BIOCHEM C785 Kaleys Comprehensive Study Guide final - 2021 Graded A

Page 1 of 39 Biochemistry: Mod 1  DNA = phosphate + deoxyribose sugar + A/T/C/G o Contains two strands. The strands are antiparallel (opposite each other). o 5’ → 3’ 3’ ← 5’  RNA = phosphate + ribose sugar + A/U/C/G o Single strand, can fold back onto itself and form pairs between itself (stem‐loop).  Each nucleic acid is made up of polymers (many monomers) that are called nucleotides. o Nucleotides contain one or more phosphates, a five‐carbon sugar, and a nitrogen base. o Nucleotides are always made in the 5’ to 3’ direction. o 5 is always the beginning of the strand, 3 is the end where nucleotides are added.  DNA organization: DNA is wrapped around proteins called histones → nucleosome → chromaƟn fiber→ chromosomes  Steps to the central dogma: o Coding DNA → template DNA → mRNA → tRNA (amino acid) o DNA → transcribed to mRNA → translated to protein o Each step is complementary (opposite) to the previous step, but if you skip a step it will be identical to the previous step. o Example  1. Coding DNA strand 5’ AAA TTT GGG CCC 3’  2. Template DNA strand 3’ TTT AAA CCC GGG 5’  3. mRNA 5’ AAA UUU GGG CCC 3’  4. tRNA Lys Phe Gly Pro  Pairing: o DNA: A → T o RNA: A → U  DNA replication: o Because DNA is a double helix, one strand can be separated and serve as a template for synthesis of a new strand. o Semi‐conservative: each copy of DNA contains a template strand and a new strand. o Steps of replication: Page 2 of 39 o 1. The DNA must be separated, creating a replication fork. This is done by helicase. o 2. Primase attaches an RNA primer, where the replication is to start. o 3. DNA polymerase adds bases to the remaining of the strand until it reaches a stop codon. This is done in fragments, called okazaki fragments.  If an error is detected, it removes the nucleotides and replaces them with correct ones, known as exonuclease. o Exonuclease removes all of the RNA primers, and DNA polymerase fills in those gaps. o DNA ligase seals the two strands forming a double helix.  DNA → transcribed → mRNA → translated → protein  Transcription occurs in the nucleus: o Initiation: RNA polymerase binds to a sequence of DNA called the promoter, found near the beginning of a gene. Each gene has its own promoter. Once bound, RNA polymerase separates the DNA strands, providing the single‐stranded template needed for transcription. Page 3 of 39 o Elongation: One strand of DNA, the template strand, acts as a template for RNA polymerase. As it "reads" this template one base at a time, the polymerase builds an RNA molecule out of complementary nucleotides, making a chain that grows from 5' to 3'. The RNA transcript carries the same information as the non‐template (coding) strand of DNA, but it contains the base uracil (U) instead of thymine (T). o Termination. Sequences called terminators signal that the RNA transcript is complete. Once they are transcribed, they cause the transcript to be released from the RNA polymerase. o Pre‐mRNA must go through extra processing before it can direct translation.  They must have their ends modified, by addition of a 5' cap (at the beginning) and 3' poly‐A tail (at the end).  Pre‐mRNAs must also undergo splicing. In this process, parts of the pre‐mRNA (called introns) are chopped out, and the remaining pieces (called exons) are stuck back together. Page 4 of 39  Translation occurs in the cytoplasm: o Initiation: The ribosome assembles around the mRNA to be read and tRNA brings in its perspective protein, decoding 3 bases at a time, beginning with the start codon, AUG. o These 3 base pairs of mRNA are called codons. The mRNA base pairs are complementary to the base pairs of the tRNA, called anticodons. o Elongation: The amino acid chain gets longer. The mRNA is read one codon at a time, and the amino acid matching each codon is added to a growing protein chain. When the complementary pairs are formed, they are added to the protein chain by peptide bonds, the result is polypeptides. o Termination: The finished polypeptide chain is released when a stop codon (UAG, UAA, or UGA) enters the ribosome.  Gene regulation o Promotor sites: can be turned off or on, enabling or disabling a gene from being replicated. o Alternative splicing: Exons are used to code for protein, introns are clipped out. The order of exons can determine different mature mRNA strands which result in different proteins. o Epigenetics: involves packaging of DNA. DNA is round around histones. These packages are called nucleosomes. How tightly packed they are determines whether or not the gene is on or off. Page 5 of 39 o Loosely packed = transcription possible. o Tightly packed = transcription impeded. o Modifications determine how tightly/loosely packed they are. Many of these modifications are determined by environment/diet.  