Biochemistry First Test Study Guide
Biochemistry First Test Study Guide CH 344
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This 4 page Study Guide was uploaded by Cody Brazel on Tuesday September 8, 2015. The Study Guide belongs to CH 344 at Southeast Missouri State University taught by dr. Jacoby in Spring 2015. Since its upload, it has received 60 views. For similar materials see Biochemistry in Chemistry at Southeast Missouri State University.
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Date Created: 09/08/15
Biochemistry First Study Guide As we all know the first coming test of the semester will be this Friday In light of this this will hold a culmination of all the knowledge I believe will be on the test First we will review the Henderson hasselbach equation How to Make a Buffer I would strongly suggest reviewing the notes I have posted for the first week of Biochemistry for even more practice Also review Dr Jacoby s video over quotwaterquot which has own take on making buffers To use the Henderson Hasselbach for his test requires a very set way If you can master this way or the way he does it in his water video you will earn full points for it on the test To summarize the process found in my first week of biochemistry notes 1 2 3 Write the Henderson hasselbach equation Note the pH that solution is going to be should be given The pKa you will plug into the equation is the pKa closest to the pH in step 2 pKa values will be given to you The compound with the chosen pKa is your acid You will be solving for the AHA portion of the equation So in short your equation should look like this pHpKaLogx Subtract the pKa from the pH giving you this equation pHpKaLogx To get rid of the Logarithm take each side and take it to the power of ten Your equation will look like this 10PH39pKax This is where my method and Dr Jacoby s method defer After step 6 you will need to find the percent composition of the acid and base X is actually equal to how much base you have for every 1 acid Therefore the total amount of ratio is equal to base1 To find the percent of the base and acid take the amount of acidampbase and then divide both of those values by base1 a For example the percent composition of the base is equal to XX1100 Multiplying by 100 gives the actual percent b The percent composition of the acid is equal to 1X1100 After finding the values given in substeps A and B multiply each value by the liters of the solution The liters of the solution should also be given Record each value for this tells you how many moles of acid and base there are in the solution a For example the amount of base in a solution is equal to XX1Liters of solution b The amount of acid in a solution is equal to 1X1Liters of solution After finding the moles of the acid and the base multiply each value by its molar mass The formula weight may be given if not just add the atomic mass of each element in the molecule together a The ending equation for how many grams of base is needed should look like this XX1Liters of SolutionFormula weight of molecule b The ending equation for how many grams of acid is needed should look like this 1X1Liters of SolutionFormula weight of molecule This concludes how to make a buffer I will give an example problem for you to work through If you have any questions or feel like you found the answer please contact via email The email will be given at the end of the study guide Example 1 The pH of the solution is 90 and it will be 4L We will be using a buffer using molecules X Y Z and P Molecules X Y39l Z392 and P393 are the same chemical but have been ionized 0 Molecule X has a molecular weight of 67 Its pKa is equal to 24 Molecule X is fully protonated 0 Molecule Y391 has a molecular weight of 90 Its pKa is equal to 79 Molecule Y has been ionized once 0 Molecule 2392 has a molecular weight of 113 Its pKa is equal to 934 Molecule 2 has been ionized twice 0 Molecule P393 has a molecular weight of 135 Its pKa is equal to 1125 Molecule P is fully ionized Using the information presented in this problem show your work and find the grams needed of the acid and base to make the correct solution Since these numbers and compounds are hypothetical the numbers may look weird In my experience as a tutor odd numbers are only problem if you do not do the process correctly If you do the process correctly in this question you can tackle any question he gives on the test Amino Acids There are quite a few different concepts that are covered with Amino Acids Things that should be studied for amino acids are 0 Recognize each amino acid by its structure 0 Memorize the single letter and triple letter code of each amino acid 0 Know how to draw each amino acid 0 Identify each amino acid by category The categories are Polar Nonpolar and Charged 0 Know the pKr values of the seven ionizable amino acids 0 DO NOT study each individual pK value of all amino acids Just the values of the amino acids referenced in the above bullet 0 Know that the carboxyl group of an amino acid usually has a pK value of 24 And that the amino group of an amino acid usually has a pK value of 911 0 Know that if the pH of a solution exceeds the value of a pKa that group is ionized Conversely if the pH of a solution drops below a pK value that group is protonated Pretty much every bullet shown here except for the last one requires selflearning Please look at pages 7779 to memorize each amino acid s structure category and codes What can be taught is the titration of an amino acid i l E I l I l I 5 1E Titrating Amino Acids As stated in my second week of notes for biochemistry titration was a simple matter in the gen chem labs It is taught that when a plateau is met on a titration curve half ofthe targeted ionizable group is 14 NI 390 12 NoznPo mo 10 8 Na HPO 2 Muse K 720 9 6 p 2 mmm aco 4 MHgmg DIM 2 9K0 39 2 HPoaca 6 ETTTGT T13 quot3926 TESTSo ml GIN NaOH added 010 ml 0 1M HgPO m 36 Malian Curve of phosphonc acid H3POQ wm soanum 39wdromde NOOHJ showmg the three bufferan plateaus Obovul me 0K0 values The weak cod and com gar 005910 eocn buffering plafeoo are Identified illlli L Ii Eiljllizlquot Jiljji III rain inn life if Lima Ell EH21 if llsIL ii is I tip it HR 3R Ht PHI fl PH ai 39l 51H El II II II Hiatitili e II l UH if i a latte I mi H3133 L I 17 in i i E t at E a interacting with the administered base In this picture provided by biologydiscussioncom the titration of phosphoric acid is shown The pKa of nonionized phosphoric acid is 212 and it can be seen that when the pH reaches 212 half of the phosphoric acid is interacting with NaOH After passing the buffering of phosphoric acid one point below and above a pK value one whole molar component of phosphoric acid has been deprotonated giving the molecule Sodium Phosphate NaHZPO4I Amino acids function the same way In each case amino acids have a carboxyl group an amino group and possibly an ionizable R group If you have an amino acid with no ionizable R group you will see two plateaus and then three plateaus if you are dealing with a triprotic amino acid The hardest part of the amino acid titration is that there are some other concepts associated with them EUH What Dr Jacoby will want us to see is the pI of an amino acid The pI of an amino acid is equal to the average of the two pK values surrounding the zwitter ion In the second picture here the titration of histidine can be seen With each passing of a pK value we subtract one hydrogen from histidine Not that eventually we reach an area where our net ionic charge is 0 THIS IS THE ZWITTERION No look at the two pK values surrounding the zwitterion and take the average of those two pK values Now you have found the pI of the amino acid easy Proteins Unfortunately Dr Jacoby is still going over protein structure and whatnot but that does not mean this section will be empty Proteins are made of an amalgamation ofamino acids Each amino acid in a protein is linked via a bond between the carboxyl group of one amino acid and the amino group of the other This bond is called a peptide bond There are four different stages in a proteins life 1 2 Primary stage The linear sequence of amino acids Secondary stage the polar and nonpolar interactions along with hydrogen bonding along the peptide backbone influence slight folding and confirmations Tertiary Stage More powerful ionic interactions begin to further fold the protein This stage is where a majority of proteins we see exist It is the lower energy state of a protein and is therefore the most stable in this stage Quaternary stage This type of stage does not exist in all proteins and is not necessary for a protein to function Via ionic bonds multiple tertiary structures may conjoin and form a larger mass of proteins Note that in a Alpha Helix there are 36 residues per turn and straight measurement per turn of 54 these units are in angstroms If you have any questions concerning the practice problem given in the HH section or would like some questions answered please feel free to contact me at cibrazdazgmailcom
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