Lecture 24: Learning objectives: Human genome

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)

2. For a typical human gene, how much of the DNA composes exons and how much introns?

• exons smaller than introns

• gene 10-12 times larger than needed to encode information

MEAN

exon numberexon size

7293 bp

intron numberintron size

65675 bp

total coding sequence 2,855 bp

protein size (amino acids)

447 aa

gene size 36,000 bp

3. Describe how one gene can code for more than one protein.

• Alternative splicing - multiple proteins from a single gene – (old notion of one gene one protein doesn’t hold)

• The rule rather than exception, estimate 75% of genes give rise to multiple proteins by

alternative splicing

• Tissue specific

• Proteome is more complex than the genome

• Other species do not carry out alternative splicing to the same degree

RNA transcript

4. ~45% of our genome is repetitive elements. Provide three examples of these.

• 45% of genome are repetitive sequences• These include clustered

structural elements (centromeres and telomeres), and repetitive sequences interspersed between and among genes.

• Transcripts – contain genes– portions of the DNA that code for proteins– Both strands (+)(-) code for proteins

• Protein coding genes are ‘unique sequences’

Transposable elements- interspersed repetitive elements

• SINES (Alu)- represent 13% of our genome

• LINES (L1) – 20%• LTRs – 8%

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

• These are highly variable among individuals and are used in DNA fingerprinting

5. Explain what SNPs are and how they contribute to our individuality. Describe a cSNP and explain why they are less common than other SNPs. What is the relationship a SNP and a mutation? & 6. Explain what an allele is.

• Every individual is genetically unique (even though we all are humans with ~23000 genes)

• What makes us unique?– In the populations there can be many different alleles

for a given gene, contributing to genetic diversity of the human species

• An allele is a different form (sequence) of a particular gene

• As diploids, we have two alleles of each gene (with the exception of genes on the X and Y chromosome)

• SNPs contribute to making alleles different

– Polymorphisms are sequence differences between individuals, which account in part for these different alleles.

– SNPs are the most common type of polymorphism

– There are ~3-5 million sequence variations between two individuals

» Note: 3.2E9 bp in total human genome

• 75% of these are SNPs

• There are ~58,000 cSNPs in human genome (the rest are in the non-coding regions)

• Most SNPs are outside of the protein coding region– Example: DYPD genes; 15 cSNPs

(SNPs in coding region), 2506 in genes overall

• cSNPs– Nonsynonymous- result in amino

acid change in protein• More likely to be associated wwith

disease phenotypes

– synonymous- do not result in change in amino acid sequence of protein

• Can be associated with disease phenotypes if they affect a process of gene expression (transcription, splicing, etc)- regulatory SNPs (rSNPS)

– These affect the quantity of the gene that is expressed

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.

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)

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

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

1. Describe the three component parts of the basic monomeric unit of a nucleic acid. In terms of these monomeric units, what are the chemical differences between RNA and DNA?

Nucleotide: component parts

• Base attached to 1’ position of sugar

•Nucleotide: 5C sugar, base, phosphate

• 5C sugar • atoms designated with (’)• 2’deoxyribose-DNA• ribose in RNA

(Monomeric unit of nucleic acids: DNA and RNA)

Bases:• Purines Adenine & Guanine (A & G)

• 9 atom (atoms designated with regular numbers)• Pyrimidines

• Cytosine (C), Thymine(T) in DNA, Uracil (U) replaces T in RNA • 6 atoms

2. Explain the key difference between a nucleotide and a nucleoside. How might that be important in using nucleotide analogs as chemotherapeutic drugs?

• A purine or pyrimidine base linked to a sugar is a nucleoside.

– Bases are linked to the sugar (ribose or deoxyribose) with a glycosidic bond between the 1' position of the sugar and the N1 (pyrimidine) or N9 (purine) of the base.

• Addition of a phosphate to the sugar (usually at the 5' C) generates a nucleoside phosphate or nucleotide.

– Nucleotides can contain one, two or three phosphates (designated α, β, γ). A single phosphate is a nucleoside monophosphate (e.g. AMP), two phosphates make a nucleoside diphosphate (e.g. ADP) and three phosphates make a nucleoside triphosphate (e.g. ATP).