Mutations‐ Mutations originate at the DNA level, but show their effects at the protein level. o Point mutations‐ when one of the DNA bases (nucleotides) are replaced with another, results in a different protein o Example:  Coding: 5’ AAA TTT GGG CCC 3’ = Lys  Mutation: 5’ TAA TTT GGG CCC 3’ = stop codon o Missense mutations‐ any mutation that causes a change in amino acid o Nonsense mutations‐ leads to a stop codon “stop the nonsense” o Silent mutations‐ doesn’t affect the protein at all, the codon results in the same protein o Frameshift mutations‐ adds one base into the protein sequence, but changes the final protein amino acids by causing a shift.  Insertion mutation‐ can change the entire read of the protein  Deletion mutation‐ can change the entire read of the protein  Conservative mutations‐ where the resulting amino acid is in the same type as the original  Non‐conservative mutation‐ amino acid is a different type than the original Page 6 of 39  Repairing Mutations o Damage to single nucleotide bases from harmful molecules (chemicals or oxygen). o Repair: Base excision‐ replace with a base that isn’t damaged.  1. DNA repair enzymes recognize the damaged base, removes it.  2. DNA polymerase fills the gap with a new base.  3. Ligase seals the gap. o Damage from UV causes multiple damaged nucleotides, ie causing two thymine’s to fuse together (called thymine dimers). o Repair: Nucleotide excision repair‐ removes 20‐30 nucleotides to fix damage.  1. DNA repair enzymes recognize damage, cuts out damage and surrounding area.  2. DNA polymerase fills in the gap with new bases.  3. Ligase seals the gap. o Base mismatch due to errors in replication. o Repair #1: DNA polymerase proof reading.  Removes incorrect base, inserts correct base. o Repair #2: Mismatch repair‐ fixes the mismatch  1. DNA repair enzymes recognize mistake, and remove several bases surrounding the mismatch.  2. DNA polymerase inserts correct bases.  3. Ligase seals the gap. o Double stranded breaks in the DNA from radiation, can lead to cell death o Repair: Page 7 of 39  Homologous recombination‐ a sister chromosome is used as a guide to recombine the strands by copying the chromosome, but this doesn’t always work. Must be done after DNA replication.  Nonhomologous end joining‐ if no sister DNA to copy (ie before DNA replication) the non‐damaged sections are joined together and the damaged DNA is lost. Last resort method d/t high risk of mutations.  Inheritance o 1‐22 are autosomal chromosomes o If a mutation is on an autosomal chromosome (1‐22) then there is no bias towards males or females. o Sex chromosomes: o Females Xx o Males Xy o X linked mutation is bias because females have Xx and males are Xy. o Allele: a copy of a gene o Genotype: complete set of genes, the genetic makeup of an individual. o Phenotype: all the observable characteristics or traits of an individual, including ones that are not easily seen, such as blood type or color blindness o The genotype (pair of genes) decide the phenotype (observable characteristics) of an individual. o Heterozygous: one allele is dominate while the other is recessive. The dominant allele is observable in the phenotype while the recessive allele is not. (Aa) o Homozygous: two identical alleles (AA or aa) o Dominant: an allele that always expresses its phenotype, even in the presence of a recessive allele, represented by a capital letter. o Recessive: an allele that is only expressed in the phenotype when both alleles of a gene are recessive. Represented by a lower‐case letter.  Determining pedigrees o Females are indicated by circles; males are indicated by squares. o Unaffected individuals are indicated by open shapes; affected individuals are indicated by filled shapes. o Recessive vs Dominant o If two unaffected parents have an affected child, they are carrier parents. o Carrier parents = recessive trait o No carrier parents = dominant trait o Autosomal vs Sex linked o Males and females affected equally = autosomal o Only males = sex linked o Autosomal dominant vs sex linked dominant Page 8 of 39 o Affected males with a sex linked dominant trait will pass it on to all of their daughters (females do not inherit father’s Y chromosome, therefore only getting affected X chromosome) Page 9 of 39  Co‐dominance‐ both equally share dominance (red and white)  Incomplete dominance‐ neither really stand out (pink)  Complete dominance‐ either one or the other is dominant (red OR white) Page 10 of 39 Visual representation with chromosomes  Remember that chromosomes come in pairs of two, eliminate the answers that don’t include pairs.  