Dilution and Lab Math Problem Set F23 STUDENT.docx-2

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BIOL200L F23 NAME__ Meah Haskins __________________ Asynchronous assignment for the week of Oct 9-13 Instructions: You may complete this document electronically in Word by typing or writing electronically, or you may print and fill in this document by hand. Please use a color other than BLACK to complete this assignment to aid in the grading. Please submit your completed assignment using a turn in link in Bb. Submissions must be either .docx or .pdf or scanned .jpg files. This assignment is due at the beginning of your lab section the week of October 16. Points will be deducted for late submissions per the syllabus. I. Enzyme Lab Prep: answer the following questions. A. Match the following numbered items with the appropriate lettered phrase. There is one correct answer per numbered item. B. What is the experimental variable your group is planning to test next week? The pH, temperature, and salinity. C. List at least ONE control variable for your experiment. 50µ of Trypsin. D. At the end of the experiment, if the absorbance reading is very high, what would that tell you about enzyme activity? If the absorbance reading is high this could mean that enzyme activity is high as well. 1 __ D __1. Trypsin a. maintains pH and salt levels during experiment __ E __2. Azocasein b. precipitates the enzyme and undigested azocasein __ B __3. TCA c. wavelength to read absorbance of undigested azocasein __ C __4. 440 nm d. the enzyme used in the experiment __ A __5. Buffer solution e. the substrate used in the experiment __ G __6. Pellet f. digested Azocasein found here g. undigested Azocasein found here

II. Metric System Review: Length, Volume and Mass The Metric System was devised by French scientists in the late 18th century to replace the disorganized collection of units of measurement then in use. To obtain a standard of length, a quadrant of the earth (one-fourth of a circumference) was surveyed from Dunkirk to Barcelona along the meridian that passes through Paris. The distance from the pole to the equator was divided into ten million parts to constitute the meter. In other words, the Meter is 1/10 millionth the distance from the pole to the equator. You are not responsible for any of the above information and have our permission to immediately forget it! It is worthwhile, however, to know that the units of volume and mass were derived from the meter. For example, the standard metric unit of volume, the liter, is defined as the volume of one cubic decimeter (10 cm on all sides) . Likewise, the milliliter, is defined as the volume of one cubic centimeter. Imagine a cube that is 1 cm in length on all sides (smaller than a cube of sugar). This is a cubic centimeter (cc or cm 3 ) and holds exactly 1 ml of liquid. Recall that in medicine, the cubic centimeter is still used as a unit of volume (for example: 0.5 cc of epinephrine). The metric unit of mass , the gram, is defined as the mass of 1 ml of water . This holds true for 4 o C temperature water, because the density of water, which is greatest at that temperature, has been designated to be 1.00 (1.00 gm/ml). Therefore, mass and volume can be independent assays for the accuracy of each other as long as you know the liquid’s density. For example, the manufacturers of graduated pipets and micropipettes calibrate their equipment by weighing a defined volume of water to determine if the mass agrees with the presumed volume. The table below describes the prefixes for metric units commonly used in molecular biology. Prefix Meaning Exponential Notation mega one million 10 6 kilo one thousand 10 3 milli one-thousandth 10 -3 micro one-millionth 10 -6 nano one-billionth 10 -9 pico one-trillionth 10 -12 Below are tables of commonly used molecular biology metric units for volume and mass. Volume /mass Symbol Equivalent Useful Conversions liter L milliliter ml 10 -3 L 1 L = 1000 ml microliter μ l 10 -6 L 1 ml = 1000 μ l gram g milligram mg 10 -3 g 1 g = 1000 mg 2