• The difference between a nucleoside and nucleotide is relevant to the clinical use of nucleoside analogs in chemotherapy, usually to inhibit DNA synthesis.

• Nucleosides are uncharged and can get into and out of cells easily.

• Nucleoside phosphates (nucleotides) cannot because the phosphate group is negatively charged at physiological pH, making it difficult for these molecules to cross the plasma membrane. Thus analog drugs that inhibit DNA synthesis as nucleotides are administered as the corresponding nucleoside, which can get across the cell membrane and be phosphorylated in the cell.

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 thymidylate synthase (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 dihydropyrimidine dehydrogenase (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 hematopoetic toxicities 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.

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’?

Phosphodiester bondLinks nucleotide units together in a nucleic acid

Negative charge

Backbone Sugar phosphates Linked by 3’-5’ phosphodiester bonds

N- Glycosidic bond

5’ phosphate (ester bond)

Phosphoanhydride bonds“high energy”

ATP

The covalent bonds within the monomeric unit in DNA

The phosphate linkage provides a net negative charge to the molecule at physiological pH

• 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:pyrimidine pair 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 hbonds and 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.

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’?

5. Explain what is meant by the two strands of DNA being ‘antiparallel’. What is the antiparallel feature 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)

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

7. Explain the notion of Tm in terms of DNA, and how it is related to the A+T or G+C content of the DNA.

• DNA can be “melted”• Base pairs separated by breaking

h-bonds with heat or chemicals

• Stability of double helix related to number of h-bonds

• Tm is temperature at which 50% of DNA single stranded

• Tm related to AT vs GC content (because of difference in # h-bonds in those base pairs)

8. Describe the fundamental basis for the hypochromic effect in double stranded DNA

• Bases are aromatic and absorb UV light

• Peak absorbance at 260 nm

• Can be used to estimate concentration of nucleic acid in a solution

• Hypochromic effect – amount of light absorbed by dsDNA ~50% less than by ssDNA

– Related to the base-stacking interactions in double-stranded DNA, which decrease absorption of light

• The change in absorbance when DNA goes from double-stranded to single stranded can be used in DNA melting experiment (figure) to monitor the % DNA that is denatured.

Lecture 30&31 Learning Objectives

Candice Nalleycandice.nalley@louisville.edu

1) a.Describe the two-step tRNA charging reaction.

• Two step reaction – catalyzed by aminoacyl-tRNA synthetase enzyme

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

• (yielding an aminoacyl adenylate and PPi)

• Hydrolysis of 2 ~P drives reaction

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

• Ester linkage – preserves energetic activation

Net reaction: amino acid + ATP + tRNA → Aminoacyl-tRNA + AMP + PPi

1) b. How is fidelity maintained and why is that important? Met

– tRNA synthetase enzymes recognize structural/sequence features unique to each tRNA (including but not limited to its anti-codon)

• proofreading activity – Aminoacyl-tRNA synthetase removes amino

acid from improperly charged tRNA– Replaces it with correct one

Fidelity 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.

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-tRNAMet

binding protein complex– eIF2B - a guanine nucleotide exchange (GEF) factor complex.

2) Initiation (continued)• Prior to initiation, 80S ribosomes dissociate into their constituent large

(60S) and small (40S) subunits.

• Initiation Steps: – Binding of tRNA: Met-tRNAii binds to eIF2 and GTP in a ternary complex, which

binds the 40S ribosomal subunit (small subunit) to form a preinitiation complex.– Binding to mRNA: The preinitiation complex locates the CAP on the 5' end of the

mRNA, mediated by the CAP binding protein complex eIF4.• AUG is the start codon (codes for methionine)• Complex slies along the mRNA (scanning) through 5’UTR to locate first AUG • Complex stops at first AUG {In bacteria, Shine-Dalgarno sequence helps align initiation

complex}- Binding of 60S subunit: large subunit binds, completing formation of the 80S complex

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

• eIF2 is a G protein (GTP binding protein) =active w/GTP, inactive w/GDP• after hydrolysis, eIF2 remains GDP bound and inactive until the GDP is exchanged for GTP

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

• Phosphorylation of eIF2 (at ser51) blocks GDP-GTP exchange, stops protein synthesis and allows for global regulation of protein synthesis

Initiation Guanine Nucleotide Exchange

inactive

P-

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 protein kinases, 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

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

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 codon and the tRNA anticodon.