Carriers have one of each allele (Rr). If the person actually has the disease, they will have both alleles (rr) or (RR).  Autosomal is chromosomes 1‐22, sex linked is chromsomes X and Y. Page 11 of 39  PCR and genetic testing‐ DNA replication in a test tube o What is needed for PCR  1. Template DNA  2. Nucleotides (dNTP’s)  3. DNA polymerase  4. DNA primers o Stages of PCR:  1. Denaturation 95 degrees C  Separates the template DNA strands to be able to copy each strand.  2. Annealing 50 degrees C  DNA Primers that match the gene were looking for attach to the ends of the piece we want to copy.  3. Elongation 70 degrees C  DNA polymerase adds on to the primers, building a copy strand. o Using PCR to detect mutation‐ must copy DNA multiple times (2^n, where n= # of cycles)  Make primers that flank the mutation and sequence the product  Use primers that stick to the mutation. If the mutation is present, it will stick, if it isn’t present, it won’t stick. Page 12 of 39  Epigenetics is the result of making different proteins and gene expression. It determines what genes function and which do not. o Increased expression means you make more proteins. o Decreased expression means you stop making proteins. o 5 required parts, the first three parts are required to turn a gene on.  Promoter‐ Start line for making the protein.  Transcription factors‐ foot blocks for the runner (RNA polymerase) to start with.  RNA polymerase‐ the runner, makes the mRNA  Need all three of the above to turn gene on (increase expression)  Nucleosomes‐ the packaging of DNA. Spacing determines whether or not the promoter is visible for a gene to be turned on or off.  If it is loosely packed, the promoter is visible and can be accessed by RNA polymerase =Increased expression.  If it is tightly packed, the promoter cannot be accessed and therefore the gene will remain turned off= Decreased expression.  Methylation‐ CH3 (Methyl) is added to DNA or nucleosomes. Turns gene off by causing nucleosome to become tightly packed. Without it, gene remains on.  Acetylation‐ Gene expression is turned on because nucleosomes are widely spaced apart and transcription factors can get in to start transcription. Page 13 of 39 Page 14 of 39 Biochem Mod 2  Amino acids o Amino acid back bone‐ same in all amino acids  Central carbon (alpha carbon)‐ carbon in the center that holds the amino acid together as other groups bind to it. C‐H, CH  Amino group‐ contains nitrogen and hydrogen. NH2, NH3+  Carboxyl group‐ has two oxygens and one carbon, gives the amino acid its acid properties. COO‐, COOH o Side groups (R group) are different which effect how the amino acid acts, will classify them as charged, polar, or hydrophobic. To determine what their classification is, first look for:  Charged amino acids‐ R group will have ‐/+ which means negatively or positively charged.  If charged, will take precedence. If not charged, look to see if its polar.  Polar amino acids‐ will have SH, OH, NH at the end of the R group. If not polar, then the amino acid is non‐polar.  Non‐polar amino acids‐ will have CH at the end of the R group.  Hydrophobic amino acids‐ will have several hydrogen molecules in the R group, each will end with an H atom. o Bonds  Amino acids will have three types of bonds  Charged amino acids make ionic bonds (only + with ‐)  Polar amino acids will make hydrogen or disulfide bonds o OH and NH make hydrogen bonds o SH makes disulfide bond (the strongest type of bond, can only bond with itself so very few of them) o “OH look it’s a Northern Hemisphere/Southern Hemisphere Polar Bear!”  Non polar amino acids will make hydrophobic interactions o CH “Can’t have water” (weakest bond, but many of them)  Strongest to weakest: disulfide, ionic, hydrogen, hydrophobic  What breaks the bonds (denaturing)  Charged: pH and salt changes  Polar: pH and salt changes, disulfide bonds have to be broken by reducing agents  Non polar: broken by heat Page 15 of 39  Alanine: be able to recognize structure. Is hydrophobic and ionized (will have + or ‐) Page 16 of 39  Protein structure‐ chains of amino acids o Linking amino acids together‐ Forming peptide bonds (the backbone of an amino acid)  Primary structure: chain of amino acids by peptide bonds, does not denature  The carboxyl group and amino group of two amino acids bond together by using 2 hydrogens and one H2O. A water molecule is lost in this process, known as dehydration. o Amino group + carboxyl group  Secondary structure: shaped that is formed when hydrogen bonds are added between carboxyl and amino groups.  Forming of alpha helix and beta sheets within the backbone, held together by hydrogen bonds.  