10pagevideoreview.UNIT 2: AMINO ACIDS/PROTEINS RECORDED COHORT NOTES, Exams of Nursing

Download 10pagevideoreview.UNIT 2: AMINO ACIDS/PROTEINS RECORDED COHORT NOTES and more Exams Nursing in PDF only on Docsity! 10pagevideoreview.UNIT 2: AMINO ACIDS/PROTEINS RECORDED COHORT NOTES 1. Identify the three types of amino acids Amino acids: building blocks of proteins Amino acid structure: ● Amino group: A nitrogen with at least one hydrogen attached ○ NH2 or NH3+ ● Carboxyl group: Carbon with two oxygens (COO) or carbon with two oxygens and a hydrogen (COOH) ○ COO or COOH *The carboxyl and amino group are held together by an alpha carbon and it is known as the backbone of the amino acid structure because it is always present* ● Alpha carbon ○ C ● R group or side chain: the unique portion of the amino acid that defines the chemical properties of that specific amino acid BE ABLE TO IDENTIFY AMINO ACID STRUCTURE Amino acid types and specific properties: 1. Hydrophobic or nonpolar amino acids: R group contains and ends in CHs (CH3 or CH2) ○ R group ends in CH of some kind ○ Forms hydrophobic interactions ○ Bond disrupted by heat ○ This Amino acid likes to hide from water causing the protein chain to turn onto itself ○ Hydrophobic interactions are the weakest type of bond but due to the nonpolar nature and turning onto self, a strong protein structure is formed 2. Polar amino acids: R group ending in OH, SH, NH ○ R group ending in SH i. Forms Disulfide bonds ii. Bond is disrupted by a reducing agent (i.e Bleach) iii. Amino acid example: Cysteine ○ R group ending in NH or OH i. Forms hydrogen bonds (very weak bonds) ii. Bond is disrupted by pH or salt ○ Misfolding and aggregation: due to protein denaturation, hydrophobic amino acids are exposed which can cause clumping of the protein structure ➢ EX: Alzheimer's disease UNIT 3: ENZYMES RECORDED COHORT NOTES 1. Understand what enzymes are and how they catalyze reactions a. Describe Enzymes and their function ■ Enzymes are proteins and they function to catalyze reactions ■ Are subject to the same denaturation techniques that other proteins are ■ EX: Lactose breakdown. Lactase enzyme is needed for lactose breakdown to glucose and galactose ■ Enzymes lower the activation energy of a reaction ★ Activation energy is the energy needed to activate a reaction. With the right amount of activation energy, substrates can be converted to products ★ When an enzyme is present in the reaction, the amount of free energy needed to create the products is decreased ● Enzymes ↓ Activation energy and ↑ Rate of reaction ■ Enzymes need to be fully folded into their tertiary or quaternary states to be functional. In their fully folded state, an active site is present that allows a binding spot. ■ How do enzymes catalyze: ★ Step 1: Substrate recognition. ● Enzyme specificity attracts the substrate to the active site of the enzyme ★ Step 2: Enzyme-substrate complex formation ● The substrate binds to the active site ● This creates an induced fit: The enzyme’s shape will change to facilitate the reaction desired ★ Step 3: Enzyme-product formation ● Products are being developed while the enzyme is attached to the active site ★ Step 4: Product release, enzyme recycled ■ Kinases are a type of enzyme that adds a phosphate molecule ■ Phosphatase is a type of enzyme that detract a phosphate molecule b. Recognize the enzymes, substrates, intermediates, and final products of a pathway ■ Substrates bind to an enzyme’s active site ★ Enzymes specificity: The active site of an enzyme fits only its specific substrates and the enzyme will not bind with a substrate that is not specifically made to bind without alterations. 