• In the case to the right, the antcodon is GCG, so the mRNA codon would be CGC by complementary base pairing

6) Which molecule is peptidyl transferase associated with?

• Peptidyl transferase catalyzes peptide bond formation, which is an activity of 28S rRNA (a ribozyme)

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 peptidyl transferase 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

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.

9) Provide two examples of how protein synthesis differs between bacteria (prokaryotes) and eukaryotes. Explain how and why polycistronic mRNAs are common in bacteria but not in eukaryotes.

• Bacteria use a formylated methionine as the initiating amino acid (referred to as fMet)

• Bacterial mRNAs do not have a 5’CAP• Selection of proper AUG is by alignment of ribosome using

Shine-Dalgarno sequence• Because bacteria use a Shine Dalgarno sequence,

frequently see internal initiation • bacteria have polycistronic mRNAs with multiple AUG initiation codons• Example the Lac operon where transcript contains

information for Lac Z, Lac Y and Lac A proteins• polycistronic mRNAs have additional initiation codons. This is not a problem in bacteria because they use the internal Shine-Dalgarno sequence to position ribosomes

10) Describe the notion of polysomes.

• multiple ribosomes translate a single mRNA at any one time

• signal amplification – one mRNA producing multiple copies of a

protein at any time.

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

– Increases efficiency

11) Explain the notion of degeneracy of the genetic code and how it can lead to silent (synonymous) mutations (polymorphisms).

• Since there are 20 amino acids but 61 amino acid encoding codons, most amino acids are represented by more than one codon. This is referred to as degeneracy of the genetic code– e.g. UUU and UUC both code for Phe

• Degeneracy allows for “silent mutations” now commonly called synonymous polymorphisms (a nucleotide change in the coding region of a gene that does not alter the amino acid sequence of the protein).– For example, an A to U change in the position representing the third

base of the codon UCA (to UCU), would NOT alter placement of a serine in the protein’s amino acid sequence.

12) Explain the concept of wobble and the role that G-U base pairs play in this phenomenon. What is the major consequence of wobble?

• 61 codons for amino acids, but fewer than 61 tRNAs needed to read the genetic code

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

• G-U base pairs are allowable.• U in the 5' position of a tRNA anticodon can

pair with either an A or G in the 3' position of a mRNA codon

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

• As a result of wobble, a minimum of 31 tRNAs are needed to read all 61 codons.

Lecture 32

Post-transcriptional and translational regulation

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

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

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

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

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

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

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

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

Lectures 37 Purine metabolism

• 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 purine and pyrimidine biosynthesis

• The enzyme, PRPP synthetase, is highly regulated by 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?

• 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.

• 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.

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?

Regulation of de novo purine 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 phosphorylated nucleotides), and activated by substrate PRPP

– a. first unique step to purine synthesis

• 3. at the branch point of IMP: AMP feedback inhibits its own synthesis; GMP feedback inhibits its own synthesis.

• 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 allosteric sites 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:PRPP amidotransferase 2 substrates, 2 independent Kms

Gln + PRPP phosphoribosylamine + glu +PPi

• 4. Describe the salvage pathway(s) for synthesis of purines. Explain the rationale behind Adenosine Kinase and why there is this separate enzyme to phosporylate this nucleoside.

A salvage pathway for purinebiosynthesis is the primary source of purine nucleotides in cell types other than liver and brain. Salvage recaptures purine bases on their way to catabolic degradation, re-attaching them to PRPP generating IMP,GMP and AMP.

Summary of Salvage Pathway• 3 reactions

– HGPRT– APRT– Adenosine kinase

• Net effect– Recover purines prior to

their complete degradation to uric acid

1

2

3

• 4. Describe the salvage pathway(s) for synthesis of purines. Explain the rationale behind Adenosine Kinase and why there is this separate enzyme to phosporylate this nucleoside.

3

Adenosine kinase• Phosphorylates adenosine

nucleoside to AMP

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

– Because there are higher levels of adenosine than other nucleosides

– SAM /SAH• SAM=S-

adenosylmethionine• These are molecules

used in methylation rxns.. So there is a lot of it

HGPRT • hypoxanthine-guanine

phosphoribosyl transferase

• 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.

