MBLG1001 Calculations Assignment #1 2015 ANSWERS Calculations (/25) Question 1. Converting units to sensible numbers (2 marks) Convert the following values to sensible numbers (between 0.1 and 1000) by changing the prefix: (i) 4 × 10-5L = 40 µL (ii) 7,000 µg = 7 mg (iii) 0.006 µmol/sec = 6 nmol/sec (iv) 0.080 µmol/sec/nmol enzyme = 80 nmol/sec/nmol enzyme Question 2. Expressing concentration in different units (7 marks) You have a 0.25 M solution of a compound (molecular weight 200). a) How many mmoles are there in the following volumes? (2 marks; 0.5 marks each) 0.25 M = 0.25 mol/L = 0.25 mmol/mL = 0.25 umol/uL (i) 1 mL 0.25 mmol (ii) 5 mL 1.25 mmol (iii) 50 µL 12.5 umol = 0.0125 mmol (iv) 200 mL 50 mmol b) Express the concentration of the solution in the following units: (2 marks; 0.5 marks each) (i) (ii) (iii) (iv) mM; µg/µL; % (w/v); mg/dL; 250 mM 0.25 mol/L*200 g/mol= 50 g/L = 50 ug/uL is g/100mL so 50 g/L = 5 g/100 mL 1dL = 100 mL. 0.25 mol/L*200 g/mol= 50 g/L = 5 000 mg/100 mL c) How much of this compound would you weigh out to make 100 mL of the solution? (1 mark) 0.25 M = 0.25 mmol/mL. So 100 mL contains 25 mmoles. (Remember 1 mmole = mol. wt in mg, and MW= 200) so you need to weigh out 25 mmol* 200 mg/mmol = 5,000 mg or 5 g. d) If you add 5 µL of this solution (0.25 M) to a PCR (polymerase chain reaction) with a final volume of 100 µL: (2 marks) (i) How many moles of this compound will the reaction contain? (Use sensible units!) You have 0.25 M = 0.25 mmol/mL = 0.25 umol/uL so 5 uL would contain 1.25 umol. Diluting it in the assay doesn’t change the number of moles (but will change the concentration) (ii) What is the final concentration of this compound in the PCR assay? You added 1.25 umol into 100 uL so the concentration is now1.25 umol/100 uL = 12.5 umol/mL = 12.5 mmol/L = 12.5 mM. Or, using the dilution factor method: The 5 uL is now diluted to 100 uL so you have made a 1 in 20 dilution (100/5=20). So the 0.25 M = 250 mM will be diluted to 250/20 mM = 12.5 mM answers page 1 MBLG1001 Calculations Assignment #1 2015 ANSWERS Question 3. Difference between #moles and concentration (2 marks; 0.5 marks each) You have been provided with the following solutions of a compound (mol. wt. = 500). What volume would you need to pipette to add 10 µmoles of this compound using these solutions? (i) 50 mM = 50 umol/mL à 10 umol in 1/5 mL = 10 umol / 0.2 mL (= 200 uL). (ii) 0.5 % (w/v) 0.5 g/100 mL à 5 g/L à 5g /(500g/mol) = 0.01 M = 10 mM = 10 umol / 1 mL (iii) 250 mg/ml in 1 mL: 250 mg/(500 mg/mmol) = 0.5 mmol in 1 mL à 0.5 umol/uL à 10 umol is in 20 uL (iv) 25 mg/dL = 25 mg/100 mL = 250 mg/L. And this 250 mg /(500 mg/mmol) = 0.5 mmol. So 250 mg/L = 0.5 mmol /L = 0.5 umol/mL à 10 umol is in 20 ml Question 4. Using molecular weight You need a 5 mΜ stock solution of insulin for an experiment. A colleague has given you 20 mg of a pure preparation of human cloned insulin (hINS) as a lyophilised (freeze-dried) powder (MW ~ 5,000). You don’t want to weigh it out and risk losing some of it so you decide to make up the whole 20 mg batch to 5mM with sterile Tris buffer. What volume of buffer would you add? First, find how many moles you have in 20 mg = 20,000 ug à 20,000 ug/(5,000 ug/umol) = 4 umol. Next, you need to make it to 5 mM = 5 umol/mL. So as you have 4 umol you will need to dissolve it in 4 umol /(5 umol/mL) = 0.8 mL or 800 uL to give a final concentration of 5 mM Question 5. Using stock solutions in a protocol; from volume to final concentration What is the concentration when you add a certain volume of a stock solution. A DNA polymerase reaction is carried out in a 20 µL reaction mix. You are going to do a number of these reactions so you decide to make up a larger volume of the reaction mixture, called