Learning Objectives for Lectures 24,25, 30&31, 32, 37,38, 40, 43, 44, 45

1. Lecture 24: Learning objectives: Human genome

2. 1. The central Dogma says that DNA contains information that codes for proteins, yet only1.5% of human DNA serves direct coding functions. Explain why this is true and whatthe rest of our DNA is. ~ 1.5% of human DNA encodes information to make proteins (information found in mRNA) (Introns do not contain protein coding information) Remainder? Junk? Structural (centromere, telomere) Regulatory (promoters, regulatory elements) Some genes don’t code for proteins (RNAs to help make proteins) rRNAs, tRNAs, snRNAs newly discovered non-coding RNAs (ncRNAs) micro RNAs (miRNA) small inhibitory RNAs (siRNA) A lot of DNA of “unknown” function transcribed into RNA (transcripts of unknown function)

5. Tissue specific

6. Proteome is more complex than the genome

8. LINES (L1) – 20%

9. LTRs – 8%

10. There are also simple nucleotide repeats (sometimes referred to as Variable Number Tandem Repeats-VNTR or microsatellites)

12. 45% of genome are repetitive sequences

14. SNPs and disease-causing mutations: Not the same! If you know what a point mutation is, then the description of a SNP might sound similar. True, both are single-nucleotide differences in a DNA sequence, but SNPs should not be confused with disease-causing mutations. The image to the right shows some tell-tale differences: First, to be classified as a SNP, the change must be present in at least one percent of the general population. No known disease-causing mutation is this common. Second, most disease-causing mutations occur within a gene's coding or regulatory regions and affect the function of the protein encoded by the gene. Unlike mutations, SNPs are not necessarily located within genes, and they do not always affect the way a protein functions.

15. 7. What is the epigenome and how does it differ from the genome? Genome Blueprint for our phenotype Epigenome Inheritable chemical marks not dependent on gene sequence that dictate how the blueprint should work to determine our phenotype Contributes significantly to phenotype The epigenome is more likely to be influence by environmental factors Epigenomic marks play role in turning gene on and off DNA methylation Methyl groups attached to DNA backbone and regulate transcription Histone modification Chemical tags attach to histones – affect chromatin structure (how tightly DNA is wound)

16. 8. Describe the differences between the human mitochondrial genome and the nuclear genome. Are there diseases associated with mutations in mitochondrial DNA? If so what are the characteristics of these diseases. Mitochondrial DNA is tightly packed with information (unlike nuclear DNA where space does not seem to pose a problem) These DNA are circular (like bacteria) Human mitochondrial DNA is ~17,000 bp (recall nuclear DNA has 3.2e9 bp) It codes for: 22 tRNAs 2 rRNAs 13 hydrophobic subunits of the ETC enzymes All other mitochondrial proteins are encoded in the nuclear genome, translated in the cytoplasm, and the proteins imported into the mitochondria Several genetic diseases of strict maternal inheritance are caused my mutations in mitochondrial DNA The hallmark of these is: 1. Maternal inheritance 2. variability in severity of symptoms Note: oxidative damage to somatic cell mitochondrial DNA has a cumulative effect with age and may be related to age-related pathologies

17. Lecture 25: Learning objectives: Overview of Molecular Genetics - DNA structure

19. 5C sugar

20. atoms designated with (’)

21. 2’deoxyribose-DNA

23. 9 atom (atoms designated with regular numbers)

24. Pyrimidines

25. Cytosine (C), Thymine(T) in DNA, Uracil (U) replaces T in RNA

28. 3. 5-flourouracil is a chemotherapeutic drug. Explain the basic reason why the patient in clinical correlation 1 had an adverse drug reaction to 5FU. 5-Fluorouracil (5-FU) is a uracil analog widely used to treat solid tumors, such as colorectal and breast cancer. This agent has is still used as a major component of cancer chemotherapy. 5-FU is a pro-drug that requires activation to 5-fluoro-2-deoxyuridine monophosphate (5-FdUMP). 5-FdUMP inhibits tumor cell replication via inhibition of thymidylatesynthase (TS) activity, an enzyme required for de novo pyrimidine synthesis (see page 113). Approximately 85% of an intravenous dose of 5-FU is inactivated in the liver by dihydropyrimidinedehydrogenase(DPYD [page 114,115]), an enzyme that exhibits up to 20-fold variation in activity among individuals due to polymorphisms in the gene. Patients with low DPYD activity cannot effectively inactivate 5-FU, leading to excessive amounts of 5-FdUMP, causing gastrointestinal, and hematopoetictoxicities that are potentially fatal. (CC1) This is a true pharmacogenetic problem, since for the most part patients with DPYD polymorphisms show no abnormal phenotype till challenged with the drug.

29. 4. Describe the covalent bond that link monomer units together in a DNA. Describe the noncovalentbonds that are the basis for the double helix and indicate the basis of their specificity and their importance. How does this feature explain ‘Chargaff’s Rules’? Phosphoanhydride bonds “high energy” ATP N- Glycosidic bond 5’ phosphate (ester bond) Backbone Sugar phosphates Linked by 3’-5’ phosphodiester bonds Phosphodiester bond Links nucleotide units together in a nucleic acid The covalent bonds within the monomeric unit in DNA Negative charge The phosphate linkage provides a net negative charge to the molecule at physiological pH

30. 4. Describe the covalent bond that link monomer units together in a DNA. Describe the noncovalent bonds that are the basis for the double helix and indicate the basis of their specificity and their importance. How does this feature explain ‘Chargaff’s Rules’? COMPLEMENTARY BASE PAIRING is the fundamental property of DNA allowing it to act as the repository of genetic information (be self-replicating, transcribed into RNA and have its information translated into the amino acid sequence of a protein) Complementary base pairs are between specific bases on opposite strands of DNA in the double helix A with T : G with C. Pairing is based on the hydrogen bonding potential of specific groups (N, NH and C=O) of the bases. Three factors contribute to stability of the base pairs. 1. The steric requirement for a purine:pyrimidinepair to fit inside the double helix. 2. The physical alignment of h-bond donor (+) and acceptor (-) groups. 3. “base stacking” interactions - negative free energy associated with maximizing hbondsand minimizing exposure of planar aromatic bases to water The result of these forces are stable, specific base pairs, between an adenine base on one strand and a thymine base on the complementary strand, or a guanine base on one strand and a cytosine base on the complementary strand. A - T base pairs contain two h-bonds, G-C base pairs three h-bonds; making G-C base pairs more thermally stable. Prior to the description of the double helix by Watson and Crick, Erwin Chargaff determined that in DNA the molar amount of A always equaled the molar amount of T, and the molar amount of G always equaled the molar amount of C. These became known as Chargaff’s rules. Watson and Crick’s model demonstrated the biochemical basis of Chargaff’s rules.

