ES complex, affects both Km and Vmax. You get different kinds of Lineweaver-Burke plots for these different types of reversible inhibitors. So you can figure out experimentally what kind of inhibitor you're looking at by using these plots. Irreversible inhibitors: combine with or destroy a functional group of the enzyme. This is a permanent change. Ex.: Penicillin irreversibly inhibits enzyme responsible for polymerizing peptidoglycans, weakening bacterial cell walls. Goes into active site and forms a covalent, irreversible complex with enzyme.

Question 122

Describe four types of enzyme regulation mechanisms and, in general, how each functions (allosteric, covalent modification, binding of another protein, proteolytic cleavage).

Answer 122

Basic principles: generally, the first enzyme in a long regulatory pathway is the one most highly regulated. Allosteric regulation: Some positive or negative modulating molecule binds to the enzyme and causes a conformational change to allow or disallow the action of the enzyme on the substrate. Sometimes this is involved in biofeedback loops where the end product of the pathway is the allosteric inhibitor/activator of the first enzyme in the pathway (neg, pos feedback). Covalent modification of the enzyme: Many enzymes can be phosphorylated to dramatically affect functionality. The enzymes that modify enzymes to turn them on or off are usually called kinases Also a variety of other modifications: adenylations, methylations, etc. Regulatory protein binding: Some enzymes are bound by proteins to activate or inactivate them. Proteolytic activation: Some enzymes are inactive until cleaved by a particular enzyme (ie trypsinogen doesn't become active until it gets to the duodenum and is cleaved to trypsin), (angiotensin-renin pathway).

Question 123

Name five sources of DNA double-strand breaks.

Answer 123

Endogenous Meiosis Generation of immune receptor diversity: both VDJ recombination and class switch recombination in B cells, and T cell receptors in T cells. DNA replication, when the replication fork encounters a nick in one template strand or stalls at fragile sites, or sites of topoisomerase failure On insertion of retroviruses or retrotransposons into genomic DNA Exogenous Damage from ionizing radiation in the environment also causes DNA DSBs: These include cosmic rays and radiation from soils. Ionizing radiation in Denver, Colorado, gives us a yearly dose of ~50 mrems, which is about twice the national average. We also get 75-140 mrem/year (2.5-5x the national average) from the enriched uranium soils of the Colorado Plateau. The damage from ionizing radiation can be both direct (deposition of energy) and indirect (clustered oxidative free radicals caused by the energy deposition in turn cause DNA breaks). Medical tests and treatments are another significant source of ionizing radiation. Many commonly used imaging techniques including X-rays and CT scans, as well as radiation treatment and some chemotherapeutic agents (e.g. bleomycin) cause DSBs.

Question 124

Distinguish between the two mechanisms used to repair DNA double-stranded breaks.

Answer 124

DNA non-homologous end joining (NHEJ) NHEJ initiates with recognition of the DS break by Ku Ku recruits DNA-PKcs NHEJ variably uses combinations of: nuclease to remove the damaged DNA if there is damage (Ku, DNA-PKcs + Artemis complex) polymerase to fill gaps repair finishes with ligase to restore the continuous phosphodiester backbone on both strands (LIG-4) Regulated by 53BPI homologous recombination (HR) Homologous recombination (HR), on the other hand, occurs during and shortly after DNAreplication, in S and G2 phases of the cell cycle. It requires nearly identical DNA strands (homologous). For perfect repair, HR uses the unbroken sister chromatid − with this perfect template, HR repairs the break perfectly, with no gain or loss of nucleotides. Multiple steps are required to repair double-strand breaks by HR. First the broken ends are resected, i.e., processed to expose single-strands ending in 3’ OHs. The single-strands then invade the homologous DNA, using it as a template for synthesis to bridge the gap caused by the break. Once the newly-synthesized DNA closes the gap, the crossover (Holliday junction) is resolved, regenerating the two separate double-strand DNAs. Typically, when dsDNA breaks are repaired by HR using the identical sister chromatid, the process perfectly restores the break with no exchange. However, it is important to realize that different resolutions of Holliday junctions can lead to exchange of genetic information between the two chromosomes undergoing HR. This exchange is actually an essential feature of meiosis and can take two forms, gene conversion or crossing over. Moreover, in some cases of mis-regulated homologous recombination, the use of the other chromosomal homolog as a template for double-strand break repair instead of the sister chromatid in somatic cells can lead to loss of heterozygosity (LOH) by the same biochemical mechanism. Regulated by BRCA1

Question 125

Explain why defects in either of these two modes of double-strand break repair can increase cancer risk.