Tertiary structure: three‐dimensional folding, the result of different secondary structures interacting with one another via their R groups/side chains.  These interactions include hydrophobic interactions, hydrogen bonds, ionic bonds, and disulfide bonds. Proteins can now function at this stage.  Disruption of its hydrophobic state is the simplest way to denature.  Quaternary structure: more than one amino acid/polypeptide/protein, held together by R groups/side chains. Not all proteins need this structure. Page 17 of 39  Protein folding o Chaperones help fold proteins o Can misfold, or take another shape (conformation), results in it being non‐functional o Denature: environmental change that causes the protein to misfold or unfold by breaking side chain bonds/secondary structure, but does NOT break the primary structure o Degradation: breaking apart of the primary structure/peptide bonds by hydrolysis o Aggregation: proteins clump together abnormally due to hydrophobic interactions either by unfolding or mutation. o Hydrophobic interactions: when a protein is exposed by unfolding, causing its hydrophobic parts to be exposed to water o Misfolding of proteins leads to Alzheimer’s: Intracellular tangles and extracellular plaques (senile plaques) are caused by aggregated amyloid‐beta fibers which accumulate in the brain. Connections between tau is lost leading to progressive neurodegeneration. Page 18 of 39  Enzymes o One of the most important types of proteins in our cells. o Known as catalysts, they speed up reactions and use less energy  ↑rate of reacƟon, ↓acƟvaƟon energy  Can do the same reaction over and over  They only act on specific substrates‐ a molecule that is specific for that enzyme  Binds via its active site, a binding platform for its specific substrate  Have a high degree of specificity‐ they catalyze only one type of reaction, and most act on only one substrate  Molecules that are different in shape or function bind to the enzyme. Therefore, the enzyme and its substrate are complementary, like a “lock and key”.  While an enzyme and its substrate are complementary, many enzymes will adjust their active site slightly to improve the fit of the substrate. Known as induced fit, “hug”. o Reusable‐ Enzyme cycle  Enzyme + specific molecule/induced fit = reaction/product, then release of product  Enzyme pathways o Enzyme reaction: substrate + enzyme = product o Enzyme pathway:  Substrate + enzyme 1 = product →  Enzyme 1 product (now substrate) + enzyme 2 = enzyme 3  The product of one enzyme can be the substrate of another. o Control of enzyme activity  An organism must be able to control its enzyme availability and activity.  1. Control of enzyme availability‐ depends on its rate of synthesis and degeneration. Changes over time.  2. Control of enzyme activity‐ can be inhibited due to too much product or presence of inhibitors, and can be modulated (modified) through structural alterations. o Control by modification: phosphorylation and dephosphorylation (attachment/removal of a phosphate group) Page 19 of 39 o Control by enzyme inhibition:  Feedback inhibition – similar to the drug induced form of noncompetitive inhibition, when excess product is detected, the pathway is stopped by inhibition.  The product at the end of the pathway binds to allosteric site on the first enzyme at the beginning of the pathway to stop the process. o Medical inhibition with drugs  Competitive inhibition‐ appears similar to the substrate, binds to the active site on the enzyme to prevent substrate from binding  Can be overcome by adding additional substrate, if there’s too much substrate and not enough inhibition the substrate will win.  Noncompetitive inhibition‐ binds to the allosteric site (away from the active site)  Changes shape of protein and therefore changes the active site, causing the substrate to not be able to fit into the enzyme. Page 20 of 39 Biochem Mod 3  Myoglobin and Hemoglobin o Myoglobin‐ storage of oxygen, holds onto it  1 subunit, one polypeptide chain with tertiary structure, has only one heme therefore can only carry one oxygen molecule  Can grab oxygen quickly, stores oxygen within muscles tightly (high affinity)  Is not affected by changes in pH  Since it is stored in muscles, can be used to detect MI, rhabdomyolysis, etc. because it will be released when muscle is damaged. o Hemoglobin‐ transport of oxygen  4 subunits, quaternary structure  Responsible for transport of oxygen from the lungs to the tissues, but not very good at grabbing oxygen (lower affinity).  