a master mix. This saves on multiple repetitive pipetting and reduces errors. You have a protocol to make 1 mL of this master mix and it says to add 5 µL of a 50 mM nucleotide mix (dNTPs). (i) What is the final concentration of the nucleotides in the master mix? The stock dNTP mix is 50 mM = 50 mmol/L = 50 umol/mL = 50 nmol/uL. So 5 uL will contain 5*50 nmol = 250 nmol. This 250 nmol is then added to 1 mL to make the master mix so the final concentration is à 250 nmol/mL = 250 umol/L or 250 uM = 0.25 mM (either is fine!). (ii) How many polymerase reactions could you get out of the 1 mL reaction mix? This is easy! 1000 µL /20 uL per reaction = 50 reactions. Question 6. Using stock solutions to make up a solution: concentration to volume Later this semester you will digest a sample of DNA with a restriction enzyme. The buffer for this reaction as specified by the manufacturer (NEB buffer 3.1) has a final concentration of 50 mM Tris, pH 7.9, 10 mM MgCl2 and 100 mM NaCl. You have the following ‘stock’ solutions: 0.5 M Tris, pH 7.9, 2.5 M MgCl2 and 5 M NaCl. How would you make up 50 ml of this buffer? answers page 2 MBLG1001 Calculations Assignment #1 2015 ANSWERS Tris: The [final] you need to make is 50 mM = 5 mmol/100 mL = so you need 2.5 mmol to make 50 mL. As you need 2.5 mmol and the stock you have is 0.5 M = 0.5 mmol/mL you need 2.5 mmol/(0.5 mmol/mL) = 5 mL of 0.5M. Or using the dilution factor method, you are diluting from 500 mM to 50 mM, a 50/500 = 1 in 10 dilution. So 1/10 of the final volume should be stock Tris (1/10 of 50 mL is 5 mL) MgCl2: The [final] = 10 mM = 1 mmol/100 mL = 0.5 mmol to make 50 mL. As the stock you have is 2.5M = 2.5 mmol/mL you will need to add 0.5/2.5 = 0.2 mL = 200 uL. NaCl: The [final] = 100 mM = 10 mmol/100 mL = 5 mmol in 50 mL. As you need 5 mmoles and the stock contains 5 mmol/mL you need 1 mL. Mix stock solutions and make to a final volume of 50 mL with water. Question 7. Working with buffers A phosphate buffer must be made to a final concentration of 50 mM. You need 500 mL. You have 2 stock solutions from which to make this buffer; 0.2 M NaH2PO4 (the acid form) and 0.2 M Na2HPO4 (the base form). You need to add 4 times more of the acid form than the base form to achieve the desired pH. How will you make up this buffer? Step 1: How many moles of total phosphate do you need for 500 mL of 50 mM phosphate? 50 mM = 50 mmol/L = 25 mmol of phosphate in total. Step 2: Work with ratios. To achieve the required pH the 25 mmol must have acid:base ratio of 4:1 which means 4 parts acid and 1 part base. This means 4/5 acid and 1/5 base. So 4/5 of 25 mmol = 20 mmol (acid) and 1/5 of 25 = 5 mmol (base). Step 3: What volume of stock do you need?: To get 20 mmol acid: stock acid is 0.2 M or 0.2 mmol/mL then 20 mmol/(0.2 mmol/mL) = 100 mL of the acid stock. For base, so using logic above 5/0.2 = 25 mL. You will then add 375 mL water to make it up to 500mL required. Alternatively using the dilution factor method: The final conc is 50 mM and the stock solutions are 0.2 M (200 mM) à 1 in 4 dilution (200 mM/50 mM)à To make 500 mL you need ¼ of the 500 mL (125 ml) to be made from the stock solutions + 375 ml water. The 125 ml has to be split 4 parts acid to 1 part base à 125/5 = 25 ml/part. Hence the acid form needs 4*25 = 100 ml and the base form = 25 ml. Question 8. Working with liquids β-mercaptoethanol (mol.wt. = 78.3) is a reducing agent used in many enzyme assays to help stabilise the enzyme. It is a liquid at room temperature with a density of 1.12 g/ml. If you needed to make up 50 ml of 10 mM β-mercaptoethanol how much would you need to add? #moles in 1 ml of β-merc = 1.12 g/(78.3 g/mol) = 0.0143 mol = 14.3 mmol à 1 µl contains 14.3 µmol. You need 50 mL of 10 mM (=10 umol/mL) , so 