• 5. Explain what happens if you don’t have the enzyme HGPRT. How does PRPP play a role in this?

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 groups\

• The key enzyme in purine catabolism is xanthine oxidase. It is a non-heme iron containing enzyme that uses FAD and molybdenum (VI) as co-factors and converts hypoxanthine to xanthine, and xanthine to uric acid.

• 6. Which enzyme of purine degradation is inhibited by the drug Allopurinol? Why is this drug given for treatment of chronic gout?

• Chronic Gout can be treated with allopurinol (trade name zyloprin), an inhibitor of xanthine oxidase.

– Allopurinol has a very high affinity for xanthine oxidase.

– It is also a weak substrate, but the product, alloxanthine, is also an inhibitor

– Allopurinol is effective in two ways.• First by inhibiting xanthine oxidase, 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.

– Allopurinol may 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?

• The pKa of uric acid is 5.4. At physiological pH, nearly all uric acid is in the form of sodium urate.

– urate is 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 synthetase resulting 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

7. Describe the factors that can contribute to gout.

Lecture 38 Pyrimidine metabolism

• de novo pyrimidine 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

1. Identify the enzyme that catalyses the first three steps in pyrimidine biosynthesis in mammals.

Steps

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

C A D

111

Steps

C A D

Reaction 2. Aspartate transcarbamoylase (ATCase)• Adds aspartate to carbamoyl phosphate• The A in CAD• A separate enzyme in bacteria

– Historical importance as first described allosteric enzyme– Bacterial enzyme inhibited by CTP, regulated enzyme in bacteria

111

Steps

• Reaction 3. Dihydrooratase• the D in CAD• Ring closing by dehydration

C A D

111

• 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.

• Thymidine synthesis from dUMP by adding a methyl group to the base at the 5 position.

– 1. The methyl group comes from N5,N10 methylene FH4 (tetrahydrofolate) – Another important use of folic acid

– 2. The reaction is catalyzed by thymidylate synthase.• 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 dihydrofolate reductase (DHFR)

– also a key enzyme for thymidine synthesis and a target for chemotherapy.

– Methotrexate is a folate analog and a competitive inhibitor of DHFR commonly used to metabolically inhibit DNA synthesis.

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 thymidine synthesis. Why is thymidine synthesis a good chemotherapeutic target? How does the compound folinic acid factor into the inhibition of thymidine synthesis by FOLFOX 4?

Deoxynucleotides:

• Made from ribonucleoside diphosphates (NDPs)

• Catalyzed by Ribonucleotide reductase (nucleoside diphosphate reductase)

– 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 (thioredoxin or

glutaredoxin)

• Oxidized to disulfides

– Regenerate -SH via thioredoxin reductase• Use NADPH – ultimate reducing agent

• Regulation– Ribonucleotide reductase 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 (ribonucleotide reductase, thymidylate synthetase, dihydrofolate reductase and thymidine kinase)are cell cycle regulated such that their expression is increased as cells pass from G1 to S phase, when DNA synthesis occurs.

4. Describe the synthesis of deoxynucleotides from ribonucleotides. What agents are used in the oxidation/reduction reaction? What inhibits the reaction?

Important enzymes in synthesis of pyrimidine

nucleotides• RR

– ribonucleotide reductase– NDP dNDP

• TS– thymidylate synthetase– dUMP dTMP– (FH4FH2)

• DHFR– dihydrofolate reductase– FH2FH4

• TK– thymidine kinase– TMPTTP

TSDHFR

TK

• As we saw in this case study, degradation of 5FU is an important consideration in its use.

• The enzyme dihydropyrimidine dehydrogenase (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 (dihydropyrimidine dehydrogenase) in pyrimidine degradation. Why are the levels of this activity only problematic when you treat a patient with 5FU?

DPYD

pyrimidine degradation

Lecture 40Folate and Vitamin B12

summary (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. – Folate deficiency 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-adenosyl methionine, 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.

1. Explain how the enzyme Dihydrofolate reductase (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

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 methylene to methyl.

– This is an important part of the methyl trap hypothesis.

• The conversion of methylene FH4 to methyl FH4 is catalyzed by methylene tetrahydrofolate reductase (MTHFR)

DHFR

3. What is the major source of 1 carbon units for tetrahydrofolate to use? Identify the tree major pathways (or reactions) that use FH4.

