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LABORATORY MATH FINAL EXAM ANSWER KEY

1. Give the whole number or fraction that corresponds to these exponential expressions. a. 25-2 = _1/625_ b. 92 = 81_ 2. Perform the operations indicated: a. 100 - 4(5 X 5) + 10 = 5 2 b. 12.2 pounds X 14.4 pounds = 176 pounds2 c. 5 X 102 cm / 2.1 cm = 238 d. (111 X 10 -3 cm) (20.9 X 102 cm2) = 232 cm3 / 3.00 = 77.3 cm3 3.00 e. (111 X 10 -3 cm) - (20.9 X 102 cm2) = 3.00 0.111cm – 2090 cm2 = 0.037 cm – 697 cm2 3

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3. Circle the larger in each pair: a. An object that is 22 mm or an object that is 220 cm b. An object that is 10,000 µm or an object that is 100 nm c. A volume that is 20 µL or a volume that is 2 dL d. 250 X 10 -4 µL or 2500 X 10 -3 µL 4. Fill in the blanks so that the numbers on both sides of the = sign are equal. For example: 2.58 X 10-2 = 25.8 X 10 -3 0.020 X 10 -5 = 0.0020 X 10_-4_ = 0.20 X 10-6_ = 200 X 10_-9___ 5. Give the common log of the following numbers without using a calculator. a. 100 __2___ b. 0.01 __-2__ c. 1000,000,000 __9__ 6. The log of 567 must lie between 2.0 and 3.0. For each of the following numbers, state the values the log must lie between. (Don't use a calculator to come up with your answers, but try checking your answers with a calculator.) a. 0.25 b. 236.8 0.1 < 0.25 < 1 100 < 236.8< 1000 -1 < log 0.25 < 0 2 < log 236.8 < 3 7. Use a scientific calculator to find the antilog of the following numbers. a. 2.0 _100__ b. 25 X 10 -3 _1.06

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8. Use a scientific calculator to convert each of the following [H+] to a pH value. a. [0.25 M] 0.6 b. [0.035 M] 1.46 pH = -log[H+] = -log(0.25) = 0.6 pH = 1.46 9. The cost of a drug is $200 for the first 100 mg and $1 for every additional 1 mg. a. How much would 2.5 grams cost? Write an equation to describe this situation and calculate the answer. = $200 + $1/ mg X (2500mg – 100 mg) = $200 + $2400 = $2600 b. If the total cost of an order is $ 800, how many mg were ordered? Write an equation to describe this situation and calculate the answer. $800 = $200 + $1/mg X (Y – 100 mg) $600 = $1/mg Y - $100 $700 = $1/mg Y 700 mg = Y

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10. (6 pts) Solve for ?. a. ? = 1.00 b. 5 mg = 1 mg 0.2 0.50 25 mL ? ? = 0.4 ? = 5 mL 11. In the following problems, you must show all the steps to get full credit. First, set up the proportion equation (with the proper units) Second, cross multiply and divide to solve for the unknown. a. If a cake requires 0.5 tsp of salt, how much salt is required to make two cakes? 0.5 tsp / 1 cake = X / 2 cakes X = 1.0 tsp b. 10 grams NaCl are required to make 250 mL of a salt solution. How much NaCl is needed to make 5.0 L of this solution? 10 g NaCl / 0.25 L = X / 5.0 L X = 200 g NaCl c. The components required to make 1 L of a solution are listed as follows: Prepare a table that shows how to prepare 0.25 liters of this solution. COMPONENT GRAMS NaCl 10.0 Na azide 0.001 10 g NaCl / 1 L (0.25 L) = 2.5 g NaCl 0.001 g azide / 1 L (0.25 L) = 0.00025 g azide