Heme‐ iron, where O2 binds, has 4 so it can carry 4 oxygen molecules at one time  Tense or "T" state  Deoxygenated, heme is bent out of shape (dome shape), its 4 subunits are spread apart, low affinity. Low affinity = decreased ability to grab oxygen, will require more and more oxygen to bind  Once one subunit binds to oxygen, the shape of all 4 subunits change to the relaxed state. This is called cooperativity, and it will bind to O2 faster now.  Relaxed or "R" state  Oxygenated, heme is flat and relaxed, subunits are closer together  Has an increase in affinity so its ability to grab oxygen is better.  Once one of the subunits grabs oxygen, the others tend to grab oxygen more quickly. Deoxygenated (tense state) Oxygenated (relaxed state) Page 21 of 39 o Oxygen binding curve ‐Dotted line is myoglobin, is hyperbolic shaped, can grab oxygen even at low levels (high affinity). Once it grabs one oxygen, it is full to capacity because it can only carry one at a time. ‐Solid line is hemoglobin, is S shaped (sigmoidal). Has 4 subunits that must cooperate together (cooperativity). Has low affinity therefore unable to grab oxygen unless there is plenty available. Page 22 of 39  Bohr Effect o How pH allows hemoglobin to function o Use CHART.  Low pH (acidic) causes the following:  C = ↑CO2  H = ↑H+ (acid)  A = acidic  R= release of O2 from hemoglobin, right shift in curve  T= tense state in the tissues  High pH (basic) causes the following (opposite of low pH):  C = ↓CO2  H = ↓H+ (acid)  A = basic  R= binding of O2 from hemoglobin, left shift in curve  T= relaxed state in the lungs ↓pH (acidic) = right shift curve ↑pH (basic) = left shift curve Think in terms of a car. The car is the hemoglobin. It has 4 tires, which are its 4 oxygen molecules. When the car goes through an acidic tunnel, its 4 tires will fall off which will result in ↑ temp, ↑ CO2, and a tense state in the Ɵssues. The lower the pH gets, the further right your S shaped graph will go. Page 23 of 39 How the above chart works, starting from the right:  Phase one 1. The tissues (muscles) generate oxygen by cellular respiration. 2. Citric acid cycle (krebs cycle) releases carbon dioxide, which must get out of the muscles. 3. CO2 cannot bind to hemoglobin. It must be converted to liquid by the enzyme carbonic anhydrase. This combines carbon dioxide with water to form bicarbonate (HCO3‐) and a hydrogen ion (H+) in the blood. It then travels back to the lungs where it can be exhaled. 4. When CO2 is released, so is H+ (acid), which makes the blood acidic therefore dropping the blood pH.  Phase two 5. In the lungs, oxygen binds to hemoglobin, putting it into the relaxed state and giving it high affinity for oxygen.  Phase three 6. Acid (H+) binds to hemoglobin, changing it to the tense state. The tense state has lower affinity which causes it to release the oxygen to the tissues either by cellular respiration or myoglobin storage. Page 24 of 39  Carbon monoxide (CO) o Binds to hemoglobin on the heme and blocks oxygen from binding. o Has 200x greater infinity than oxygen, which makes it a competitive inhibitor. o Puts hemoglobin in the R state, causing it to pick up even more CO. o Will not allow oxygen to be dropped off. o Patients will appear be very pink in color as poisoning starts off with flu like symptoms, dizziness and confusion.  2,3 BPG o Produced in the body naturally. o Stabilizes the T state, similar to H+. No oxygen binding, mom becomes deoxygenated. o Produced under two circumstances:  Produced at high altitudes when there is less oxygen available, to help oxygen get released to tissues.  Produced during pregnancy in the placenta, in order to help maternal hemoglobin pass to the fetus. Mom’s affinity for O2 is lower so that baby’s can be higher. Maternal Hbg binds to BPG and is passed to baby.  Fetal hemoglobin o Has higher affinity for oxygen than maternal hemoglobin, allowing it to pick up more oxygen. o Has alpha and gamma forms of hemoglobin (adults have alpha and beta) Page 25 of 39 Biochem Mod 4  Anabolic pathways: small molecules are assembled into large ones. Energy is required.  Catabolic pathways: Large molecules are broken down into small ones. Energy is released.  Metabolism consists of all of the anabolic and catabolic pathways, and these pathways require nutrients to proceed.  Adenosine triphosphate (ATP) is the main molecule used to provide energy to metabolic pathways in the cell.  5 Carbohydrate metabolism pathways o 1. Glycolysis o 2. Citric acid cycle (aerobic) o 3. Electron transportation (aerobic) o 4. Fermentation (anaerobic) o 5. Glucogenesis (anaerobic)  1. Glycolysis‐ the breaking down of glucose in the cytoplasm o Can occur under both aerobic and anaerobic conditions because it does not require oxygen. o 4 steps:  1. The transfer of phosphate groups from ATP to glucose.  