1.Understand how information is transmitted in the cell from DNA to RNA to proteins a. Basics of DNA replication: Must follow complementary rules ● Antiparallel: Coding and template are in opposite directions ★ Template strand: 3’--5’ ★ Coding strand: 5’--3’ ● Base pairing: Nucleotides fit together ★ A+T, C+G ● DNA replication creates complementary DNA sequences between the coding and template strands ● Remember: 1. Coding strand (5’--3’) 2. Template strand (complementary to coding strand, 3’--5’) 3. mRNA (Complementary to the template strand 5’--3’) 4. tRNA (complementary to the mRNA strand, 3’--5’) b. Structure of nucleic acids: DNA is made of a sugar-phosphate backbone and has bases adenine, guanine, cytosine, thymine, uracil. c. DNA and RNA work together to make proteins ● Transcription: Formation of complementary mRNA from template DNA ★ The cell will only use the template DNA to make mRNA. The actual DNA is not used! ★ RNA will substitute uracil base in place of thymine base ★ mRNA will follow the complimentary rules and will be in 5’-- 3’ format because template DNA is 3’--5’ format ● Translation: mRNA codes for amino acid sequences ★ The Ribosome is responsible for reading the mRNA strand to translate it into protein language ● The ribosome always reads mRNA in the 5’--3’ format and in 3 nucleotide codon sets at a time. ● Each codon translates to an amino acid ● The ribosome and tRNA work together to create a protein peptide chain ★ tRNA: has two ends. One end with an anti-codon that will base pair with the codon of the mRNA and the other end with an amino acid ● When translation occurs, each tRNA will pair using an anti-codon (a base pair with the mRNA codon set being read) ● When two tRNA amino acids are next to each other, a peptide bond can be created to begin or continue the polypeptide chain 2. Identify the factors that control gene expression (transcription) a. Factors that affect gene expression : ● Promoters: Indicates the start of a gene. This tells the RNA polymerase where to start making RNA ● Transcription factors: Small proteins that act like little foot blocks that help the RNA polymerase get to the correct start position ● RNA Polymerase: The enzyme that makes mRNA. Connects to the transcription factors at the promoter site and begins transcription. ● Nucleosomes: quaternary structures made of proteins called histones. Used to keep the DNA organized and combined. ★ When the nucleosomes are tightly packed the gene expression is low or off. The promotor is hidden. Transcription factors are not present. ★ When the nucleosomes are widely spaced gene expression is turned on. Promotor is exposed. Transcription factors are in place. RNA polymerase is ready to move. ● Chemical markers: ★ Methylation: Chemical marker on the histone that signals the nucleosomes to tightly combine. ● High methylation = No gene expression ★ Acetylation: Chemical marker present on the histone that signals the nucleosomes to space out ● High acetylation= gene expression b. DNA/RNA Splicing: ● Introns: sequences of nucleotides that do not code for protein ● Exons: Sequences of nucleotides that contain the codons for proteins ● RNA Splicing: the process of cutting out introns from RNA and joining together the exons ★ Splicing makes it statistically less likely that a mutation will occur so splicing is a protective mechanism ★ Alternative splicing: the process of cutting out exons on top of introns in splicing. This can be done to create different proteins from the same genes. ■ Damage to several nucleotides → Nucleotide Excision Repair (NER) ★ Nucleotide Excision Repair: Multiple Base issue ○ 1. DNA repair molecules recognize the damage and cut out the problem area breaking the backbone in the process ○ 2. DNA polymerase will fill in the resected portion ○ 3. Ligase repairs the break in the DNA backbone ■ Mistakes in DNA replication → Proofreading, mismatch repair (MMR) ★ Proof-reading: ○ 1. DNA polymerase “proof-reads” the DNA strand simultaneously as it is creating it ○ 2. DNA polymerase inserts correct based ★ Mismatch repair: ○ 1. DNA repair enzymes recognize the mistake made by DNA polymerase during replication and remove it ○ 2. DNA polymerase inserts the correct bases ○ 3. Ligase comes and seals the break in the backbone ■ Double-stranded breaks → Homologous recombination (HR), Non- homologous end joining (NHEJ) ★ Homologous Recombination: ○ 1. Double-stranded breaks are aligned with homologous DNA ○ 2. The undamaged DNA provides the needed sequence to repair the damage ★ Non-homologous end joining: ○ 1. broken ends of the DNA are glued together without regard for the sequence resulting in some genetic information being lost 3. Understand how PCR is used in genetic