10 µmoles/mL * 50 mL= 500 µmol. Volume required = 500 umol/(14.3 µmol/ µL) = 35 uL answers page 3 MBLG1001 Calculations Assignment #1 2015 ANSWERS Question 9. Working with concentrated acids You have an acid supplied as a liquid. A bottled was assayed and found to have a density of 1.5 g/mL and a purity of 48%. If the molarity of the solution is 12 M what is the molecular weight of the acid? Density is 1.5 g/mL à 1500 g/L. If it is 48% pure then the 1 L actually only contains 1500*0.48 = 720 g. Its concentration is 12 M = 12 mol/L. So in 1L we have 720g = 12 mol. So the MW (the weight of 1 mol) = 720/12 = 60. The molecular weight is 60. Question 10 putting it all together with common molecular biology buffers. TAE is a commonly used buffer for DNA electrophoresis, a technique you will do later this semester. This buffer contains 40 mM Tris, 20 mM Acetate, 1 mM EDTA as final ‘working’ concentration. A 50× stock solution (50 times more concentrated) is normally prepared. You have the Tris powder (mol. wt. 121.14), glacial acetic acid (a liquid with a density of 1.05 g/mL and a mol. wt. of 60) and 0.5 M EDTA stock solution. How would you make 2.5 L of 50× TAE? The 50× TAE buffer would contain 50 x 40 mM Tris = 2000 mM or 2M. In 2.5 L this would mean 121.14 g/mol *2 mol/L *2.5 L = 605.7 g Tris powder. The 50× TAE buffer would contain 50*20 M acetate = 1000 mM or 1 M . So for 2.5L: 1 mol/L* 2.5 L à need 2.5 mol. Acetate. Glacial acetic acid has a density of 1.05 g/mL or 1050 g/L à (1050 g/L)/(60g/mol) = 17.5 M. So for 2.5 mol/(17.5 mol/L)= 0.14285 L à 143 mL. The 50× TAE buffer would contain 50*1 mM EDTA = 50 mM. The stock solution is 0.5 M or 500 mM à 1 in 10 dilution à 250 mL OR We need 2.5 L of 50 mM EDTA so 50 mmol/L * 2.5 L = 125 mmoles of EDTA. The stock solution is 0.5 M = 0.5 mmol/mL so 125/0.5 mL = 250 mL. Add the 605.7 g Tris powder, 143 mL acetic acid and the 250 mL 0.5 M EDTA, make up to ~2.3L adjust pH then make up to 2.5 L with water. How would you use your 50 x TAE stock to prepare 500 mL of TAE to use in electrophoresis? The 50× stock is, by definition, 50 times more concentrated than needed. So to make 500 mL of ‘1×’ working buffer, you must dilute the 50× stock 1 in 50. So 1/50 of 500 mL is 10 mL. Dilute 10 mL of 50× TAE in 490 mL water. answers page 4
MBLG1001 Theory Assignment Assignment # A Name: ANSWERS 1 SID: Prac Group: Total 38 marks, scaled down to a mark out of 10 Question 1 (1 mark each/ 8 marks in total) Using the five amino acids (A – E) presented on page 3: A. Arginine, B. Phenylalanine, C. Aspartate, D. Glutamine, E. Glycine (i) (ii) Which amino acid has no chiral carbon? glycine; option E Which amino acids (A – E) could form a salt bridge? Arginine (option A) and Aspartate (option C) At what pH range would this occur? Salt bridges form between positive and negative charges, eg between Arginine (+) and Aspartate (-) when side chains of both are ionized, so above pH 4 and below pH 12 These five amino acids join to form a pentapeptide. (iii) (iv) (v) (vi) Which residue (A – E) would lower the pI of the peptide the most? Explain your choice. Aspartate, option C because to get the side chain of aspartate to be neutral you must drop the pH to ~3. What would be the overall charge of this peptide at pH 1? Explain your answer. Charge would be +2. At pH 1, ionisable groups will be protonated. The arginine and the N-terminal will be protonated and positive (~NH3+), the C terminal and the aspartate will be protonated and neutral (~COOH). How many charged groups would this peptide have at pH 7? What are they? There will be 4 charged groups at pH 7: the N terminal (+ve), the C terminal (-ve) the aspartate (-ve) and the arginine (+ve) Would this peptide absorb light at 280 