31. 5. Explain what is meant by the two strands of DNA being ‘antiparallel’. What is the antiparallelfeature of the double helix really based on? Two strands of opposite polarity (anti-parallel) [implications in replication and transcription] Allows proper alignment of the complimentary base pairs– Essential for forming double helix (recall analogy from lecture of putting pointer and middle finger together.. Don’t fit.. Flip one hand upside down, now they fit)

32. 6. Describe how complementary base pairs in RNA differ from those in DNA. RNAs have secondary structure mediated by complementary base pairing, but unlike DNA, the base pairing in RNA is intramolecular; the molecules still contain single 5’ and 3’ ends

34. Stability of double helix related to number of h-bonds

35. Tm is temperature at which 50% of DNA single stranded

37. Lecture 30&31 Learning Objectives Candice Nalley candice.nalley@louisville.edu

39. Addition of the amino acid’s COOH group to the alpha phosphate of ATP

40. (yielding an aminoacyl adenylate and PPi)

41. Hydrolysis of 2 ~P drives reaction

42. Transfer of amino acid from the aminoacyl adenylate to the 3’ (or 2’) OH at the CCA end of the cognate tRNA

43. Ester linkage – preserves energetic activationNet reaction: amino acid + ATP + tRNA -> Aminoacyl-tRNA + AMP + PPi

45. proofreading activity

46. Aminoacyl-tRNAsynthetase removes amino acid from improperly charged tRNA

47. Replaces it with correct oneFidelity of aminoacylation is important. Attachment of an incorrect amino acid to a tRNA (mis-charging) results in mis-incorporation of an amino acid in any protein using the mischarged tRNA.

48. 2) Describe the initiation of protein synthesis. Where does protein synthesis start (what codon), and how does the protein synthesis machinery know to start there? What are the roles of eIF2; eIF4, eIF2B? Initiation is the rate-determining step and assembles the components needed for translation (mRNA, initiator tRNA, ribosomes) Initiation is mediated by initiation factors (IF in prokaryotes; eIF in eukaryotes) that mediate the assembly process Three major eukaryotic initiation factors eIF4 - the CAP binding protein complex eIF2 - the Met-tRNAMetbinding protein complex eIF2B - a guanine nucleotide exchange (GEF) factor complex.

50. accompanied by hydrolysis of the eIF2 bound GTP to GDP and release of the eIFs from the complex.

51. eIF2 is a G protein (GTP binding protein) =active w/GTP, inactive w/GDP

52. after hydrolysis, eIF2 remains GDP boundand inactive until the GDP is exchanged for GTP

53. Exchange catalyzed by initiation factor eIF2B (also called GEF – guanine nucleotide exchange factor)

55. 3) Explain how eIF2 phosphorylation can globally regulate protein synthesis. after the large subunit binds, GTP is hydrolyzed to GDP, and eIF2 leaves with it, tightly bound to GDP (eIF2 bound to GDP is inactive) phosphorylation of eIF2 (at serine 51) by specific eIF2 proteinkinases, prevents the exchange of GTP for GDP and inhibits eukaryotic translation. stops protein synthesis by depleting active eIF2-GTP Thus, eIF2 is the common target for global protein synthesis regulation using this mechanism

56. 4) Describe the P site, the A site, and the E site and their respective roles in protein synthesis. The growing polypeptide chain rests in the P-site (on the Peptidyl—tRNA) The incoming aminoacyl-tRNA enters into the A site A peptide bond forms between the COOH group of the polypeptide and the NH2 group of the amino acid on the aminoacyl-tRNA The ribosome moves (translocates) one codon so that the polypeptide (currently in the A site) is positioned in the P site, and the now-uncharged tRNA moves into the E site, from which it exits The A site is now available for entry of a new aminoacyl-tRNA

57. 5) Explain the role of complementary base pairing in mRNA translation (protein synthesis). EF1α (Elongation Factor 1 alpha), a GTP binding protein (G protein) positions the aminoacyl-tRNA in the ribosome’s ‘A’ site using antiparallel complementary base pairing between the mRNA codonand the tRNAanticodon. In the case to the right, the antcodon is GCG, so the mRNA codon would be CGC by complementary base pairing

58. 6) Which molecule is peptidyl transferase associated with? Peptidyl transferase catalyzes peptide bond formation, which is an activity of 28S rRNA (a ribozyme)

59. 7) Explain the directionality of protein synthesis and mRNA translation. Synthesis of a protein is from the protein’s amino (N) terminus to its carboxy (C) terminus The peptidyltransferase reaction attaches the COOH group of the nascent polypeptide (on Peptidyl-tRNA) in the P site to the NH2 group of the next amino acid (on the Aminoacyl-tRNA) in the A site. This mechanism dictates directionality N to C terminus (i.e. step wise addition of amino acids from the amino (N) terminus to the carboxyl (C) terminus of the protein). translation of the mRNA is from its 5’ to 3’ end

60. 8) Describe the process of protein synthesis termination. Termination of protein synthesis is triggered when one of three termination codons (UAG, UAA, UGA) are positioned in the ribosome’s A site. there are no corresponding aminoacyl-tRNAs for these codons. (we will see an exception to this later for selenocysteine) a releasing factor (RF) binds in the A site the completed polypeptide chain is hydrolyzed and released from the peptidyl-tRNA The tRNA, ribosome and mRNA dissociate, which requires GTP hydrolysis.

62. Bacterial mRNAs do not have a 5’CAP

63. Selection of proper AUG is by alignment of ribosome using Shine-Dalgarno sequence

64. Because bacteria use a Shine Dalgarno sequence, frequently see internal initiation

65. bacteria have polycistronic mRNAs with multiple AUG initiation codons

66. Example the Lac operon where transcript contains information for Lac Z, Lac Y and Lac A proteins

68. signal amplification

69. one mRNA producing multiple copies of a protein at any time.