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12. Express each of the following per cents as fractions AND as decimals. a. 25% _25/100, 0.25____ b. 0.0005 % __0.0005/100; 0.000005 13. Express the following fractions as percents. a. 3/5 __60%__ b. 5/35 __14.3%__ 14. State the following as percents: a. 3 parts in 100 total parts b. 35 parts in (25 parts + 60 parts) 3% 35/85 = X/100 X = 41% 15. How much ethanol is present in 500 mL of a 3.5 % (v/v) solution of ethanol? 3.5 mL ethanol / 100 mL solution = X / 500 mL solution X = 17.5 mL ethanol 16. A laboratory disinfectant solution is made of water and ethanol. How could one prepare 5 L of a 30% disinfectant solution? 30 mL ethanol / 100 mL solution = X / 5000 mL solution X = 1500 mL ethanol BTV 5 L with water 17. A veterinarian orders for Billy the goat 100 mL of 2.5% (w/v) dextrose in sterile water for injection (WFI). You have on hand 1000 mL of 25% (w/v) dextrose. How should you prepare this solution? Dilute 1/10 1/10 = X/100 X = 10 mL Take 10 mL of 25 % stock and BTV 100 mL with WFI

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18. You are about to begin crucial molecular biology experiments involving cloned sheep. The media for storing eggs requires 0.1% (v/v) glycerol in highly purified water. You have on hand 30% glycerol and purified water. How should you proceed? Dilute stock 0.1 to 30 with WFI = 1/300 Take 1 mL 30% glycerol and BTV 300 mL with WFI 19. You need to prepare 0.9% (w/v) NaCl solution to use for preparing medications. You have a container with NaCl crystals and you have WFI (water for injection) in which to dissolve the NaCl. The molecular weight of NaCl is 58.45. You need to make 1500 mL in WFI. How should you proceed? 0.9 g NaCL / 100 mL = X / 1500 mL X = Take 13.5 g NaCl and BTV 1500 mL 20. You are working in a cancer research laboratory and you need to prepare media in which to grow cancer cells isolated from patients. The media requires 0.1 % antibiotic to keep bacteria from contaminating your cancer cells. The antibiotic comes as a powder. You need to make 5 liters of media to last a few weeks. How would you proceed? 0.1 g / 100 mL media = X / 5000 mL media X = Take 5 g powder and BTV 5 L 21. The density of glycerol at 20 oC is 1.26 g/mL. What is the volume of 150.0 g of glycerol? 1.26 g / mL = 150 g / X X = 119 mL 22. 26 weeks = 182 days = 4368 hours = 262080minutes = 15724800 seconds

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23. 0.55 km = 550 m = 55,000 cm = 550,000 mm = 550,000,000 µm=550,000,000,000 nm Relationships (These units relate to the decay of radioactivity.) 1 Ci = 3.7 X 1010 dps (disintegrations per second) 1 Ci = 1000 mCi = 103 mCi 1 mCi = 1000 µCi = 103 µCi 1 Bq (Becquerel) = 1 dps 24. 2.7 X 10-8 Ci = 2.7 x 20-5 mCi = 103 dps = 60,000 dpm (disintegrations per minute) 25. Suppose a particular enzyme must be added to a nutrient solution used to grow bacteria. The enzyme comes as a freeze-dried powder. The manufacturer of the enzyme states that every gram of enzyme powder actually contains only 950 mg of enzyme. The rest is an inert filler that has no effect. If a recipe calls for 10.0 ounces of this enzyme for every 20.0 liters of broth and if you prepare 1500.0 liters of broth, how much of the enzyme powder will you need to add? Remember to compensate for the inert filler. You need 10 oz / 20 L X 1500 L = 750 oz enzyme 750 oz enzyme / X = 1 oz / 28 g X = 21,000 g enzyme X / 21,000 g enzyme = 1 g powder / 0.95 g enzyme X= 22,105 g powder

