(Test Bank) Nuclear Medicine and Pet CT

Waterstram-Rich, 8

Nursing

Test Bank

8th Edition

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Test Bank for Nuclear Medicine and Pet CT, The downloadable file consists of 177 pages, and the total size of the file is 1.49 MB. Purchase now only for highly discounted price.

Author(s): Waterstram-Rich

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Nuclear Medicine and Pet CT 8th Edition Waterstram - Rich Test Bank Chapter 01: Mathematics and Statistics Waterstram -Rich: Nuclear Medicine and PET/CT: Technology and Techniques, 8th Edition MULTIPLE CHOICE 1. The radiation intensity from a point source of 99mTc is 9 mR/hr at 3 m from the source. If the distance is changed to 9 m, what is the new radiation intensity? a. 0.5 mR/hr b. 1.0 mR/hr c. 2.0 mR/hr d. 3.0 mR/hr ANS: B The inverse square law results in the radiation intensity quadrupling if the distance from the source is halved, or it results in the radiation intensity decreasing to one fourth of its value if the distance is doubled. REF: p. 4 2. A technologist is sitting near a bone mineral densitometer, which is a point source of x -rays. If the x -ray intensity a t 1 m from the x -ray beam is 0.20 mR/hr, to what distance from the x -ray beam should the technologist move to decrease total weekly exposure to the occupational limit of 2 mR for a 40-hour workweek? a. 0.5 m b. 1.5 m c. 2.0 m d. 2.5 m ANS: C The inverse square law results in the radiation intensity quadrupling if the distance from the source is halved, or it results in the radiation intensity decreasing to one fourth of its value if the distance is doubled. The unit of intensity is mR/hr, but a 40-hour workweek must be taken into consideration when calculating the final answer. REF: p. 4 3. 10 mCi is equal to how many becquerels? a. 370 Bq b. 370 kBq c. 370 MBq d. 370 GBq ANS: C 1 curie = 3.7  1010 disintegrations per second [dps], and in Système International d’Unités (SI units), 1 Bq = 1 dps; therefore, 1 Ci = 37 GBq and 1 mCi = 37 MBq. In this problem, activity in Bq = REF: p. 5 4. 20 mCi is equal to how many becquerels? a. 0.74 GBq b. 0.37 GBq c. 0.54 GBq d. 0.20 GBq ANS: A 1 curie = 3.7  1010 disintegrations per second [dps], and in SI units 1 Bq = 1 dps; therefore, 1 Ci = 37 GBq and 1 mCi = 37 MBq. In this problem, activity in Bq = REF: p. 5 5. The dose equivalent for occupational whole -body exposure is commonly limited to 50 mSv. How many rem is this? a. 5000 rem b. 500 rem c. 50 rem d. 5 rem ANS: D To convert the dose equivalent in the new SI units of 1 cSv into the old unit of rem: Dose equivalent = . REF: p. 6 6. A source of is delivered to the nuclear medicine department calibrated for 100 mCi at 8:00 AM on Monday. If this radioactivity is injected into a patient at noon on Tuesday, what radioactivity will the patient receive? a. 80 mCi b. 85 mCi c. 90 mCi d. 95 mCi ANS: C Solving the differential equation yields the radioactive decay law: A = A 0e–t where A = activity at time t, A 0 = activity at starting time,  = decay constant, t = time since starting time. The decay constant  is the fraction of atoms that decay per (small) time interval, and has units of 1 over time (e.g., 1/hr) or inverse time (hr–1). The typical radioactive decay calculation required in nuclear medicine specifies three of the four variables (A, A0, , t) in the decay equation, requiring that the fourth unknown variable be solved for. REF: p. 10 7. A patient was injected with 131I on Monday at 10:00 AM. On Tuesday at 10:00 AM, the thyroid probe was placed over the thyroid and produced 100,000 counts. On Thursday at 10:00 AM, the probe showed 25,000 counts. What is the effective half -life in this patient’s thyroid? a. 12 hours b. 24 hours c. 48 hours d. 72 hours ANS: B Typically the patient’s organ excretes the radiopharmaceutical with some biologic half -life t B while the radioactivity also decays physically with a physical half -life that is denoted as t P. The counts observed by the gamma camera follow an exponential decay law based on the effective half-life tE, where: or, in a format that is much easier for calculation