• Major source of 1C units for thransfer thru FH4 is serine• 3 major recipients that us FH4:

– 1. purines (rxn 5)– 2. pyrimidine (just thymidine)

• deoxyuridine to form deoxythymidine (using N5,N10 methylene FH4) (rxn 6)

– 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

Thymidine sythesis(thymidylate synthase)

TS

MTHFR

CH3

4. Describe the methyl trap and explain how it ties into deficiency of vitamin B12.

• S-adenosyl methionine (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 [adenylate kinase] to salvage this nucleoside and we have no mechanism to salvage any other nucleosides).

• Methylation of Vitamin B12 only 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 (methylene tetrahydrofolate reductase) 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)

5. Describe how a vitamin B12 deficiency can occur, and explain why dietary deficiencies of B12 are rare.

• 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:

B12 binds to R-bindersProtein made by salivary gland and/or gastric cells

(transcobalamin I)

B12 transferred from R-binder to Intrinsic Factor glycoprotein made by gastric parietal cellsR-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

B12 deficiency not dietary, due to failure to absorb

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/folate trap

• Pernicious anemia is treated with injections of B12

– Folate supplementation treats anemia but can mask neurological deficits.

• B12 deficiency can also cause neurological symptoms

Lecture 43: Introduction to signal transduction

Ligands Ligand transport

General receptor structure

Mechanism of action

Intracellular effects

Chemical messengers that act via cell membrane receptors

Neurotransmitters, neuropeptides, cytokines, growth factors, insulin

Do NOT require transport proteins (these are water soluble proteins, unlike lipophilic hormones)

structure of cell surface receptors: 3 domainsa. ligand binding domain (extracellular)- N-terminal domain; often highly glycosylatedb. transmembrane domain – hydrophobic α helices that span the membrane, anchor receptor in membrane; hydrophilic aa in the loops;c. cytoplasmic domain- C-terminal domain; may have catalytic function, e.g. tyrosine kinase; may have sites for phosphorylation

hormone-cellsurface receptor interaction is fast,high affinity, and reversible; thenumber of receptors/cell varies with tissue type; in general, the more receptors, the > theResponse (but can be saturated). For cells of a given tissue, the number of receptors is finite.

second-messenger concept ofhormone action

ligand binding to the extracellular domain causes a conformational change that is communicated to the intracellular domain through the transmembranedomain. The activated intracellular domain initiates a characteristic cascade of biochemical activities that activate intracellular effector systems. This usually involved the binding of a specific intracellularsignal transduction protein.

Chemical messengers that act via intracellular receptors

Steroid hormone/nuclear receptor ligandsLipophilic hormones include the steroidhormones (cortisol is a glucocorticoid, aldosterone is amineralocorticoid), thyroid hormone, retinoic acids and vitamin D

Transport of steroid hormones in blood requires transport proteins- CBG, TBG, SHBG, Albumin

3 domains (see LO#2)a. N-terminal variable region b. central DNA binding domain c. C-terminal ligand binding domain (LBD)

See next slide Activate nuclear receptors (transcription factors)- bind to specific DNA sequences and regulate the rate of gene transcription (mRNA synthesis)

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

• 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 coactivators that have

histone acetyltransferase (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

• 1. N-terminal variable region containing transcriptional activation function 1 (transactivation domain (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 “ligand binding 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

• 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 kinase phosphorylates 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.

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 Ras causing a conformational change in Ras that promotes Raf binding to

Ras. • Raf is a serine protein kinase that is a MAPKKK (mitogen activated protein kinase kinase kinase). • Raf initiates a sequence of successive phosphorylations of 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.

3. Describe the pathway activated by a chemical messenger binding to receptors that are tyrosine kinases or that bind tyrosine kinases.

KNOW that activation of Tyr-kinaseReceptors leads to activation ofRas (a guanine nucleotide exchg. factor)And consequently of the MAPK pathway

• 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

• Insulin Receptor is a member of the tyrosine kinase family 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 transmembrane region plus tyrosine kinase activity.

• When insulin binds to the α subunits, the β subunits become active and autophosphorylate each other.