70. PABP (poly-A-binding proteins, bound to the 3’ poly-A tail) interact with eIF4 to form circular polysome

73. Wobble is alternative base pairing in the third position of codon / anti-codon recognition

74. e.g. the codons UCA and UCG both code for Ser, but a single Ser-tRNAser with an anticodon 3'-AGU-5' could read both codons

75. G-U base pairs are allowable.

76. U in the 5' position of a tRNAanticodon can pair with either an A or G in the 3' position of a mRNA codon

77. G in the 5' anticodon position can pair with either a C or U in the 3' position of a mRNA codon

79. 1. Explain how the absence of heme regulates protein synthesis in a red cell/reticulocyte Global Protein Synthesis: Phosphorylation of eIF2 at Ser 51 prevents exchange of eIF2-bound GDP for GTP, therefore INHIBITING translation Heme-Regulated Inhibitor (HRI): specific eIF2 Kinase Heme binding protein and “heme sensor” has a short half-life and is degraded rapidly; good control In the PRESENCE of Heme: HRI is INACTIVE Hemoglobin synthesis is normal In the ABSENCE of Heme: HRI is ACTIVE Phosphorylates eIF2 at Ser 51 to prevent the exchange of GTP for GDP Protein Synthesis is INHIBITED

80. 2. Describe how this regulation affects someone like Pricilla Twig, who is anorexic. Twig’s anorexia has limited her iron intake; she is anemic Unable to synthesize Heme due to lack of Fe In absense of Heme: HRI is activated Activation » phosphorylation of eIF2 at Ser 51 Prevents exchange of GTP for GDP Inhibits Protein Synthesis

81. 3. Describe the molecular basis for regulation of Ferritin and Transferrin receptor response to iron. Ferritin: stores Fe inside the cell Transferrin Receptor: mediates Fe import into cell located on cell surface Iron concentrations control production of each protein High Fe » Increased Ferritin for storage Low Fe » Increased Transferrin receptor to bring in more Fe Control mediated by IREBP

82. IREBP IREBP: Iron Response Element Binding Protein Is a RNA secondary structural element, a Stem Loop Binds to an Iron Response Element (IRE) on mRNA in the ABSENCE of Iron In Ferritin mRNA: IRE is in 5’ UTR IREBP binding » DECREASED translation of Ferritin mRNA The Stem Loop is stabilized preventing ribosome scanning and initiation of transcription In Transferrin receptor: IRE is in 3’ UTR IREBP binding » INCREASED translation Transferrin receptor mRNA The IREBP blocks the decay of the mRNA » increasing translation

83. 4. Identify two differences between miRNA and siRNA. siRNAs Accelerate decay of specific mRNAs EXOGENOUS Derived from double stranded RNA precursor 100% complementarity between siRNA and target miRNAs Inhibit translation of specific mRNAs ENDOGENOUS Genome encoded Synthesized as stem-loop containing precursors <100% complementarity between miRNA and target

84. 5. What are the roles of Drosher, Dicer, and RISC in miRNA action? Drosher: endoribonuclease in the nucleus that processes Pre-miRNA Dicer: Endoribonuclease in the cytoplasm to generate mature miRNA RISC: RNA-Induced Silencing Complex Targets specific mRNAs by complementary base pairing between the miRNA usually in 3’ UTR of target mRNA

85. 6a. Describe the general structure of the proteosome. 26S Proteosome: Macromolecular structure with 4 cylindrical rings and a cap on each end Each ring has 7 protein subunits 19S regulator (Cap) regulates entry into 20S core Interior of cylinder contains proteasee enzymes that degrade proteins to peptides as they pass through

86. 6b. Explain how E3 ligases work. Proteins targeted to proteosome by attachment of multiple copies of Ubiquitin Proteosome cap contains receptor for Ubiquitin Ubiquitin molecules are added sequentially once the first is added Generates a Polyubiquitin tag Ubiquitination is catalyzed by ubiquitin ligases (E1, E2, E3) Target specificity is located in E3 ubiquitin ligase enzymes Final Substrate Specificity for tagging proteins

87. Lectures 37 Purine metabolism

88. R-5-P comes from pentose phosphate pathway PRPP (5'-phosphoribosyl 1'-pyrophosphate) is made by adding two phosphates from ATP to the C1' position energetically activates at C1' of the ribose where the purine ring will be assembled PRPP is the immediate sugar precursor for both purineand pyrimidinebiosynthesis The enzyme, PRPP synthetase, is highly regulatedby feedback inhibition by purine nucleotides (GDP and ADP) Levels of PRPP are an important determinant of rate of purine biosynthesis 1. Describe where PRPP comes from and how it is used in nucleotide synthesis and its importance in determining the overall rate of de novo purine biosynthesis. What regulates PRPP synthetase?

89. precursors Amino acids Glycine Glutamine Aspartate (aspartic acid) Tetrahydrofolate (FH4) Activated form of folate CO2 (respiratory) 2. Identify the specific precursor molecules that contribute atoms to the purine ring.

90. 3. Identify the rate limiting enzyme in de novo synthesis of purines. How is it regulated? What role does PRPP play in its activity and regulation? First step unique to purine synthesis The amide N from glutamine is attached to the C1' of ribose-5'-phosphate, displacing pyrophosphate to form phosphoribosylamine, glutamic acid and PPi (the N becomes N9 of the purine ring). The enzyme, Glutamine: PRPP amidotransferase, is rate limiting and the regulated enzyme for de novo purine synthesis.

91. Regulation of de novopurine synthesis The single most important regulator of de novo purine synthesis is the AVAILABILITY of PRPP!!!!! Pathway is regulated at 3 points: 1. PRPP synthetase is inhibited by GDP and ADP, end products of purine biosynthesis a. availability of R-5-P from the pentose phosphate pathway and ATP are also important for nucleotide synthesis b. PRPP used for both purine and pyrimidine biosynthesis 2. Glutamine: PRPP amidotransferase is feedback inhibited by GMP/IMP or AMP (other phosphorylatednucleotides), and activated by substrate PRPP a. first unique step to purinesynthesis 3. at the branch point of IMP: AMP feedback inhibits its own synthesis; GMP feedback inhibits its own synthesis.