purposes, . The effective half-life is always less than or equal to the smaller of tP or t B. REF: p. 15 8. A 99Mo-99mTc generator is eluted Monday at 7:00 AM, producing 1.8 Ci of 99mTc in the eluate vial, in 20 ml saline. WhatTvEolSumTBe AofNeKluSatEeLshLoEuRld.bCe OwMithdr awn from the eluate vial into a syringe in order to inject a patient with 20 mCi of 99mTc at 3:00 PM? (Given the half-life of 99mTc is 6.02 hours.) a. 0.56 ml b. 0.66 ml c. 0.76 ml d. 0.86 ml ANS: A The radioactive decay law can be algebraically rearranged (dividing both sides of the decay equation by A 0) as: . After solving for the half -life, the concentration (radioactivity per volume) in the eluate vial is then determined and the volume needed to be withdrawn into the syringe can be calculated from the equation activity = concentration  volume: A = C  V. REF: p. 10 9. If the HVL for some radionuclide in lead is 0.30 mm, what thickness of lead shielding is necessary to reduce the radiation exposure from 8 mR/hr to 1 mR/hr? a. 0.30 mm b. 0.45 mm c. 0.60 mm d. 0.90 mm ANS: D It is common to follow the method used in radioactive decay and define a half -value layer (HVL) as that thickness of material that absorbs 50% of the photons. The HVL is the analog of t /2in radioactive decay. One HVL transmits 50% of the photons. The equat ion of photon attenuation then can be expressed as: I = I 0e–x = I0e–0.693  (x/HVL) and a relationship exists between  and HVL, given by  = 0.693/HVL. Remember the units of x and HVL must be the same, and remember that this equation calculates the transmitted intensity. REF: p. 16 10. The linear attenuation coefficient in lead for 99mTc gamma rays (140 keV) is 23 cm–1. What percentage of these gamma rays will be absorbed by a lead apron that contains 0.60 mm of lead? a. 75% b. 50% c. 25% d. 12.5% ANS: A The intensity of the transmitted radiation is given by: I = I0e–x where  is the linear attenuation coefficient, or the fraction of the beam absorbed in some (very small) thickness x. The linear attenuation coefficient  is the analog of the decay constant  in radioactive decay. The linear attenuation coefficient  depends on the type of absorbing material and the energy of the photons. REF: p. 15 11. A new gamma camera/computer system that uses a new method of calculating cardiac ejection fraction (EF) is installed in a nuclear medicine department. The department decides to calculate EF for the next 25 patients on both the old gamma camera and the new gamma camera before discontinuing the use of the old camera. In the future, if it is desirable to convert t he new EF value to that which would have been obtained on the old gamma camera (e.g., to assess if the patient’s EF had changed), the mathematical analysis to be used is called a(n) a. independent t-test. b. linear regression. c. standard error. d. chi-square. ANS: B The least squares method, or linear regression, calculates the best -fit values for y-intercept ( a) and slope ( b) in the best -fit straight line: y = a + bx. The line of identity is often drawn on regression graphs when the same parameter, here the ejection fraction, is being plotted on both the x and y axes. The line of identity facilitates an evaluation of whether the two methods (the x and y axes) are predicting the same result for ejection fraction. If the two methods produce the same v alue for ejection fraction, then the regression line should be the same as the line of identity. REF: p. 20 KSELLER.C 12. What is the standard deviation of 40,000 counts? a. 4000 counts b. 2000 counts c. 400 counts d. 200 counts ANS: D Counting statistics, meaning the number of counts expected from a sample, follow the Poisson distribution. In the Poisson distribution, the standard deviation (C) for any number of counts (C) is fixed at the square root of C: This fixed definition of standard deviation does not exist in Gaussian distributions; essentially only counting statistics are Poisson. REF: p. 24 13. What is the coefficient of variation of 40,000 counts? a. 2% b. 1% c. 0.5% d. 0.25% ANS: C The standard deviation can be expressed as a percentage of the mean value, which