nm? (yes/No)Why? Yes because of the phenylalanine (aromatic ring in side chain). What would be the effect on pI of the following modifications? (vii) (viii) If the C terminal was converted to an amide. The C-terminal which was acidic and lowering the pI is now neutral so the pI will be raised when compared to the unamidated structure. If a variant of this peptide also naturally exists with an extra residue of amino acid E? Amino acid E is glycine, which has a hydrogen for a side chain (definitely not charged at any pH) so it would have no effect on the pI. MBLG1001 Theory Assignment ANSWERS 2 Question 1 continued (7 marks) Draw the pentapeptide as it would predominate at pH 13. Present the amino acids in the order presented on page 3 i.e. amino acid A is residue 1 through to amino acid E as residue 5. Diagram /3 The diagram should have the 5 amino acids joined by peptide bonds (red circles below), an N terminal and C terminal with correct ionization. The N terminal should be –NH2 and the C terminal should be –COO-. Check the ionization of each side chain. At pH13 all ionisable side chains will be unprotonated. Amino acid C (aspartate) should be ~COO- and amino acid A should be ~NH-CNH2NH On your diagram label the following: (1 mark each feature/4) • A peptide bond (red circle) • An amide nitrogen eg the one that participates in the peptide bond (green arrow) or the N of the glutamine side chain (D) • A backbone ionisable group the C terminal or the N terminal • A hydrogen atom which DOES NOT participate in hydrogen bonding many, but the safest is the alpha carbon H (blue arrow). this can be in any of the residues For example: Note this is a generic diagram at pH7 to show structures. It does NOT indicate the charges at pH 13 MBLG1001 Theory Assignment ANSWERS 3 MBLG1001 Theory Assignment ANSWERS 4 Question 2 (8 marks) Below is the structure of Arginine as shown in the textbook. The pKas quoted in the textbook are as follows: Group pKa α-COOH (pKa1) 1.8 α-NH2 (pKa2) 9.0 Side chain (pKa3) 12.5 O H 2N CH C OH CH2 CH2 CH2 NH C NH NH2 Arginine, R i) What is the pI of Arginine? Show working. (1 mark) pI is the pH where the overall charge on the amino acids is 0: ie where the number of positive charges = the number of negative charges. It might help you to draw the ionization state either side of the pKas).The pI is between 9 and 12.5 = 10.75 ….10.8 will do! ii) Draw the titration curve for this amino acid, labeling the 3 pKas and the pI. (2 marks) Note the pKas are in the flat parts of the curve, meaning these pHs are where this amino acid acts as a buffer. MBLG1001 Theory Assignment ANSWERS iii) Draw the predominant ionic form(s) at the following regions: (4 marks) • • • • At pHs < pKa1 At pHs between pKa1 and pKa2 At pKa3 At the pI iv) Would the side chain of this amino acid be predominantly ionic at physiological pH (~7.4)? If so what charge would predominate? (1 mark) The positive charge would very much predominate 5 MBLG1001 Theory Assignment ANSWERS 6 Question 3 (4 marks) Below is the structure of glutathione, a tripeptide involved in keeping the intracellular environment of the red blood cell reduced.. OO C CH O H2 C H2 C O C HN CH +NH3 C O NH CH2 C O- CH2 SH i) What are the 3 amino acids that make up this peptide? Glutamate, cysteine, glycine ii) What is unusual about the way these amino acids are joined together? The peptide bond formed between the glutamate and cysteine is formed through the carboxyl group of the glutamate side chain (red circle), known as the γ carboxyl not the alpha carboxyl (blue circle). iii) Which amino acid would be responsible for maintaining the reduced environment? The cysteine in the ~SH form. iv) Predict what this structure would look like when oxidized. Note the red circle where