92. Glutamine: PRPP amidotransferase has two allosteric sites allowing synergistic inhibition by GMP/IMP and AMP. With substrate glutamine, a graph of reaction rate vs. substrate concentration is hyperbolic. The Km for glutamine is close to normal physiological concentration of glutamine. With substrate PRPP, the dependence of reaction rate to substrate concentration is sigmoidal. When PRPP levels change, there is a more abrupt change in reaction rate. Activation by substrate represents feed forward regulation by PRPP. Normal physiological concentrations of PRPP are below its Km . Binding of AMP, GMP or IMP to allostericsites shifts the curve to the right, and increases the Km for PRPP. at high GMP/IMP or AMP and normal physiological concentrations of PRPP, very little enzymatic conversion to phosphoribosylamine occurs and the pathway is essentially shut down. However, at high PRPP concentrations, inhibition by nucleotides is overcome at the branch point of IMP: AMP feedback inhibits its own synthesis; GMP feedback inhibits its own synthesis. Glutamine:PRPPamidotransferase2 substrates, 2 independent Kms Gln + PRPP phosphoribosylamine + glu +PPi

94. HGPRT

95. APRT

96. Adenosine kinase

97. Net effect

98. Recover purines prior to their complete degradation to uric acid3 1

100. Why do we have an enzyme to salvage adenosine, but not the other base nucleosides?

101. Because there are higher levels of adenosine than other nucleosides

102. SAM /SAH

103. SAM=S-adenosylmethionine

104. These are molecules used in methylationrxns.. So there is a lot of it3

105. 5. Explain what happens if you don’t have the enzyme HGPRT. How does PRPP play a role in this? HGPRT hypoxanthine-guanine phosphoribosyltransferase Attach Hx or G to PRPP The importance of HGPRT is evidenced by the pathology associated with Lesch-Nyan syndrome (genetic defect in HGPRT), severe hyperuricemia. results in breakdown of excess Hx and G to uric acid failure to use PPRP stimulates synthesis of more purines, which are ultimately degraded to uric acid.

106. End product of purine degradation is uric acid Pathway eliminates purines from RNA/DNA degradation Excess synthesis Dietary sources Overall Remove phosphates Remove sugars Oxidize bases removes exocyclic amino groupsbr />The key enzyme in purine catabolism is xanthineoxidase. It is a non-hemeiron containing enzyme that uses FAD and molybdenum (VI) as co-factors and converts hypoxanthine to xanthine, and xanthineto uric acid. 6. Which enzyme of purine degradation is inhibited by the drug Allopurinol? Why is this drug given for treatment of chronic gout?

107. Chronic Gout can be treated with allopurinol(trade name zyloprin), an inhibitor of xanthineoxidase. Allopurinolhas a very high affinity for xanthineoxidase. It is also a weak substrate, but the product, alloxanthine, is also an inhibitor Allopurinolis effective in two ways. First by inhibiting xanthineoxidase, it favors excretion of the more soluble products hypoxanthine and xanthine. Second, allopurinol is converted by HGPRT to the corresponding nucleotide that feedback inhibits PRPP synthesis and purine synthesis. Allopurinolmay decrease frequency of gouty attacks, but does nothing to lessen the pain of an acute attack. Also addresses overproduction, not under-excretion 6. Which enzyme of purine degradation is inhibited by the drug Allopurinol? Why is this drug given for treatment of chronic gout?

108. 7. Describe the factors that can contribute to gout. The pKa of uric acid is 5.4. At physiological pH, nearly all uric acid is in the form of sodium urate. urateis insoluble at concentrations near the normal physiological range. Excess uric acid (hyperuricemia), occurs in Gout. Urate crystals precipitate, particularly in the joints, causing severe inflammation and pain. Etiology of gout is not always clear. Familial Xlinked gout may be caused by 1. mutations resulting in mis-regulation of PRPP synthetaseresulting in failure to decrease activity in presence of excess purines 2. partial inactivation of HGPRT 3. factors governing uric acid re-uptake and excretion (80 % of uric acid is excreted in the urine, 20 % in the feces). Secondary gout can be caused by tissue destruction (which increases purine degradation and uric acid production) or diuretics for used for treatment of hypertension excess consumption of alcohol exacerbates gout by acting as a diuretic, and causing lactic acidosis, which interferes with the ability to excrete uric acid. diet rich in organ meats

109. Lecture 38 Pyrimidine metabolism

110. 1. Identify the enzyme that catalyses the first three steps in pyrimidine biosynthesis in mammals. de novopyrimidine biosynthesis-Make the pyrimidine ring first, then attach to PRPP C A D catalyzes first three steps in pyrimidine ring synthesis CPS II ATCase Dihydrooratase See next three slide taken from lecture

111. Steps 111 C A D Reaction 1: Carbamoyl phosphate synthetase II (CPS II) First, and regulated step in eukaryotes Condensation of CO2, an NH2 from glutamine, and a phosphate from ATP to form carbamoyl phosphate Feedback inhibited by UTP Similar reaction catalyzed by CPS I in mitochondria – entry of NH3+ into urea cycle: CPS II in the cytoplasm In mammals, CPS II part of a single multifunctional protein called CAD catalyzes the first three reactions in pyrimidine biosynthesis. Named for the first three activities, C = CPSII

113. The A in CAD

114. A separate enzyme in bacteria

115. Historical importance as first described allosteric enzyme

117. Atoms of base from Glutamine (N3) Aspartate (N1through C4) CO2 (C2) Sugar from PRPP 2. Identify the specific precursor molecules that contribute atoms to the pyrimidine ring.

118. 3. Identify the two enzymes that are critical to make thymidine from dUMP. Describe how the drugs methotrexate and 5-FlouroUracil used in cancer chemotherapy, inhibit thymidinesynthesis. Why is thymidine synthesis a good chemotherapeutic target? How does the compound folinic acid factor into the inhibition of thymidine synthesis by FOLFOX 4? Thymidine synthesis from dUMP by adding a methyl group to the base at the 5 position. 1. The methyl group comes from N5,N10 methyleneFH4 (tetrahydrofolate) – Another important use of folic acid 2. The reaction is catalyzed by thymidylatesynthase. key enzyme because thymidine is uniquely needed for DNA synthesis making it an excellent target for drugs used to inhibit DNA synthesis and cell growth. (5-fluorouracil (5FU) clinical correlation 1) 3. N5,N10 methylene FH4 is converted to dihydrofolate (FH2) in the reaction. FH4 must be regenerated for continued thymidine synthesis. catalyzed by dihydrofolatereductase (DHFR) also a key enzyme for thymidine synthesis and a target for chemotherapy. Methotrexateis a folate analog and a competitive inhibitor of DHFR commonly used to metabolically inhibit DNA synthesis.