Answer 125

Mis-repair is the source of genetic instability with translocations being particularly bad. Loss of HR leads to gross genomic rearrangements and thus genome instability. Translocations most often occur when DNA DSBs are healed between different chromosomes by NHEJ. Failure to repair a DSB will lead to cell death, a fact exploited by cancer therapies (ie radiation and drugs that cause DSBs like bleomycin). DSB repair deficits, which increase cancer susceptibility, can be caused by loss of proteins involved in the signaling (both at the level of sensors and transducers) of the DSBs and in the enzymes that mediate their actual repair

Question 126

Describe three classes of proteins with distinct functions that, when mutated, lead to failure to repair DNA double-strand breaks. (Add to this one)

Answer 126

``` Sensing Ku, MRE11 complex, Parp1 Signal Transduction ATM DNA repair HR, cNHEJ, aNHEJ ```

Question 127

Describe the major types of next-generation DNA sequencing.

Answer 127

Short-read sequencers. DNA sequencers that produce millions to several billion of short ~100 base-pair (bp) reads in a single run (Mardis 2013). The machines use ~1000 bp DNA templates as input, and can produce sequence reads from one or both sides of the molecule. In general, these platforms have very low error rates that approach 10-6 (i.e., there is a 1 in a million chance that a given base call is wrong). Short-read sequencers are commonly used to provide high coverage (“read depth”) over specific subsets of DNA templates from a complex mixture (e.g., exome sequencing). Current companies are Illumina and Ion Torrent. Long-read sequencers. DNA sequencers that produce ~10,000 sequencing reads that can reach 10,000+ bp in length (Mardis 2013). These platforms have high error rates, around 10-1 (i.e., there is a 10% chance that a given base call is wrong). Long-read sequencing approaches are useful in human genetics for linking contiguous polymorphisms that occur on the same haplotype. In addition, long reads are useful for measuring the diversity of mRNA isoforms in human cells that arise from alternative splicing of pre-mRNA transcripts. These machines focus on single-molecule approaches to extract information from individual DNA templates. For example, the Pacific Biosciences platform can “watch” single DNA polymerases incorporating fluorescent nucleotides into DNA chain, and the Oxford Nanopore platform extrudes DNA templates through tiny pores and measures diagnostic changes in electrical current across the pore that report on the specific DNA sequence moving through. Because both of these platforms measure the properties of single DNA molecules in close to real time, very long sequences can be generated.

Question 128

Compare and contrast the types of data that next-generation DNA sequencers produce, in terms of read number and error rates.

Answer 128

``` Read depth (“coverage”). The number of times a base in the reference genome was independently sequenced. For example, if 100 unique reads align to and cover a position in a reference genome, then coverage of that base is 100X. When this position contains a variant relative to the reference sequence, higher coverage improves the confidence that a true variant was identified, and is not due to errors in the sequencing reaction. Error rate. A platform-dependent measure of the quality of converting a signal into an observed base call. Short read sequencers usually sequence many copies of the same molecule in a given reaction, and are therefore more robust to small fluctuations in noise, leading to low error rates (10-6). In contrast, long read sequencers typically sequence single molecules, and are therefore more susceptible to noisy measurements that lead to higher error rates (10-1). High error rate can be offset by higher coverage. Contiguity. The genome of a given individual is derived from recombination events between the DNA of male and female gametes. These recombination cross overs tend to be interspersed over large distances. Any variant between two crossovers can be inherited or lost, and the pattern of inherited variants is therefore associated with one or the other of the haploid gametes, leading to “haplotypes” of inherited variation. Short-read sequencers only interrogate ~1,000 bases at a time, and so a variant observed with a short-read sequencer cannot be linked to other variants in the same haplotype block. Long reads (104 – 106 bases) can sometimes be used to link variants together, providing a picture of contiguity of variation in haplotype blocks. ```

Question 129

Explain the key criteria that must be considered during analysis of next-generation DNA sequencing data for SNP identification.