is frequently called the percent standard deviation, or coefficient of variation (CV): TESTBAN OM In this problem, the counts would be substituted for the mean in the equation. REF: p. 25 14. How many counts should be acquired into each pixel of a nuclear medicine flood image if it is desired to be 95% confident that the true count in each pixel is within 1% of the measured counts in each pixel? a. 100,000 counts b. 40,000 counts c. 10,000 counts d. 4000 counts ANS: B Counting statistics, besides being Poisson, are also described by a Gaussian distribution as long as the number of counts is greater than about 30. Hence, given some number of counts C, the standard deviation is automatically known. It is also known that 68% of repeat measures of the sample fall within C ± and 95% of repeat measures of the sample fall within C ± and so on (95% confidence = C ± 2c = C ± ). Sometimes this type of problem is expressed by saying that a certain number of counts are needed to be 95% confident that the true count is within 2% of the measured value. The presence of two different percentage values can seem confusing. The 95% confide nce interval on 10,000 counts is ±200 counts, and this figure of 200 counts represents 2% of 10,000 counts. The general formula expressing the number of counts needed to be “n ” sure that the true answer is within some percent (p) of the measured counts is given by: In this formula, n is replaced with a 1 for 68% confidence, a 2 for 95% confidence, and a 3 for 99% confidence. REF: p. 25 15. A patient’s thyroid is counted with the thyroid probe and produces 8000 counts. Then the patient is removed and background is found to be 2000 counts. The (net counts) ± (standard deviation in the net counts) in this patient is a. 10,000 ± 100 counts. b. 10,000 ± 77 counts. c. 6000 ± 77 counts. d. 6000 ± 100 counts. ANS: D Background counts are a problem in most measurements of counts from a radioactive sample. Background arises from naTtuEraSlTteBrrAesNtrKiaSl EanLdLcEosRm.iCc OsoMurces of radioactivity or from other nearby sources of radioactive material. The sample is usually counted in the presence of some background radiation, yielding a gross count for sample plus background, denoted by the letter C. The sample is then removed from the counter, a nd the background B is counted. The net, or true counts, which represents the sample only, is denoted by N and given by: N = C – B The standard deviation of the net counts is given by: This is an example of the rules for combinations of errors, which states that for addition or subtraction, errors add in quadrature within the square root: Additionally, since we obtain and REF: p. 26 16. The gamma camera seems to be producing erratic results. A 57Co flood source is counted 10 times, producing the following count values: 1000, 975, 1032, 1096, 982, 997, 1012, 1090, 994, 977. What is the chi -square value for these counts? a. 19.3 b. 18.3 c. 17.3 d. 16.3 ANS: C Too little or too much variation in repeat measurements indicates equipment may not be functioning properly. The chi-square test is used to determine an acceptable range of variability in the repeat measurements. The mean of the 10 measurements is determined ( = C/n), and chi-square is calculated as: . REF: p. 26 17. Which expression describes the operation of the gamma camera in question 16? a. Operating properly b. Showing too much variation c. Showing too little variation d. Not enough information to answer the question ANS: B The definition of acceptable chi -square is given by looking up the probability (P) of this chi-square value for the n repeated measures in a table. Too little varia tion is shown by a chi-square smaller than the P = 0.9 value; an amount of variation exactly as expected from the statistical nature of the decay process would yield P = 0.50; and an unacceptably large chi-square would yield a PTvEalSuTe BgrAeaNteKrSthEaLn L0.E10R..SCo,OPM= 0.90–0.10 range is acceptable. REF: p. 26 18. One group of 20 patients is given a drug that is thought to have an effect on kidney function, and another group of 20 patients is given a placebo (i.e., sugar