testing, including the steps of PCR, Components, and the number of DNA molecules produced. a. What is PCR : A laboratory procedure used to synthesize copies of DNA in a machine called the thermocycler. Through using PCR many copies of DNA sequences can be synthesized (up to millions of copies) b. Components of PCR: ■ Target DNA: the DNA that is to be copied ■ Heat stable DNA polymerase: Enzyme that copies DNA that tolerates high temperatures ★ Taq DNA polymerase is used and is able to stand high temperatures ■ Nucleotides (dNTP): nucleotides needed to make a DNA strand ■ Primers: short complementary strands of DNA that allow DNA polymerase to bind to the target DNA ★ Provides a start site c. What makes PCR different from normal DNA replication? ○ Binds to oxygen and stores it within the muscles ● Myoglobin does not have cooperativity ● Myoglobin is a storage protein b. Hemoglobin structure and function: ● Hemoglobin is a 4 subunit protein (2 alpha, 2 beta) containing four heme, four iron and can bind to four oxygen ○ It is a quaternary structure ● Hemoglobin is found in the blood ○ Delivers oxygen to tissues in need ○ Hemoglobin has a low affinity for oxygen (likes to give oxygen away) ● Hemoglobin is dynamic with changes in shape due to oxygen binding. ○ Hemoglobin is said to have cooperativity–Meaning that if one oxygen bind to one hemoglobin subunit, oxygen will be attracted to the other 3 subunits. ○ In graph form, we would see a slow rise to the curve because of initial oxygen binding. As more oxygen binds the curve spikes because of cooperativity. ● Delivery protein 2. Given the oxygen binding curve (and an oxygen concentration), identify the saturation of myoglobin or hemoglobin a. Oxygen binding curve: ● Hemoglobin has a lower affinity for oxygen making the hemoglobin curve more to the RIGHT ● Myoglobin has a higher affinity for oxygen making the myoglobin curve more to the LEFT ● Shifts to the LEFT = increased affinity for oxygen ● Shifts to the RIGHT = decreased affinity for oxygen 3. Describe the structural and functional properties of R-State and T- state a. Oxygenated vs deoxygenated hemoglobin: ● OXYGENATED HEMOGLOBIN: ○ Heme is planar or flat ○ R state is oxygenated. R state of hemoglobin promotes oxygen binding. ○ Subunits move closer together ● DEOXYGENATED HEMOGLOBIN: ○ Heme is bent ○ T state is deoxygenated. T state of hemoglobin does NOT promote oxygen binding and has a lower affinity for oxygen in this state. ○ Subunits move farther apart b. Why are the shape changes of hemoglobin important? ● These shape changes are how hemoglobin subunits exhibit cooperativity and aid in the efficiency of the hemoglobin molecule 4. Understand what causes pH to change and what constitutes an acidic pH in the blood. Explain how pH affects the oxygen binding of hemoglobin. a. The Bohr effect: describes how pH affects hemoglobin’s ability to bind or release oxygen ● What is pH: the concentration of hydrogen ions in a specific solution ○ Normal blood pH = 7.4 ○ Lower pH= acidity, higher pH= alkalosis ● CO2 conversion: ○ CO2 is very nonpolar and requires conversion to bicarbonate via the carbonic anhydrase enzyme. The conversion creates bicarbonate and free hydrogen. ○ CO2 concentration is high in tissues because there is constant use of oxygen within the tissues ○ When hemoglobin comes to the tissues, the free hydrogen binds to the hemoglobin. This in turn encourages the oxygen to be released off the hemoglobin and into the tissues. The hydrogens/protons then continue to bind to the hemoglobin and catch a ride to the lungs where the CO2 is exhaled and the hydrogen is repurposed to water. LOW PH HIGH PH -Carbon dioxide is high -low carbon dioxide -Excess Hydrogen ions -low hydrogen ions -Hemoglobin will release oxygen at low pH -Basic -Right shift on the oxygen dissociation -Hemoglobin will bind to oxygen curve due to the hemoglobin having less -Left shift on the oxygen dissociation curve affinity for oxygen because hemoglobin has a higher affinity -Low pH will be in the tissues for oxygen -Low pH will cause a T state because -High pH in the lungs 2. Describe how sugars, amino acids, and fatty acids are used to create new ATP by aerobic