• The activated tyrosine kinase phosphorylates proteins 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 MAPK signaling via Grb2 activation, DAG and IP3 through activation of PLCγ, and downstream targets of PI3K including protein kinase B (also called AKT) which in turn, phosphorylates other 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

• 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 ligand binds to a heptahelical receptor, 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 adenylate cyclase which produces Camp, the second messenger molecule.

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.

E = effector: e.g., adenylate cyclase, phospholipase C,

7 transmembrane regions

C-terminus in cytoplasm

GDP inactive GTP active

G-protein coupled receptors (GPCRs)

Trimeric G protein:, -a b g subunits

Clinical Relevance: 27% of all FDA-approved drugs target GPCRs

• Activation of Protein Kinase A (PKA) by cAMP results in PKA activation and PKA phosphorylates intracellular target proteins. cAMP is hydrolyzed to AMP (inactive) by phosphodiesterase.

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.

recallCFTR

cAMP activates PKA (protein kinase A)PKA phosphorylates intracellular targets

cAMPphosphodiesterase

know 3 aa that can be phosphorylated- Ser, Thr, Tyr

•know that phosphorylation is reversible

•know that phosphorylation can either activate or inhibit the activity of an enzyme or other protein

Phosphatidyl inositol pathway:

This diagram shows a G-protein coupled with phospholipase C

know this pathway

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 4’,5’ bisphosphate (PI-4,5-bisP, PIP2) is cleaved by phospholipase C to generate 2 second messengers in the cytoplasm: diacylglycerol (DAG) and inositol trisphosphate (IP3)

• IP3 activates release of intracellular Ca++ that activates Ca++- calmodulin kinase that phosphorylates intracellular proteins

• DAG activates protein kinase C that phosphorylates intracellular proteins.

5. List mechanisms by which chemical messenger signaling is terminated.

Signal Termination 1. stimulus abates: messenger

no longer released2. messenger may be rapidly

degraded, e.g., peptide hormones

3. Receptor may be desensitized by hydrolysis of phosphate residues by phosphatases

4. Degradation of 2nd messenger, e.g., cAMP, IP3 – GTPase – GDP by phosphodiesterases

5. Internalization of receptor- degradation- recycling, e.g. insulin receptor

How are signals terminated?

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.

Lecture 44

Signal transduction for the regulation of fuel metabolism

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 diagramsPointOut theReceptorlocations

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 adenylyl cyclase activation– G-protein linked receptor

• Epinephrine– 2nd messenger (both via G protein linked receptors):

• Beta receptor: cAMP (via adenylyl cyclase activation)• Alpha receptor: DAG/IP3

• Cortisol– Intracellular steroid hormone receptor; binding of cortisol turns

glucocorticoid receptor into transcription factor (see pg 5)

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

Long-winded table

Insulin

Glucagon

Epinephrine

cortisol

Lecture 45

Glycolysis

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)

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.

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

Glycolysis: know substrates, enzymes, cofactors, regulated stepssee pg 6

Glucose-- hexokinase/glucokinase (uses Mg2+) -> glucose 6 phosphate + ADP

<- (phosphoglucose isomerase) -> 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<- phosphoglycerate dehydrogenase -> 1,3 bisphosphoglycerate + NADH

<- phosphoglycerate kinase -> 3 phosphoglycerate + ATP<- phosphoglyceromutase -> 2 phosphoglycerate

<- enolase -> phosphoenolpyruvate-- pyruvate kinase -> pyruvate + ATP

FYI: kinases phosphorylate/dephosphorylate isomerases, mutases rearrange aldolases split dehydrogenases reduce/oxidize (see NADH) enolases remove a water

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, glucokinase translocated 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) (+)

• Pyruvate kinase– Phosphorylation (-) Fructose 1,6 Bisphosphate (+)

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)– Phosphoglycerate kinase– Pyruvate kinase

• 4 ATP – 2 ATP = 2 total ATP produced per glucose

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 Dehydrogenase oxidizes NADH to NAD+ by reducing pyruvate to lactate. This makes NAD+ available to continue glycolysis

7. Describe the effect of pyruvate kinase deficiency on anaerobic glycolysis

• Pyruvate kinase– Last step of glycolysis, produces 1 ATP– If no mitochondria utilizing NADH to make extra ATP

via oxidative phosphorylation (as in RBCs), pyruvate kinase 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

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