microgram μ g 10 -6 g 1 mg = 1000 μ g nanogram ng 10 -9 g PRACTICE: Fill in the equivalent conversions. a. 0.18 ml = ___ 180 ____ μ l b. 150 mg = ___ 0.15 ______ g c. 465 μ l = __ 0.465 ___ ml d. 12 ng = ___ 0.012 ____ μ g e. 1.02 ml = __ 1020 ______ μ l f. 2.5 μ g = __ 2500 ___ ng g. 0.85 L = ____ 850 _____ ml h. 0.05 g = ____ 50 _______ mg III. Dilutions and Dilution Factors As we saw in the vitamin C lab, we sometimes must dilute or reduce the concentration of a sample in order for a particular assay to be effective. Dilution is the process of adding a solvent (often water, but occasionally other things such as ethanol) to reduce the concentration of a sample. However, we do not simply mix a random amount of solvent with a random amount of sample. That would mean we’d have no idea what the concentration of our diluted sample was! Instead, we can calculate specific volumes of solvent and sample to mix together in order to create a specific dilution, such as 1:10, 1:50, 1:100, 1:1000, or even greater dilutions. Note that the bigger the number in the denominator of the dilution ratio (the number to the right of the colon) the smaller the concentration you will end up with after you perform the dilution. So, a 1:10 dilution will give you a higher diluted concentration than a 1:1000 dilution. But both will be a smaller concentration than the original undiluted sample. The values given above (for example 1:10) are known as dilution factors or dilution ratios . Dilution factors tell you how many parts of a concentrated sample is used per how many parts of solvent. In other words, in a 1:10 (read as one-to-ten) dilution you would have one part of your concentrated sample and ten total parts of diluted solution. Because there are ten total parts of diluted solution but only one part of your original sample, you would need nine parts of solvent (such as water) to make this dilution. The denominator of the dilution factor (10 in the example above) can be used to calculate the initial volume of a sample add or the final volume of a diluted sample through the following equation: DF=V f /V i where DF is the dilution factor, V f is the final total volume and V i is the initial volume of sample added. Once you know both the volume of sample added and the total final volume, you can determine how much water must be added to the sample to get to the total final volume. (Note you can also use the final and initial concentrations in place of volume but then your equation would be DF=C i /C f where C i is the initial or starting concentration and C f is the final concentration). Example: What is the dilution if you add 4 ml of water to 1 ml of orange juice? Answer: The final volume is 4 ml + 1 ml = 5 ml The initial volume of juice is 1 ml DF = V f /I v or 5ml / 1ml or 5 So this is a 1:5 dilution of orange juice with water. If you know the starting and ending concentrations of your undiluted and diluted samples and the final volume or the volume of the initial solution you will use in your dilution, you can use the equation 3

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C 1 V 1 =C 2 V 2 to calculate the unknown concentration. C 1 is the initial concentration of a solution V 1 is the volume of the initial solution you will use in your dilution C 2 is the diluted or final concentration V 2 is the final volume of the solution You can also use the same equation if you know both the initial and final volume and the concentration of only one of the solutions (undiluted or diluted). To calculate your unknown value, simply plug in your known concentrations and volumes into the correct places in the equation and solve for your unknown. (Note the sometimes C 1 V 1 =C 2 V 2 is written as M 1 V 1 =M 2 V 2 instead). Example: You have a 2M (molar = moles/L) sodium acetate solution. To make fresh DCIP, you need 500 ml of a 0.5 M solution of sodium acetate. How do you make a 500 ml solution of 0.5M sodium acetate? Answer : Use C1V1 = C2V2 Since you know the initial concentration (C1 = 2M), the final concentration (C2 = 0.5M), and the final volume (V2 = 500 ml), you can use the formula to find V1: ● (initial concentration)(initial volume) = (final concentration)(final volume) ● (2M)(V1 ml) = (0.5M)(500 ml) ● V1 ml = (0.5M)(500 ml) / 2M ● V1 ml = 125 ml of 2M sodium acetate solution ● Then, to calculate the amount of water needed, use the following formula: ● final volume - initial volume = volume of diluent ● 500 ml total - 125 ml of 2M sodium acetate = 375 ml water We have covered dilutions briefly in class and in the descriptions above, but to complete this assignment, you will need a better understanding of dilutions and get practice calculating them. To do this, you will read the articles and watch the videos below to help you understand the process of setting up dilutions in a mathematical way: READ AND VIEW ● Quansys Biosciences: Dilutions: Explanations and Examples of Common Methods https://www.quansysbio.com/support/dilutions-explanations-and-examples/#:~:text=To%20make%20a%20fi xed%20amount,Final%20volume%20of%20new%20solution ● Dilution Problems, Chemistry, Molarity & Concentration Examples, Formula & Equations (Only watch 0:00-5:33) https://www.youtube.com/watch?v=FPidlCmymVg ● Dilution Factor Equation https://byjus.com/chemistry/dilution-factor-equation/#:~:text=Because%20M1V1,2%20is%20the%20final%2 0volume . This problem set will help you practice dilution calculations as well as understand the applications of the C 1 V 1 = C 2 V 2 equation. Some of the problems below will also require you to convert between metric units, for example, from mL to 4