119. 4. Describe the synthesis of deoxynucleotides from ribonucleotides. What agents are used in the oxidation/reduction reaction? What inhibits the reaction? Deoxynucleotides: Made from ribonucleosidediphosphates (NDPs) Catalyzed by Ribonucleotidereductase(nucleoside diphosphatereductase) Removal of O at 2’ position Reduction reaction; something must be oxidized Immediate reducing agent for the reaction Pair of sulfhydril group on small redox proteins (thioredoxinor glutaredoxin) Oxidized to disulfides Regenerate -SH via thioredoxinreductase Use NADPH – ultimate reducing agent Regulation Ribonucleotidereductase inhibited by dATP Hydroxyurea (drug), also inhibits RR There is little need for dNTPs when cells are not actively synthesizing DNA. The key enzymes for DNA synthesis, (RR, TS, DHFR, TK (ribonucleotidereductase, thymidylatesynthetase, dihydrofolatereductase and thymidinekinase)are cell cycle regulated such that their expression is increased as cells pass from G1 to S phase, when DNA synthesis occurs.

120. Important enzymes in synthesis of pyrimidine nucleotides RR ribonucleotidereductase NDP  dNDP TS thymidylatesynthetase dUMP dTMP (FH4FH2) DHFR dihydrofolatereductase FH2FH4 TK thymidinekinase TMPTTP TS DHFR TK

121. As we saw in this case study, degradation of 5FU is an important consideration in its use. The enzyme dihydropyrimidinedehydrogenase (DPYD) is an enzyme in the pyrimidine degradation pathway. The gene is polymorphic in the population with a significant number of people containing mutations that lead to loss of function. However, since failure to degrade pyrimidines does not ordinarily lead to pathology, the polymorphism goes unnoticed until a person is challenged with high doses of the pyrimidine 5FU. 5. Describe the role of DPYD (dihydropyrimidinedehydrogenase) in pyrimidine degradation. Why are the levels of this activity only problematic when you treat a patient with 5FU? pyrimidine degradation DPYD

122. Lecture 40Folate and Vitamin B12summary (from notes) Folate(the salt of folic acid) water soluble vitamin found in green leafy vegetables. participates in a select set of 1 carbon transfers as the active co-factor tetrahydrofolate (FH4) One of its major uses is in the synthesis of nucleotides and deoxynucleotides to make RNA and DNA. Folatedeficiency in pregnant women is the major cause of neural tube defects. This may be related to decreased ability to make DNA and RNA, but could also be caused by a decreased ability to methylate DNA. Although folate does not directly donate methyl groups to DNA, it does donate a methyl group to Vitamin B12 which is used to generate S-adenosylmethionine, the direct DNA methylating agent. Vitamin B12, (cobalamin), is another water soluble vitamin. One of its two uses is in re-methylation of homocysteine to methionine, a reaction requiring methylated vitamin B12. The ultimate source of the methyl group is N5-methyltetrahydrofolate. Formation of N5-methyltetrahydrofolate is irreversible and the only way to regenerate FH4 is for methylated-FH4 to donate its methyl group to vitamin B12. Pernicious anemia is the major symptom of vitamin B12 deficiency and results from a secondary deficiency in folate related to its being trapped in the otherwise unusable form, N5-methyltetrahydrofolate. This is referred to as the methyl trap. Since we store large amounts of B12 in the liver and B12 is abundant in the food supply, dietary deficiencies per se are rare. However, there are defects in uptake of B12 secondary to stomach or intestinal surgery that result in B12 deficiency. More commonly there is an age-related decrease in B12 uptake that results in B12 deficiency in the elderly.

123. 1. Explain how the enzyme Dihydrofolatereductase (DHFR) is used in converting the vitamin folic enzyme to the active co-factor Tetrahydrofolate (FH4). What other pathway is DHFR important for? Folate= folic acid (a vitamin!) DHFR required to convert folic acid to its active co-factor form FH4 2 rxns, both utilized NADHP + H+ as a source of reducing equivalents DHFR is also important in pyrimidine synthesis Thymidine synthesis from dUMP by adding a methyl group to the base at the 5 position. N5,N10 methylene FH4 is converted to dihydrofolate (FH2) in the reaction. See Lecture 38 #3 DHFR reduces these 2 double bonds

124. 2. Explain the importance in the reaction that converts Methylene FH4 to Methyl FH4, and identify the enzyme that carries out the reaction. FH4 carries carbons in several oxidation states. In humans, the oxidation states are reversible up to methylene, but not from methyleneto methyl. This is an important part of the methyl trap hypothesis. The conversion of methylene FH4 to methyl FH4 is catalyzed by methylenetetrahydrofolatereductase (MTHFR)

125. 3. What is the major source of 1 carbon units for tetrahydrofolateto use? Identify the tree major pathways (or reactions) that use FH4. Thymidinesythesis (thymidylatesynthase) TS DHFR Major source of 1C units for thransfer thru FH4 is serine 3 major recipients that us FH4: 1. purines(rxn5) 2. pyrimidine (just thymidine) deoxyuridineto form deoxythymidine (using N5,N10 methylene FH4) (rxn6) 3. vitamin B12 (cobalamin) to form CH3 - Vit B12 (methylcobalamin) (using N5-methyl FH4). (rxn 8) This is the ONLY reaction which uses methyl FH4 A deficiency in B12 traps folate as N5-methyl FH4 this is sometimes referred to as the methyl trap