Answer 129

Ploidy need enough coverage to determine heterozygosity | Also see above

Question 130

Describe exome sequencing experiments and explain types of mutations that can be detected with this strategy.

Answer 130

Sequence the ~1% of the human genome that codes for proteins Used to identify novel coding SNPs among e.g. patients with same genetic disorder

Question 131

Describe which types of diseases are best characterized by exome sequencing and which types of diseases are likely not diagnosable by exome sequencing.

Answer 131

Single-gene Mendelian diseases Variants are RARE, and not likely to be in existing databases of variation (e.g. dbSNP, 1000 genomes) Variants are non-synonymous (i.e. change an amino acid residue) = likely to have functional consequence E.g Miller Syndrome- Embryonic exposure to methotrexate

Question 132

Explain the relationship between the structure of the prion protein (PrP) and the infectious agents known as prions.

Answer 132

PrP is mostly alpha helices, infectious prions are full of beta-sheets. These beta-sheets are more hydrophobic and clump into lesions.

Question 133

Describe how prion disease can be sporadic, inherited or infectiously acquired.

Answer 133

Sporadic- random misfolding that infects the rest of the prions Inherited- familial strains that lead to early onset Creutzfeldt-Jakob Diseas e, fatal insomnia Infectious- can be from diet (vCJD or Kuru) or iatrogenetic (from medical treatment- pituitary hormones and dura mater)

Question 134

Describe the relationship between bovine spongiform encephalopathy and the disease of humans known as "variant Creutzfeldt-Jakob disease."

Answer 134

BSE is mad cow disease and vCJD is a strain of prion disease that originated in cows and through dietary ingestion led to vCJD in human (much lower incidence and 5-10 year lag compared to CJD).

Question 135

Explain what is meant by the concept of "prion strains."

Answer 135

A prion strain has distinct characteristics such as incubation time in defined host, clinical signs, distribution of protease resistant PrP in the brains of affected animals. These features are stable on serial propagation. Note that all strains are have the same sequence (as PrP) Ex: Hyper and Drowsy strains of PrPsc as tested in hamsters. The varying characteristics are incubation time (70 v 148 days), symptoms (hyper v drowsy), distribution of histological lesions, distribution of prions in brain

Question 136

Explain the possible relationship between prion disease and other human neurodegenerative diseases, at the level of protein structure.

Answer 136

Many common neurodegenerative diseases share similari.es with prion diseases on a protein level Parkinson’s Alzheimer’s Amyloidosis

Question 137

Comprehend the pathophysiology of Alzheimer's Disease.

Answer 137

A-42 is the incorrectly cleaved protein that incorrectly folds and becomes insoluble and leads to the amyloid plaques Tau protein also forms fibrillary tangles The tau and A plaques lead to inflammation and neuronal loss Sources: Apolipoprotein E4 increases amyloid deposition ApoE2 decreases amyloid Mutations can preferentially lead to A-42 cleaving by blocking alpha or gamma secratase sites, or easier to cleave the beta site Presinilin 1 mutation on chromosome 14 increases A-42 production via gamma secretase Presinilin 2 mutation on chromosome 1 also increases A-42 Extra chromosome 21 increases all APP products

Question 138

Describe the central role of post-translational amyloid precursor protein processing in the pathogenesis of Alzheimer's Disease.

Answer 138

see card 137 (past card)

Question 139

Consider interventions that might help delay onset or prevent the development of Alzheimer's Disease.

Answer 139

Inhibit all APP products | Antibodies that attack A-42