2. Breaking of a six‐carbon molecule (glucose) into two three‐carbon molecules (pyruvate).  3. The transfer of two electrons to the coenzyme NAD+ to generate two NADH molecules.  4. The capture of energy in ATP via phosphorylation (the adding of phosphate to a molecule). o What goes in: glucose, 2ATP, 4 ADP, 2 NAD+, 2H+, 4 phosphate = 4 pyruvate. o What comes out: 2 phosphate, 2 ADP, 2 NADH = 2 pyruvate  Aerobic Respiration: When oxygen is present (aerobic) the pyruvate from glycolysis is converted to a two‐ carbon molecule called acetyl‐CoA in the matrix of the mitochondria. o Before pyruvate can enter the citric acid cycle, it must first move to the mitochondrial matrix and be converted to acetyl‐CoA. o The enzyme pyruvate dehydrogenase is located in the mitochondrial matrix and converts pyruvate to a two‐carbon unit known as acetyl‐CoA. o This reaction removes one molecule of CO2, transfers electrons to NAD (to form NADH), and adds on coenzyme A (CoA). The resulting acetyl‐CoA molecule can then enter the citric acid cycle. Page 26 of 39  The citric acid cycle (Krebs Cycle) is a cycle of eight enzymatic reactions that convert acetyl‐CoA to CO2 and transfers the electrons to the electron carrying molecules NADH and FADH2. o Ins: Acetyl CoA, NAD+, H+, FAD o Outs: CO2, NADH, FADH2  The electron transport chain  The NADH and FADH2 generated from the citric acid cycle move to the inner membrane of the mitochondria where they donate their electrons to the electron transport chain.  Oxygen is the terminal electron acceptor that ultimately receives the electrons from the electron transport chain. Upon accepting the electrons, oxygen combines with protons to create water. The role of oxygen is crucial for the generation of ATP in the cell and is the sole use of the oxygen we breathe.  ATP synthase uses the proton gradient created by the electron transport chain to generate ATP from ADP and phosphate in a process known as oxidative phosphorylation. Page 27 of 39  The electrons are passed throughout the chain via each protein complex (as illustrated, left to right)  Protein complex 1 (1st complex) takes NADH, passes to protein complex 2, which accepts it as FADH2, etc.  The energy passed between complexes moves protons (H+) across the membrane into the intermembrane space (proton gradient).  Only complexes 1, 3 and 4 can pump protons. Complex 2 can only pass electrons to the next complex.  H+ evens itself out on either side by going through the channel of ATP synthase (as illustrated on the right). 30 ATP are made per 1 glucose. H+ goes through channel, turns into ATP.  Oxygen is the final electron acceptor. Without oxygen (anaerobic) the electrons have no where to go and they begin to back up, which in turn causes the citric acid cycle to back up as well. The mitochondria shuts down and changes to anaerobic metabolism mode.  Anaerobic Metabolism: Page 28 of 39 o When there is no oxygen available, cells can convert pyruvate to lactate under anaerobic conditions to allow glycolysis to continue producing some ATP. This is done via fermentation.  Two hydrogen atoms are added to pyruvate to form lactic acid (lactate).  Formation of lactate regenerates the NAD+ that was used during glycolysis, thus allowing glycolysis to continue making small amounts of ATP for the cell.  Lactate is eventually converted back to glucose via gluconeogenesis when the cell is no longer without oxygen. This whole process is known as the Cori Cycle. o The Cori Cycle  Glycolysis → pyruvate → fermentaƟon → lactate → gluconeogenesis in liver (uses 6 ATP) → glucose → glycolysis  Inefficient because uses 6 ATP to produce 2 ATP  Takes place in muscles when over worked, and in the red blood cell because they have no mitochondria.  Gluconeogenesis—when the cell has ample ATP it shifts from making ATP to storing it as glucose or glycogen via glucogenesis‐ essentially the opposite of glycolysis. Gluconeogenesis takes excess pyruvate and converts it back into glucose for storage. Page 29 of 39 IN OUT +Net gain Glycolysis Glucose 2 ATP 2 NAD+ 4 ADP 2 pyruvate 4 ATP 2 NADH 2 ADP 2 pyruvate 2 ATP Citric Acid Cycle Acetyl CoA NAD+ FAD H+ CO2 NADH FADH2 ATP The Cori Cycle Pyruvate NADH H+ 6 ATP Lactate NAD+ 2 ATP Loss of 4 ATP Electron transport chain NADH FADH2 NAD FAD ATP + 30 ATP Page 30 of 39  G word break down: o Glyco/gluco = glucose o Glycogen = stored glucose o Genesis= make o Lysis= break o Neo= new  G words o Glucose= sugar o Glycogen‐ stored glucose o Glycolysis‐ break down glucose to make ATP o Gluconeogenesis‐ make new glucose o Glycogenesis= make stored glucose in the liver for short term use o Glycogenolysis‐ break down (stored) glycogen into new glucose o Glut4= door that allows insulin into cells  G words and their