126. 4. Describe the methyl trap and explain how it ties into deficiency of vitamin B12. MTHFR CH3 S-adenosylmethionine (SAM) is the major methyl donor used in methylating DNA and RNA When SAM donates a methyl it is converted to SAH which loses its adensoine (see Lec 37 #4- why purine salvage pathway has a separate enzyme [adenylatekinase] to salvage this nucleoside and we have no mechanism to salvage any other nucleosides). Methylation of Vitamin B12only reaction that uses N5methyl FH4 Formation of N5methyl FH4 from N5N10methylene FH4 is irreversible B12 deficiency – (acceptor for methyl group), traps folate in an unusable form, leading to secondary deficiency of folate (diminished nucleotide synthesis) MTHFR (methylenetetrahydrofolatereductase) is regulated (inhibited) by SAM B12 deficiency- lots of homocysteine, cant make nucleotides, cant regenete SAM, MTHFR is not inhibited--- methylFH4 accumulates and can’t be used for anyhting (secondary deficiency to folate)

127. 5. Describe how a vitamin B12 deficiency can occur, and explain why dietary deficiencies of B12 are rare. B12 deficiency not dietary, due to failure to absorb VitB12 is only needed in small amounts It is very abundant in food supply Also, mammals store large amounts (mg quantities ) in the liver True dietary deficiencies rare Severe, chronic malnutrition (alcoholics) Strict vegans?? B12 deficiency not dietary, due to failure to absorb Process of absorption of B12 in ileum leads to its storage in the liver: B12binds to R-binders Protein made by salivary gland and/or gastric cells (transcobalamin I) B12 transferred from R-binder to Intrinsic Factor glycoprotein made by gastric parietal cells R-binder degraded by pancreatic proteases, releases B12 B12 binds to IF (protease resistant) B12 absorbed in the ileum, in complex with IF, [by IF receptor] B12 transferred to transcobalamin II for transport in the blood Stored in large amounts in liver

128. 6. Describe Pernicious Anemia, how it might occur, what specific populations seem to be at most risk, and how it is best treated. Pernicious anemia from defects in B12 absorption Surgical resection of stomach or ileum Malabsorption disorder Crohn’s disease; Celiac Sprue High intestinal pH Decreased degradation of R-binder and B12 release Low acid secretion in stomach (long term use of proton pump inhibitors for acid reflux) Age related decrease in synthesis of Intrinsic Factor pernicious anemia common in the elderly: 10% of those over 75 clinical deficiency Pernicious anemia Symptoms - megaloblastic anemia Maturation of RBC requires adequate folate for RNA and DNA synthesis B12 deficiency results in secondary folate deficiency due to methyl/folatetrap Pernicious anemia is treated with injections of B12 Folate supplementation treats anemia but can mask neurological deficits. B12 deficiency can also cause neurological symptoms

129. Lecture 43: Introduction to signal transduction

130. 1. Differentiate between chemical messengers that act via cell membrane receptors versusintracellular receptors in terms of ligand transport, general receptor structure, mechanism ofaction, and intracellular effects.

131. Summary steps in steroid hormone/nuclear receptor activation: 1. ligand binds receptor. 2.Receptor activation- dissociation of chaperone proteins, Heat Shock Proteins, e.g., hsp90, hsp70, and dimerization 3. receptor binds to specific DNA sequences called hormone response elements (HREs)- usually located in 5’ enhancer region of a target gene 5. hormone-receptor-DNA complex is recognized by coactivatorsthat have histoneacetyltransferase (HAT) activity that help “open” chromatin and interact with components of the RNA polymerase II transcription initiation complex (see Dr. Geoghegan’s lectures 27-9) 6. increase in transcription of target gene – mRNA synthesis(5’-3’) 7. translation of mRNA into protein 8. protein activates cellular response(s) – inside cell or communication with adjacent cells Location of intracellular receptors: Cytoplasmic and/or nuclear – then act in the NUCLEUS

132. 1. N-terminal variable region containing transcriptional activation function 1 (transactivationdomain (TAF-1)- hormone-independent; 2. central DNA binding domain (DBD)– receptor has 2 zinc fingers that physically interacts directly with DNA in the major groove and along the phosphate backbone; 3. C-terminal ligand binding domain (LBD) – where hormone binds non-covalently in the “ligandbinding cavity” or “pocket” – LBD has AF-2 that is hormonedependent 2. Name the 3 major domains of steroid/nuclear receptors and describe the function. Steroid receptors are hormone-dependent transcriptional activators

133. 3. Describe the pathway activated by a chemical messenger binding to receptors that are tyrosine kinases or that bind tyrosine kinases. Receptors that are kinases or that bind kinases: Growth factor and cytokine receptors share the common feature that the intracellular domain of the receptor or an associated protein is a protein kinase that when the growth factor/cytokine (ligand) binds the extracellular domain, becomes active. The activated receptor kinasephosphorylates an amino acid residue on the receptor itself (autophosphorylation) or an associated protein. The message is propagated through signal transducer proteins that bind the activated receptor complex, e.g., STAT or Smad. Tyrosine Kinase Receptors are monomers in the cell membrane with a single transmembrane spanning region). When the ligand, i.e., a growth factor, binds the receptor, the receptor dimerizes which allows activation of the intracellular tyrosine kinase domain in each receptor monomer which then phosphorylate each other on certain tyrosine residues (autophosphorylation). The tyrosine-phosphorylated intracellular domain forms specific binding sites for signal transducer proteins.

134. 3. Describe the pathway activated by a chemical messenger binding to receptors that are tyrosine kinases or that bind tyrosine kinases. Ras and the MAP kinase pathway. The activated tyrosine kinase domain forms a binding site for proteins with a SH2 domain. The adaptor protein Grb2, which is bound to a membrane phophoinositide, has an SH2 domain. When the SH2 domain of Grb2 binds the activated (tyrosine-phosphorylated) domain of an activated growth factor receptor, a conformational change occurs in another domain of the Grb2 protein called the SH3 domain which is a binding site for the protein SOS – do not worry about this name. SOS is a guanine nucleotide exchange factor (GEF) for Ras, a monomeric G protein located in the plasma membrane. SOS activates exchange of GTP for GDP on Rascausing a conformational change in Ras that promotes Raf binding to Ras. Rafis a serine protein kinasethat is a MAPKKK (mitogen activated protein kinasekinasekinase). Rafinitiates a sequence of successive phosphorylationsof intracellular protein kinases called a phosphorylation cascade. The MAP kinase cascade terminates at the phosphorylation of a number of transcription factors, which one or ones depends on the cell type. The phosphorylation of the transcription factor regulates its activity, for example stimulating its activity and increasing transcription of genes involved in cell growth and survival. You do not need to memorize every protein in this pathway, but know the concept. KNOW that activation of Tyr-kinase Receptors leads to activation of Ras (a guanine nucleotide exchg. factor) And consequently of the MAPK pathway