metabolism a. Aerobic Metabolism : the goal is to create ATP ➢ ATP: The chemical energy that is used to drive many of the necessary reactions within the body ● ADP (adenosine diphosphate) with the addition of a phosphatase created ATP (adenosine triphosphate). ● ADP +¿ Phosphate ↔ ATP b. Stages of Aerobic metabolism: 1. Glycolysis: Breakdown of a six-carbon glucose molecule into 2 three-carbon pyruvate molecules. ● occurs in the cytoplasm of cell ● Important end products: 2 Pyruvate, NET 2 ATP, 2 NADH ● The pyruvate that is produced in glycolysis will be converted to acetyl -COA. ● DOES NOT REQUIRE OXYGEN ● Steps of glycolysis: ○ 1. Consume an ATP molecule. Achieved through hydrolysis of ATP to ADP. Hydrolysis produces energy. ○ 2. Consume another molecule of ATP through hydrolysis. ○ 3. For each molecule of glucose, consume a molecule of NAD+ and convert it to NADH. Two NADH are produced for two pyruvate molecules. ○ 4. Removal of a phosphate on the 2 pyruvate molecules and transfer to ADP. Creating 2 ATP. ○ 5. Removal of another phosphate on the 2 pyruvates, transferring to ADP to create 2 more ATP. *Left with 2 pyruvate molecules which will travel on to the citric acid cycle* 2. Citric acid cycle: Acetyl-COA is fed into the cycle and stripped of electrons which are then transferred to electron carriers NADH and FADH2. ● occurs in the mitochondrial matrix ● Important end products: 6 NADH, 2 FADH2 ● What is happening : Stripping of high-energy electrons away from the carbon atoms that came in two at a time as the acetyl-CoA. Stripping the high-energy electrons away from those carbon atoms and passing those high-energy on to two molecules call NADH and FADH2 ● REQUIRES OXYGEN ● NADH delivers electrons to complex 1 of the electron transport chain. FADH2 delivers electrons to complex 2 of the electron transport chain. *Do not focus on the steps of the citric acid cycle so much as the function of NADH and FADH2* 3. Oxidative phosphorylation: Uses the energy from the electrons to pump protons to create the proton gradient, which is the energy source for ATP synthase to make ATP. ● consists of the electron transport chain, the proton gradient, and ATP synthase ○ Electron transport chain: Series of large protein complexes embedded in the inner mitochondrial membrane. Accepts electrons from NADH and FADH2. Uses the energy provided by NADH and FADH2 to pump protons (hydrogen ions) from the mitochondrial matrix to the intermembrane space. ○ Proton gradient: A concentration difference between the mitochondrial matrix and the intermembrane space. A form of stored energy that provides the energy source for making ATP in the mitochondria. ○ ATP synthase: necessary machinery needed to harvest energy created by the proton gradient to create ATP. ● Occur in the inner mitochondrial membrane ● Important end products: ATP ● REQUIRES OXYGEN ● Steps to Oxidative Phosphorylation: ○ 1. NADH delivers an electron to complex 1. The energy pumps a proton into the intermembrane space. ○ 2. The electron then moves to Coenzyme Q10. ○ 3. FADH2 delivers its electrons to complex 2. The electrons then are passed to coenzyme Q10. Complex 2 does not pump protons! ○ 4. Coenzyme Q10 transports the electrons it has accumulated from complexes 1 and 2 to complex 3. Complex 3 then uses some energy from the electrons to pump protons into the intermembrane space. ○ 5. Complex 3 passes its “less” energetic electrons to Cytochrome C. ○ 6. Cytochrome C passes the electrons it receives to complex 4. ○ 7. Complex 4 then uses the rest of the energy that the electrons have to pump protons into the intermembrane space. ○ 8. Oxygen molecules accept the used electrons from complex 4 which prevents a blockage in the electron transport chain. Hydrogen molecules are taken as well to create water. ■ Metformin: ➢ Decreases gluconeogenesis in the liver ➢ Increases glucose uptake in the muscle cells ➢ Overall, lowers blood glucose through the above processes ➢ Lowering blood glucose increases beta-oxidation resulting in overproduction of acetyl-CoA which leads to overproduction of ketones ➢ Ketones are acidic in nature and excess keeps in the blood leads to diabetic ketoacidosis due to a drop in blood