processes: o Insulin is released when there’s high glucose in the blood. Lets glucose into cells, signals glut4 to open its door to allow glucose into cell.  1. Once in the cell‐ glycolysis= glucose → pyruvate → ATP (energy)  2. If too much glucose, short term storage in liver  3. Fatty acid synthesis‐ make fat from acetyl CoA, stored in triglycerides within adipose tissue o Glucagon= released when there’s low glucose, (glucose is gone)  Glycogenolysis‐ break glucose out of glycogen  Gluconeogenesis‐ make new glucose from  Lactate  Acetyl CoA → pyruvate →glucose  Glycerol (breakdown fats)  Amino acids (proteins, as last resort)  Beta Oxidation‐ break fats into acetyl CoA Page 31 of 39 Biochem Mod 5  Fatty acids‐ the simplest of lipids. All lipids are nonpolar and hydrophobic. o Alpha, beta, omega bonds  Alpha is attached to carboxyl  Beta is attached to alpha‐ breaks when we break down fatty acid for energy (beta oxidation)  Omega is complete opposite (end) of carboxyl, count left to right from omega carbon to determine which # carbon is the double bond o Fluidity  More double bonds always mean more fluid (except trans)  If they have the same number of double bonds, shortest chain has the most fluid  Saturated fatty acid‐ solid at room temp, from animal sources (ie. butter)  No double bonds between carbons  Straight or linear, long chain, stays compacted together, can stack well  Contain maximum number of hydrogens  Unsaturated fatty acid‐ oils that are liquid at room temp, plant based  At least one carbon that is not paired with hydrogen  Has at least 1 double bond between carbons  Short chain, kinked, can move around easily, doesn’t stack well (d/t double bonds)  Lose 2 hydrogens per double bond  Unsaturated fatty acids melt at cooler temperatures than saturated fatty acids of the same chain length  Monounsaturated fatty acid‐ A fatty acid containing one double bond  Polyunsaturated fatty acid‐ A fatty acid with more than one double bond in its carbon chain Page 32 of 39 Page 33 of 39 o Essential fatty acids‐ Must be eaten in diet. There are different categories of unsaturated fatty acids depending on the location of the first double bond in the carbon chain.  If the first double bond occurs between the third and fourth carbons, counting from the omega end (CH3) of the chain, the fat is said to be an omega‐3 fatty acid (EPA/DHA, fish oil)  If the first double bond occurs between the sixth and seventh carbons (from the omega end), the fatty acid is called an omega‐6 fatty acid; this type of fatty acid is obtained from vegetable oils such as corn and safflower oil.  Omega 3, Omega 6 have to be eaten in our diet‐ essential  Omega 9 is not essential, body can make it o Rules for fatty acids  All fatty acids only have 2 oxygens  If it is saturated, there will be twice as many hydrogens as carbons.  If unsaturated, every double bond loses 2 hydrogens.  More double bonds = more fluid, if they have the same # of double bonds then the shorter chain has more fluid. o Cis vs trans  Cis fats: have both hydrogen atoms on the same side of the double bond  Have a kink or bend in the carbon chain, making it difficult for the fatty acids to pack together  Unsaturated natural oils  Trans fats: when the hydrogen atoms are on opposite sides of the double bond  Trans fatty acids can pack tightly like saturated fatty acids so they have a higher melting point  Has double bonds but acts saturated  Unsaturated fats treated with chemicals (margarine)‐ body cannot process  When consumed, trans fats raise blood cholesterol levels and increase risk of heart disease. Page 34 of 39 Page 35 of 39  Lipid structure and function o Fatty acid structures  Four types:  1. Full structure  2. Structural formula  3. Zig‐ zag structures‐ every peak represents a carbon  4. Chemical formula  Counting carbons: Shaded yellow is omega carbon group. Count from it to the right until you reach the double bond to determine its omega number. This example is an omega 6 fatty acid.  Structural formula: Parentheses mean that that group is repeated, the number to its immediate right is how many times. In this example, CH2 is repeated 4 times.  Zig zag structure: Each point on the zig zag structure represents a carbon. If any of the carbons are missing their bonds, you can assume that the remaining bonds not shown are to hydrogen. Page 36 of 39  Sterol family‐ lipids with 4 rings, include cholesterol and testosterone. Be able to recognize them. o Cholesterol is needed to synthesize vitamin D in the skin; cholic acid, a component of bile; and steroid hormones. o The steroid hormones include testosterone and estrogen, which promote growth and the development of sex characteristics, and cortisol, which is released in response to stress and promotes glucose synthesis in the liver  Triglycerides‐ fat and oils within the body, consist of a backbone of glycerol with three fatty acids. o Glycerol + 3 fatty acids = triglyceride o Two glycerols can make glucose o If only one fatty acid is attached to the glycerol, the molecule is called a monoglyceride o When two fatty acids are attached, it is a diglyceride. o Triglycerides store energy‐ fatty acids can be oxidized to produce ATP.  