135. Fibroblast growth factor receptor (FGF-R) is a tyrosine kinase receptor. FGF-R mutations are involved in a variety of craniosyntoses and chondrodysplasias Diseases associated with mutations in FGF-R receptor Apert’s syndrome Craniofacial abnormalities Mental retardation Achondroplasia Most common form of dwarfism

136. Insulin Receptor is a member of the tyrosine kinasefamily of receptors, but it exists as a pre-formed dimer in cell membranes. Each monomer consists of 2 domains: α and β with the α forming the binding surface for insulin and the β subunits having a single transmembraneregion plus tyrosine kinase activity. When insulin binds to the α subunits, the β subunits become active and autophosphorylate each other. The activated tyrosine kinasephosphorylatesproteins with IRS (insulin receptor substate) at multiple sites thus creating binding sites for other proteins with SH2 domains, e.g., Grb2, PLCγ, and PI3 kinase. Thus, insulin binding to the insulin receptor can activate MAPKsignalingvia Grb2 activation, DAG and IP3 through activation of PLCγ, and downstream targets of PI3K including protein kinase B (also called AKT) which in turn, phosphorylatesother proteins Activation of the insulin receptor leads to phosphorylation of target proteins that either activate or inhibit the activity of these proteins, resulting in physiological responses. insulin receptor: present in virtually all tissues, but the major sites of insulin action are liver, muscle, and adipose tissue insulin has distinct effects in other tissues including pancreas, kidneys, brain, lungs, immune system, platelets, nervous system, and bone. You do not need to memorize every intermediary protein, but know the concept, i.e., that insulin activates an intracellular tyrosine kinase pathway. ras MAPK

137. 4. List the components of G-protein coupled receptors linked to activation of protein kinase A and phospholipase C and how cAMP, DAG, and IP3 activate intracellular signaling. Clinical Relevance: 27% of all FDA-approved drugs target GPCRs G-protein coupled receptors (GPCRs) 7 transmembrane regions C-terminus in cytoplasm E = effector: e.g.,adenylatecyclase, phospholipase C, Trimeric G protein: a, b-g subunits GDP inactive GTP active Heptahelical Receptors (7 transmembrane spanning α- helices): These are also called G-protein-coupled receptors and they are the most common type of membrane receptor. Binding of the hormone activates the receptor via a G-protein-coupled mechanism to generate second messengers, which are small non-protein compounds such as cAMP inside the cell’s cytoplasm. Second messengers are present in low concentrations so that the message can be rapidly initiated and terminated. Different heptahelical receptors bind different G proteins that exert different effects on their target proteins. G-proteins are heterotrimers: α,β,γ subunits When ligandbinds to a heptahelicalreceptor, G- proteins, which are anchored in the cell membrane, are activated causing the release of GDP and binding of GTP to the Gα-protein that activates adenylatecyclase which produces Camp, the second messenger molecule.

139. know that phosphorylation can either activate or inhibit the activity of an enzyme or other proteinrecall CFTR

140. 4. List the components of G-protein coupled receptors linked to activation of protein kinase A and phospholipase C and how cAMP, DAG, and IP3 activate intracellular signaling. Phosphatidylinositol pathway: Phosphatidylinositol 4’,5’ bisphosphate (PI-4,5-bisP, PIP2) is cleaved by phospholipase Cto generate 2 second messengers in the cytoplasm: diacylglycerol (DAG) and inositoltrisphosphate (IP3) IP3 activates release of intracellular Ca++ that activates Ca++- calmodulinkinase that phosphorylatesintracellular proteins DAG activates protein kinase C that phosphorylatesintracellular proteins. This diagram shows a G-protein coupled with phospholipase C know this pathway

141. 5. List mechanisms by which chemical messenger signaling is terminated. How are signals terminated? stimulus abates: messenger no longer released messenger may be rapidly degraded, e.g., peptide hormones Receptor may be desensitized by hydrolysis of phosphate residues by phosphatases Degradation of 2nd messenger, e.g., cAMP, IP3 – GTPase – GDP by phosphodiesterases Internalization of receptor- degradation- recycling, e.g. insulin receptor Signal Termination Some signals needed to regulate metabolic responses or transmit nerve impulses must be rapidly turned off when the hormone is no longer being produced or released. On the other hand, signals that stimulate cell proliferation may turn off more slowly. Many chronic diseases are caused by failure to terminate a response at the appropriate time.

142. Lecture 44 Signal transduction for the regulation of fuel metabolism

143. 1. List the cellular location of the receptors for the following hormones: cortisol, insulin, glucagon, and epinephrine Insulin receptors (pg13-15) Liver Adipocytes Skeletal muscles Glucagon receptors (p16) Adiopocytes liver Epinephrine receptors (p17) Beta receptors: liver, skeletal m, other tissues Alpha receptors: liver Cortisol receptors (p17) Liver, adipose, muscle, pituitary, hypothalmus Diagram not in notes but on powerpoint These diagrams Point Out the Receptor locations

144. 2. Describe the signaling mechanisms and “second messengers” by which glucagon, epinephrine, insulin, and cortisol regulate glucose homeostasis Insulin 2nd messenger: Insulin receptor substrates(IRS), PI2P, MAPK tyrosine kinase receptor: alpha and beta subunits (pg 14 ) Glucagon 2nd messenger: cAMP via adenylylcyclase activation G-protein linked receptor Epinephrine 2nd messenger (both via G protein linked receptors): Beta receptor: cAMP(via adenylylcyclase activation) Alpha receptor: DAG/IP3 Cortisol Intracellular steroid hormone receptor; binding of cortisol turns glucocorticoid receptor into transcription factor (see pg 5)