pill, which is known not to have an effect on kidney function). The nuclear medicine gamma camera is used to calculate the glomerular filtration rate (GFR) in these two groups of patients. Wh at would be the proper statistical test to use to test the hypothesis that no difference in GFR exists between these two groups of patients? a. Chi-square test b. Paired t-test c. Independent t-test d. Linear regression analysis ANS: C The t-test is used to test for differences between mean values. The two forms of the t-test are for independent samples and paired samples. For independent t-tests, two independent groups of data are used. The two groups are completely independent. No relations hip or correlation exists between the two groups, typically because the groups represent different patients. The two groups have mean and standard deviations denoted by , SD1, , and SD2. TBANKSELLER.COM REF: p. 28 19. A nuclear medicine test produces a positive test result in only 80 of the 100 patients known to be ill. Similarly, the test produces a negative test result in only 190 of the 200 patients known to be not ill. Which is correct? a. Sensitivity = 80%, specificit y = 95%, accuracy = 90%, prevalence = 33% b. Sensitivity = 90%, specificity = 95%, accuracy = 90%, prevalence = 33% c. Sensitivity = 90%, specificity = 80%, accuracy = 90%, prevalence = 33% d. Sensitivity = 90%, specificity = 80%, accuracy = 70%, prevalence = 33% ANS: A The sensitivity, or true positive fraction (TPF), is the percentage or fraction of ill patients who have a positive test: The specificity, or true negative fraction (TNF), is the percentage or fraction of well patients who have a negative test: TES The prevalence is the fraction or percentage of ill persons in the study population: REF: p. 31 20. One radiologist is known to be a lax reader compared with another radiologist who is known to be a strict reader. In comparing these radiologists, how would one expect their sensitivity and specificity to compare? a. The lax reader would have lower sensitivit y and higher specificity. b. The lax reader would have higher sensitivity and lower specificity. c. The readers would be expected to have the same sensitivity and specificity. d. Not enough information is given to answer the question. ANS: B The sensitivity , or true positive fraction (TPF), is the percentage or fraction of ill patients who have a positive test: The specificity , or true negative fraction (TNF), is the percentage or fraction of well patients who have a negative test: 1/2 2006? 1/2 For an example, data that show a sensitivity of 78%, less than the specificity of 95%, mean that this test does a better job at correctly diagnosing well people than it does at correctly diagnosing ill people. High sensitivity = correctly fin ding ill people; high specificity = correctly finding well people. REF: p. 31 21. A 99Mo-99mTc generator is calibrated for 1.00 Ci of 99Mo at 7:00 AM Monday. The generator is eluted daily Monday through Friday at 8:00 AM, but the workload is especially heavy on Friday so the generator is eluted again at 3:00 PM to obtain more 99mTc. Assuming 100% elution efficiency, how many mCi of 99mTc will be eluted at 3:00 PM Friday? a. 930 mCi b. 710 mCi c. 160 mCi d. 80 mCi ANS: C The 99mTc radioactivity expected to be eluted from the generator can be calculated by knowing three values: (1) the activity of 99Mo in the generator, (2) the time since the last elution of the generator, and (3) the ratio of 99mTc to 99Mo in the generator. Using the equation, A = Ci  e –0.693  (t/65.9 hr), enter in the known values and solve for the unknown. REF: p. 14 22. A radioactive source of 137Cs (t 1/2= 30 y) was calibrated on October 23, 2000, to contain 10 Ci. This is used as a daily accuracy check source in the dose calibrator. Presuming the dose calibrator is working pr opeTrlEyS, wThBaAt aNcKtivSiEtyLsLhoEuRld.tCheOdMose calibrator show on April 23, a. 9.6 Ci b. 8.8 Ci c. 5.4 Ci d. 2.2 Ci ANS: B The radioactive decay law can be algebraically rearranged (dividing both sides of the decay equation by A 0) as A/A 0 = e–0.693(t/t ). REF: p. 11 23. A source of 18F (t1/2 = approximately 2 hr) is noted to contain 3 mCi at noon. What was the radioactivity at 8:00 AM that same day? a. 24 mCi b. 18 mCi c. 12 mCi d. 6 mCi ANS: C The radioactive decay law can be algebraically rearranged (dividing both sides of the decay equation by A 0) as: A/A 0 = e–0.693  (t/t ). REF: p. 11 24. The biological half -life of 131I in a particular patient is 30 days. The physical half-life is 193 hours. If the patient’s thyroid is counted with the thyroid probe detector, what effective half-life will be observed? a. 26 days b. 6.3 days c. 5.1 days d. 1.2 days ANS: B Typically the patient’s organ excretes the radiopharmaceutical with some biological half -life tB while the radioactivity also decays physically with a physical half -life that is denoted as t P. The counts observed by the gamma camera follow an exponential decay law based on the effective half-life tE, where: or in a format that is much easier for calculation purposes, The effective half-life is always less than or equal to the smaller of tP or t B. REF: p. 15 25. A radioactive source decays from 20 mCi to 2.5 mCi in 18 hours. What is the physical half-life? a. 8 hours b. 7 hours c. 6 hours d. 5 hours ANS: C The , which depends on the radioactive material involved, is the time in which the activity is decreased to half its original value. The radioactive decay law may alternatively be expressed as The factor 0.693 is actually ln 2, which is commonly written with three significant figures. Each half-life of radioactive decay causes the activity level to drop by 50%. REF: p. 10 26. A sample shows a count rate of 36,000 cpm during a 3-minute counting period. Express this as the count rate ± standard deviation. a. 36,000 ± 220 cpm b. 36,000 ± 110 cpm c. 12,000 ± 55 cpm d. 12,000 ± 28 cpm ANS: B Counting statistics, meaning the number of counts expected from a sample, follow the Poisson distribution. In the Poisson distribution, the standard deviation (C) for any number of counts (C) is fixed at the square root of C: This fixed definition of standard deviation does not exist in Gaussian distributions; essentially only counting statistics are Poisson. In this proble m, cpm must be calculated first. REF: p. 24 27. A long -lived radioactive source is counted for 1 minute and yields 10,000 counts. If this source is counted immediately again, there is a 95% probability that the result will be in the range: a. 9950 to 10,050 b. 9900 to 10,100 c. 9800 to 10,200 d. 9700 to 10,300 ANS: C Counting statistics, besides being Poisson, are also described by a Gaussian distribution as long as the number of counts is greater than approximately 30. Hence given some number of counts C, the standard deviation is automatically known. It is also known that 68% of repeat measures of the sample fall within C ± and 95% of repeat measures of the sample fall within C ± , and so on: 95% confidence = C ± 2c = C ± . REF: p. 25 28. A 5-ml sample of a standard diluted 1:10,000 produces 27,200 counts in the well counter. A 5-ml sample of patient plasma, counted for the same time as the diluted standard sample, produc es 99,100 counts. What is the plasma volume? a. 10.8 liter b. 2.74 liter c. 1.73 liter d. 0.91 liter ANS: B The dilution principle can also be used to measure an unknown volume. Using the dilution principle, it is possible to estimate a patient’s plasma volume, C 1V1 = C 2V2. Because the standard may be too concentrated, a portion of the standard is diluted and counted while the full-strength undiluted standard can be injected into the patient. The standard can be diluted by adding 1 ml of it to a flask, which is then filled to 500 ml with distilled water. It could then be expressed as a 1:500 diluted standard. The standard counts have to be multiplied by the dilution factor of 500 to obtain the true standard concentration that was injected into the patient. Taking this into consideration, the new equation to be used to calculate the plasma volume would be C1V1 = (DF)C 2V2. REF: p. 5 29. Readers are encouraged to use Microsoft Excel to solve this problem. What is the mean of the following five numbers: 6.40, 7.20, 3.50, 9.20, 