μL and vice versa. ( Note: You do not have to be familiar with the units of concentration to solve the problems and some problems may require a calculator). Show your work for full credit. PRACTICE: Solve the following problems – show your work for full credit. Add spaces or additional sheets as necessary. 1. You are measuring enzymatic activity using the spectrophotometer, but the absorbance value is too high. To obtain a more accurate reading, you perform a 1:20 dilution using water. If you want a final volume of 1000 μL in your cuvette, how much of your experimental sample must you add and how much water must you add to achieve the desired dilution? Report your answer in both μL and mL units. C1V1 = C2V2 (1)V1 = (0.5)1000 V1 = ((0.5)1000) / 1 = 50µ 1000µ - 50µ = 1950µ You must add 50µ of experimental sample and 1950µ of water. 2. You take 3L of a 5M NaCl stock solution and add 7L of water. What is the concentration of your new NaCl solution? If you converted your unit of volume to mL instead of L would that change the concentration of the new NaCl solution? Write the equation using both L and mL to support your answer. C2V2 = C1V1 C2(7L) = 5M(3L) C2 = (5L(3l)) / 7L = 2.143M 3. You have a stock of streptomycin, an antibiotic, with a concentration of 200 mg/mL. You add water until the concentration of streptomycin is 8 mg/mL. What dilution have you performed? Hint: Think of how many times less concentrated your final concentration is compared to your initial. C1V1 = C2V2 200mg/mL(1mL) = 8mg/mL(V2) V2 = ((1mL)200mg/mL) / 8mg/mL = 25mL 25mL is the final volume solution from 1mL; so dilution factor is 24. 4. After combining 25 mL of HCl and 975 mL of water, you find that your new concentration is 0.4M HCl. What was the initial concentration of the HCl you added and what dilution was performed? C1V1 = C2V2 C1(25mL) = 0.4M(975mL + 25mL) C1(25mL) = 0.4M(1000mL) 5

C1 = (1000mL(0.4M)) / 25mL = 16M 5. You have a 95% EtOH stock solution and want a 70% EtOH solution for sterilization purposes. If you want a final volume of 250 mL, how much 95% EtOH must you add, and how much water must you add? (You can treat percentage here as a concentration.) Report your answer to the nearest mL, μL, and L. C1V1 = C2V2 95%(V1) = 70%(250mL) V1 = (70%(250mL)) / 95% = 184.21mL 250mL - 184.21mL = 65.79mL 6. You perform a 1:50 dilution on colored water during lab and use an initial volume of 25 μL. Assuming the dilution was done correctly, what is your final volume? Report your final volume in both uL and mL. V1 = 25μL V2 = 50 V1/V2 = 1/50 25/V2 = 1/50 V2 = 25(50) = 1250μL 1250μL + 25μL = 1275μL 7. If you dilute penicillin to decrease the concentration from 400 mg/mL to 50 mg/mL, what dilution has been performed? If you perform the same dilution on a sample of amoxicillin that is initially at a concentration of 1600μg/mL what would the final concentration be? Which final concentration is larger, penicillin or amoxicillin? Hint: Think of how many times less concentrated your final concentration is compared to your initial and remember to take into account your units! (400mg/mL) / 50mg/mL = 8:1 1600μL/mL = 1.6mg/mL (1.6mg/mL) / 50mg/mL = 0.032:1 The amoxicillin concentration is larger. 8. You have an initial volume of 50 μL of orange juice and add 1450 μL of water for a final volume of 1500 μL. What dilution was performed? Hint: Think of what proportion of volume the initial is compared to the final. C2 = V1 / V2 C2 = 50 / 1500 C2 = 1/30 = .033 1:30 9. How would you make 400 ml of 1X PBS (phosphate buffered saline) from a stock of 10X PBS? How much stock and how much water will you need? C1V1 = C2V2 (10)V1 = 1(400mL) V1 = ((1)400mL) / 10 = 40mL 400 - 40 = 360mL 6

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(Note: many stock solutions for research are made up in a concentrated form to last for long periods of time and take up less space. 10X = 10 times concentrated, 50X = 50 times concentrated. You can consider “X” like a concentration when solving problems like this.) 7