the disulfide bond has formed in the oxidized state. O- O C CH O H2 C H2 C C O HN +NH3 CH O C NH CH2 C O- C NH CH2 C O- CH2 S S +NH3 CH O C O- CH2 H2 C H2 C C O HN CH O O MBLG1001 Theory Assignment ANSWERS 7 Question 4 (3 marks) L-Dopa (L-3,4 dihydroxyphenylalanine) is a drug commonly administered to Parkinson’s patients in order to increase the levels of dopamine in their brain. L-Dopa is synthesized biologically from one of the 20 naturally occurring amino acids. It is the first step in the synthesis of adrenalin and as such is a very important pathway in our body. i) Which amino acid is L-Dopa made from? Highlight the differences in the two structures. L-Dopa comes from tyrosine which can originally come from phenylalanine. ii) Comment on whether L-Dopa could be considered to be included in the list of naturally occurring amino acids. L-Dopa is a naturally occurring alpha amino acid BUT not one used in translation. There are many more amino acids made by our body or modified. Only 20 are used in translation O H3N+ CH C O- CH2 HO OH L- Dopa Tyrosine MBLG1001 Theory Assignment ANSWERS 8 Question 5 (8 marks) Below are the results obtained for an enzyme catalysed reaction, performed at a number of substrate concentrations. The enzyme, alkaline phosphatase, cleaves the phosphate from a variety of substrates. In this series of reactions you use a synthetic substrate, p-NitroPhenyl Phosphate (pNPP) which, when dephosphorylated, releases p-nitrophenol, a yellow product that can be monitored at 420 nm spectrophotometrically. The first column of velocity results (Velocity1) shows the data obtained from a series of experiments with alkaline phosphatase (1 µg) and p-NPP. The second series of results (Velocity2) were also obtained using alkaline phosphatase, however a different amount of enzyme was added. [S] (mM) 0 0.1 0.2 0.5 1 2 5 10 100 500 Velocity1 (nmol/min) Velocity2 (nmol/min) 0.0 16.7 25.0 35.7 41.7 45.5 48.1 49.0 49.9 50.0 0.0 8.3 12.5 17.9 20.8 22.7 24.0 24.5 25.0 25.0 i) Find the Vmax and Km for both sets of data? (2 marks) Series 1 Vmax (nmol/min) Km (mM) 50 0.2 Series 2 Vmax (nmol/min) Km (mM) 25 0.2 ii) From this information determine how much enzyme was added to obtain the second series of results. (2 marks) The Vmax is half the original series (50 nmol/min à 25 nmol/min) so half as much enzyme has been added. If 1 ug was added to the original series 1 then 0.5 ug have been added to each assay in series 2 MBLG1001 Theory Assignment ANSWERS 9 Arctic phosphatase, isolated from the arctic shrimp, is used extensively in molecular biology. It also catalyses the removal of the phosphate from compounds such as p-NPP. This enzyme is used to remove the terminal 5’ phosphate from RNA, DNA and nucleotides. It has a much lower optimal temperature for activity but is heat labile and so can be inactivated at temperatures >60oC unlike “traditional” alkaline phosphatase. iii) Would the source of the phosphatase enzyme alter the ΔGo’ and the Keq for this reaction (p-NPP)? Explain your answer. (2 marks) The thermodynamics of the reaction won’t change if a different phosphatase is used. The ΔGo’ and the Keq for the reaction therefore won’t change. These parameters are based on the energy involved in the removal of the phosphate. Whichever enzyme is used or even no enzyme won’t change these values. iv) If these phosphatase enzymes were used to remove the phosphate from substrates such as DNA and RNA would the thermodynamic parameters (ΔGo’ and Keq) of the de-phosphorylation reaction be altered? Explain your answer. (2 marks) The thermodynamics (ΔGo’ and Keq) will change with different substrates. The reaction is different if you are removing the phosphate from the 5’ end of DNA or RNA to removing it from p-NPP.