If the body has no immediate need for energy, they are stored as triglycerides in adipose tissue and in the liver.  Consist of 98% of our energy reserves. Page 37 of 39  Vitamins‐ fat soluble, must have some sort of fat in order to be absorbed. Recognize them‐ they do not have 4 rings o Vit A: for the eyes o Vit K: blood clotting o Vit E: antioxidant o Vit D: promotes absorption of calcium  Arachidonic acid is the precursor to many eicosanoids. o Eicosanoids are another class of lipids derived from essential fatty acids that also act as local hormones.  Generated as needed  Are involved in the production/regulation of pain and fever, blood pressure, blood coagulation, and reproduction.  Treatment with aspirin inhibits the production of PGH2, which decreases the production of several eicosanoids involved in the development of inflammation, fever, and blood clotting. Page 38 of 39  Phospholipid structure and function o Have a backbone of glycerol with two fatty acids attached, as well as a phosphate group which attaches to a variety of other molecules  Glycerol + two fatty acids + phosphate group  Fatty acid end is hydrophobic (lipid soluble)  The phosphate end is hydrophilic (water soluble) o Considered amphipathic‐ are both lipid and water soluble. o The lipid bilayer‐ phospholipids have polar heads that interact with the watery environment and nonpolar tails that remain inside together to avoid water. Can form tightly packed bilayers, where fatty acids and triglycerides cannot and are leaky.  Controls the way things go in and out of cell.  Things that control membrane fluidity:  number of double bonds (more = fluid, less = solid)  cholesterol  length of tail (longer = solid, shorter = fluid)  Fatty acid synthesis‐ takes place in the cytoplasm o Acetyl‐CoA from digestion is intercepted before it goes through the citric acid cycle. It is taken out of the mitochondria to build fat. o Acetyl‐CoA cannot leave the mitochondria by itself, must pair with oxaloacetate. o Acetyl‐CoA + oxaloacetate = citrate → crosses into cytosol → unjoin, releases Acetyl‐CoA o Acetyl‐CoA is made of two carbons, attached to coenzyme A. o Linking Acetyl‐CoA together, process repeats over and over: Page 39 of 39  Activation by biotin (below picture) then malonyl + Acetyl‐CoA = Butyryl Lipid Metabolism  Beta Oxidation‐ the breakdown of fats by breaking the beta bond that occurs in the mitochondria. o Fats are stored as triglycerides‐ three fatty acids attached to triglycerol o Beta oxidation is a major source of cellular energy, especially during a fast when carbohydrates are not available. o Each round of beta oxidation produces one FADH2, one NADH, and one acetyl‐CoA  Ins: FAD, NAD+, Coenzyme A, fatty acid  Outs: 2 carbon shorter fatty acid chain, acetyl CoA, FADH2, NADH o The citric acid cycle oxidizes the acetyl‐CoA to produce an additional three NADH, one FADH2, and one GTP, further oxidation occurs in the electron transport chain, making the yield of ATP much higher than that of glucose. o To determine the # of acetyl CoA made: # of carbons divided by 2 o To calculate number of rounds “chopping down a tree”: # of acetyl CoA ‐ 1, or # of carbons divided by 2 – 1. o A defect in the enzyme used in beta oxidation is called medium‐chain acyl‐CoA dehydrogenase deficiency (MCADD), which usually presents in infants as vomiting and/or lack of energy due to their inability to utilize medium‐chain fatty acids. The disease is often fatal if not detected early, and newborn infants are routinely screened for this disease. Individuals with MCADD must avoid fasting for prolonged periods and need diets rich in slow‐release carbohydrates to maintain their energy levels. Filename: Kaleys comprehensive study guide final Directory: C:UsersKaley DuniganDocumentsbsn Template: C:UsersKaley DuniganAppDataRoamingMicrosoftTemplatesN Title: Subject: Author: Kaley Dunigan Keywords: Comments: Creation Date: 7/16/2019 7:45:00 PM Change Number: 41 Last Saved On: 8/9/2019 2:20:00 AM Last Saved By: Kaley Dunigan Total Editing Time: 13,206 Minutes Last Printed On: 8/29/2019 8:24:00 AM As of Last Complete Printing Number of Pages: 39 Number of Words: 5,419 (approx.) Number of Characters: 30,891 (approx.)