pH 3. Identify the impact on glucose production and storage given different scenarios. a. Insulin: ● Responds to high blood sugars ● A hormone that signals the storage of surplus glucose as glycogen ● In response to insulin release, blood sugar levels will decrease to normal ● Encourages Glycogenesis and Glycolysis b. Glucagon: ● A hormone that responds to low blood glucose levels ● A hormone that signals the body to start breaking down stored molecules ● In response to glucagon, blood glucose levels will elevate ● Encourages Glycogenolysis and gluconeogenesis UNIT 7: LIPIDS RECORDED COHORT NOTES Functions of lipids: ● Energy storage ● Hormonal signaling ● Neruotransmission ● inflammation/pain regulation ● Makes up components of cell membranes Lipids: ● Hydrophobic molecules ● Fatty acids: ○ Type of lipid ○ Contains a hydrocarbon chain(Hydrophobic) and a polar carboxylic acid group ○ Two main classes: saturated and unsaturated 1. Explain the differences between saturated and unsaturated fatty acids and how they affect membrane fluidity a. Saturated fatty acids : ■ No double bonds between carbons in the hydrocarbon chain ■ Long-chain saturated fatty acids are mostly solid at room temperature ■ Many saturated fats are from animal products b. Unsaturated fatty acids: ■ One or more double bonds between carbons in the hydrocarbon chain ★ Monounsaturated: contains one double bond in the hydrocarbon chain ★ Polyunsaturated: contain more than one double bond in the hydrocarbon chain ■ Mostly liquid at room temperature ■ Many are from plant products ■ Will be in a Cis or Trans configuration ★ Cis: Hydrogen around the double bond is both pointed up, they are both on the same side of the double bond. The structure bends at the cis double bond ➢ A more common configuration in nature ★ Trans: Hydrogens around the double bond are on opposite sides (one pointing up and one pointing down). No bend in the structure. c. Fatty acid nomenclature: Right to left in numbering ■ Alpha carbon: The carbon within the ACID group ■ Alpha bond: Follow the alpha carbon connecting the acid group to the chain structure ■ Beta carbon: Present next to the beta bond ■ Beta bond: Next to the beta carbon. This bond is acted upon during beta-oxidation. ■ Omega-9 double bond: between omega carbons 9 and 10 ■ Omega 6 double bond: between the omega carbons 6 and 7 ■ Omega 3 double bond: Between omega carbons 3 and 4 ■ Omega carbon: Surrounded by three hydrogens d. General fatty acid formula: Cn❑ H 2 n O2❑ 4. Recognize components of the fluid mosaic model: a. Membranes: Important because it helps compartmentalize b. Fluid mosaic model: ■ Moves like a fluid ■ Composed of phospholipids and are flexible within the cell/membrane ★ Phospholipids form a bilayer ★ Amphipathic molecules that are part hydrophilic (head) and part is hydrophobic (tail) ★ Two fatty acids tail attached to a glycerol molecule. A phosphate group is present as well. ➢ The hydrophilic head contains glycerol and phosphate i. Face outward and have contact with the cell environment ➢ Hydrophobic tails contain the two fatty acid tails that can be saturated or unsaturated. i. Located in the middle of the bilayer away from water ■ The composition of the fatty acids within the membrane will affect the fluidity of the membrane ★ Factors Increasing membrane fluidity: ➢ The presence of a kink in the tail helps to prevent packing and this increases fluidity ➢ Increasing the number of double bonds ➢ Shorter fatty acid chains lead to higher fluidity and lower melting points ★ Decreasing factors to fluidity: ➢ Packing of the fatty acid tails within the bilayer decreases fluidity ➢ Lack of double bonds and more single bonds decreases fluidity ➢ Longer fatty acid chains increase viscosity and cause higher melting points 5. Identify different types of lipids such as essential fatty acids, triglycerides, phospholipids, steroids, vitamins, eicosanoids: a. Essential fatty acids: ■ Essential fatty acids contain an omega-3 and/or an omega-6 double bond b. Triglycerides: ■ Triglycerides: Molecule that contains 3 fatty acid chains and one glycerol molecule. ■ Stored in adipose tissue and are the main source of energy storage in the body