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26. A study suggests that disposable diapers may pick up as much as 25 mg/kg of dioxin, that is a byproduct of a bleaching process. If each diaper weighs 2 ounces, and a baby wears 9 diapers per day, how many grams of dioxin may a baby be exposed to from its diapers in a year? Per day: 18 oz / X = 1 oz / 28 g X = 504 g diapers or 0.504 kg diapers Dose per day = 25 mg dioxin / kg diaper (0.504 kg diaper) = 12.6 mg dioxin Dose per year= 12.6 mg dioxin / day X 365 days = 4599 mg dioxin / year 27. You have 2000 µL of a drug solution with a concentration of 15 mg/L. How much drug do you have? 15 mg / 1000 mL X 2 mL = 0.03 mg drug 28. Suppose you dilute 10 mL of concentrated Kool Aid with 1 L of water.

a. Express this dilution with a ":" 10 mL:1000mL b. Express this dilution with a “/” 10mL:1010mL

29. a. How would you prepare 25 mL of a 1/15 dilution of apple juice? 1/15 = X/25 mL X = Take 1.7 mL apple juice BTV 25 mL b. How would you prepare 5 mL of a 1/75 dilution of blood? X / 5mL = 1/75 X = Take 0.067 mL blood and BTV 5 mL

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30. You have a diluted drug solution. The concentration of drug in this solution is 2.5 mg/L. The drug was diluted by taking 0.1 mL of drug stock solution and adding 99.9 mL of saline solution. What was the concentration of drug in the original, undiluted stock solution? Dilution = 0.1/100 or 1/1000 Original conc. = 2.5 mg / L X 1000 = 2.5 g drug / L 31. You have 200 µL of a protein stock solution. This protein stock solution is diluted by taking 1 µL of the sample and adding 99 µL of buffer. A protein assay was performed on the diluted sample. The concentration of protein in the diluted sample was 30 µg/mL. a. Draw a sketch of this. b. What was the concentration of protein in the original, undiluted sample? Dilution = 1/100 Original conc. = 30 µg / mL X 100 = 3000 µg / mL c. How much protein was present altogether in the 200 µL of original, undiluted sample? 3000 µg / mL X 0. 2 mL = 600 µg protein 32. What is the slope and the Y intercept for each of these equations? a. Y = 2 mg/L X + 3.5 mg b. Y = ( 5.5 /sec) X slope = 2 mg/ mL slope = 5.5 / sec intercept = 3.5 mg intercept = 0

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33. a. What are the slope and Y intercept for the line plotted on this graph? m=1.8 mg / sec y = 0 b. What is the equation for this line? y = 1.8 mg/sec X

ENZYME PRODUCT FORMED

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0 2 4 6 8 10 12Time (in sec.)

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34. A biotechnology company is testing a drug to cure Scrapie in sheep. They test different dosages of the drug in sheep and find the following results. a. Graph the data that follow. Remember to label the axes and the graph completely. b. What do the data tell us about the drug dose? Explain your graph in words. Initially no effect, then linear dose effect, then drops off c. For the linear part of the graph, what is the slope? Remember that slope has units. Explain in your own words, what does the slope mean? Slope = 2 effect units / mg/kg Means You get 2 effect units per every mg/kg increase in dose Dose of Drug in mg/kg Effect in Sheep 10 10 20 10 25 9 30 20 35 71

40 82 45 89 50 98 55 110

60 121 65 114 70 111

75 109

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35. a. Graph the data in the table below. Label your graph completely. b. Draw the line that best fits the values.

c. What is the equation for the line? _________y = 0.025 / mg/mL or y = 0.025 mL / mg_______ CONCENTRATION OF STANDARD INSTRUMENT RESPONSE (in mg/mL) 0 0.00 2.0 0.05 4.0 0.09 6.0 0.14 8.0 0.17 10.0 0.27 20.0 0.53 40.0 1.04

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36. The half-life of the radioisotope 22Na is 2.6 years. If there are initially 75 µCi of this radioisotope, how much will be left after: a. 0.55 years b. 45 years (1/2))(.55/2.6)(75 µCi) = 64.8 µCi (1/2))(45/2.6)(75 µCi) = 0.00046 µCi