145. 3. List the major carbohydrate metabolic pathways regulated by insulin, glucagon, epinephrine, and cortisol and describe how each hormone regulates each pathway Insulin Anabolic affector of glucose metabolism Promotes uptake of glucose into organs, synthesis of glycogen, fatty acids, amino acids Glucagon Catabolic affector of glucose metabolism Promotes synthesis of glucose, breakdown of glycogen, fatty acids Epinephrine Catabolic affector of glucose metabolism Promotes synthesis of glucose, breakdown of fatty acids, glycogen; lesser extent than glucagon Cortisol Promotes synthesis of glucose; release of FA from adipose Long term regulator, mostly catabolic actions

146. Long-winded table Insulin Glucagon Epinephrine cortisol

147. Lecture 45 Glycolysis

148. 1. List the types of polysaccharides ingested in a normal diet. Water-insoluble fiber (add bulk to stool) Cellulose Water-soluble fiber (form gels in GI tract, slows absorption) Pectin Starch (glucose polymers MW 100 kDa+; potatoes, rice, bread) Amylose (linear, alpha 1-4 linkages) Amylopectin (branched, alpha 1-4 and alpha 1-6) Glycogen (glucose polymer, highly branched, alpha 1-4 and 1-6; MW 4 mDa+; meat, liver)

149. 2. List the tissue compartment in which starch and glycogen digestion begins and the names and activities of enzymes involved in carbohydrate digestion in the small intestine Salivary alpha-amylase hydrolyzes alpha 1-4 linkages in starch and glycogen in the mouth, and activity is halted in the acidic stomach Pancreatic alpha-amylase continues that of the salivary counterpart in the duodenum In the brush border of the intestine,di- and oligo- saccharidases complete the digestion to simple sugars.

150. 3. Describe 2 mechanisms for glucose transport across a cell membrane and the location of each type Na+-dependent glucose transporters Secondary active transport; uses E from Na+ gradient to pull glucose into cells Facilitative glucose transports GLUT2: pancreas, liver, intestine, etc; high capacity, low affinity GLUT4: adipose, skeletal muscle, etc; high affinity

151. Glycolysis: know substrates, enzymes, cofactors, regulated stepssee pg 6 Glucose --hexokinase/glucokinase (uses Mg2+) ->glucose 6 phosphate + ADP <- (phosphoglucoseisomerase) -> fructose 6 phosphate -- (phosphofructose kinase-1) ->fructose 1,6 bisphosphate+ ADP <- (aldolase) -> dihydroxyactone phosphate (DHAP) + glyceraldehyde 3 phosphate (note: glucose split from here on) DHAP <- triose phosphate isomerase -> Glyceraldehyde 3 phosphate <- phosphoglyceratedehydrogenase -> 1,3 bisphosphoglycerate+ NADH <- phosphoglyceratekinase -> 3 phosphoglycerate+ ATP <- phosphoglyceromutase -> 2 phosphoglycerate <- enolase -> phosphoenolpyruvate --pyruvatekinase->pyruvate+ ATP FYI: kinasesphosphorylate/dephosphorylateisomerases, mutases rearrange aldolases split dehydrogenases reduce/oxidize (see NADH) enolases remove a water

152. 4. Describe which enzymatic steps in the conversion of glucose to pyruvate are regulated and how each regulator impacts enzyme activity, whether positively or negatively Note: all important regulated enzymes listed below are irreversible steps in Glycolysis Hexokinase (located in skeletal mm) Glucose-6-phosphate (-) Glucokinase (isozyme of hexokinase; liver, pancreas) Fructose-6-phosphate (F6P) indirect (-) through activity of glucokinase regulatory protein (GKRP) With the presence of F6P, glucokinasetranslocated into nucleus where GKRP binds tightly, thus inactivating (glycolysis occurs in the cytosol) Glucose reverses activity. Insulin (+) via transcription upregulation Phosphofructokinase (GLYCOLYSIS RATE LIMITING STEP) AMP (+) ATP (-) in liver citrate (-) Fructose 2,6 bisphosphate (gluconeogenesis metabolite) (+) Pyruvatekinase Phosphorylation (-) Fructose 1,6 Bisphosphate (+)

153. 5. Calculate how many ATP are used and produced from 1 molecule of glucose in glycolysis 2 ATP are needed in the “investment phase” of glycolysis One to “trap” glucose in the cell as Glucose-6-phosphate (hexokinase/glucokinase) One to set up glucose to be split into two 3-carbon phosphorylated sugars (phosphofructokinase) 4 ATP are produced through substrate level phosphorylation steps. ATP is made at 2 different enzymatic reactions that make 1 ATP each per split molecule of glucose (1x2 + 1x2) Phosphoglyceratekinase Pyruvatekinase 4 ATP – 2 ATP = 2 total ATP produced per glucose

154. 6. Describe the differences between aerobic and anaerobic glycolysis and in what tissues lactate dehydrogenase is important and why. Aerobic glycolysis (NADH produced to perform oxidative phosphorylation in mitochondria) Glucose, NAD, 2 ADP, 2 Pi  2 pyruvate, NADH, 2ATP, 2 H2O Anaerobic glycolysis (important in erythrocytes, lymphocytes, kidney, eye, skeletal mm) Glucose, 2 ADP, 2 Pi  2 lactate, 2 ATP, 2H2O If mitochondria are overwhelmed or not present(as in RBCs), there is overproduction of NADH Lactate Dehydrogenaseoxidizes NADH to NAD+ by reducing pyruvate to lactate. This makes NAD+ available to continue glycolysis

155. 7. Describe the effect of pyruvatekinase deficiency on anaerobic glycolysis Pyruvatekinase Last step of glycolysis, produces 1 ATP If no mitochondria utilizing NADH to make extra ATP via oxidative phosphorylation (as in RBCs), pyruvatekinase deficiency results in only 1 ATP being produced per molecule of glucose (instead of 2) half as much energy made in RBCs  inefficiency leads to RBC lysis and hemolytic anemia Largely related to inability to maintain activity of Na-K pumps

156. 8. Name the cell type that produces 2,3-bisphosphoglycerate, the substrate from which 2,3 BPG is produced, the enzyme responsible for this conversion, and the role of 2,3 BPG in cell function Erythrocytes 1,3-bisphosphoglycerate –(1,3 BPG mutase)-> 2,3-bisphosphoglycerate 2,3 BPG binds hemoglobin to decrease Hb affinity for oxygen  release of oxygen to the tissues