5.10? NKSELLER.CO a. 9.01 b. 6.28 c. 5.97 d. 4.83 ANS: B Precision refers to the spread, or range, of data values obtained when some parameter is measured many times or in many patients. A small precision means that little variation exists in the data values. Then, the mean (or average) of the n values of the parameters x is defined by the symbol: The symbol x means the sum of all the measurements of x. REF: p. 22 -23 30. Readers are encouraged to use Microsoft Excel to solve this problem. What is the standard deviation of the following five numbers: 6.40, 7.20, 3.50, 9.20, 5.10? a. 4.02 b. 3.26 c. 2.15 d. 1.92 ANS: C The standard deviation , a measure of the precision of the data, is given by the symbol , defined as: TESTBA M Recall that an exponent of means the same as the square root . The standard deviation is a measure of the deviation or spread of the data points from the mean value. The standard deviation is often expressed in a format of the mean value ± . A large  value indicates data with a large range and, therefore, poor precision. REF: p. 23 31. Readers are encouraged to use Microsoft Excel to solve this problem. What is the coefficient of variation (CV) of the following five numbers: 6.40, 7.20, 3.50, 9.20, 5.10? a. 34% b. 31% c. 28% d. 12% ANS: A The standard deviation is a percentage of the mean value, which is frequently called the percent standard deviation, or coefficient of variation (CV): REF: p. 23 32. Readers are encouraged to use Microsoft Excel to solve this problem. In a n uclear medicine technology training program, the students wonder if their final exam grade in their training program is related to their subsequent board score on the NMTCB exam. The five students attain the following pairs of scores (exam score, board sco re) = (72, 60), (84, 74), (88, 71), (68, 60), (91, 79). What is the linear correlation coefficient, r, between exam score and board score? a. 0.99 b. 0.95 c. 0.86 d. 0.75 ANS: B The value of goodness -of-fit parameters helps determine whether it is reasonable that the data points are truly represented by a straight line. The linear correlation coefficient, r, is calculated from the sums used to find the intercept and slope, along w ith one additional sum from the data points ( y2): The sign of the correlationTcEoeSffTicBieAnNt iKs SmEerLeLlyEthRe.sCamOeMas the sign of t he slope. REF: p. 21 33. Using Table 1 -9 in the book, what is the P value for the linear correlation coefficient r in question 31? a. P = 0.001 b. 0.01 < P < 0.05 c. 0.05 < P < 0.1 d. P > 0.10 ANS: B A correlation r = 1 is a perfect fit, with the straight line passing exactly through each data point. The closer r is to 1, the more highly correlated are the x and y data. A chance, or probability ( P), always exists that data that are truly not linearly related to each o ther will randomly appear in a fairly linear fashion. The statistical significance of correlation values less than 1 must be evaluated from statistical tables of probability. REF: p. 21 34. Using Table 1 -9 in the book, does a linear correlation coefficient r = 0.95 for n = 5 data points indicate a statistically significant linear relationship between the x and y variables? a. Yes b. No c. There is not enough information to answer the question. d. None ANS: A A correlation r = 1 is a perfect fit, with the straight line passing exactly through each data point. The closer r is to 1, the more highly correlated are the x and y data. A chance, or probability ( P), always exists that data that are truly not linearly related to each other will randomly appear in a fairly linear fashion. The statistical significance of correlation values less than 1 must be evaluated from statistical tables of probability. REF: p. 21 35. Readers are encouraged to use Microsoft Excel to solve this problem. What is the slope and intercept in the regression equation for problem 31? (board score) = intercept + (slope)  (exam score) a. Intercept = 7.15, slope = 0.502 b. Intercept = 6.15, slope = 0.602 c. Intercept = 5.15, slope = 0.702 d. Intercept = 4.15, slope = 0.802 ANS: D The intercept and slope are calculated REF: p. 20 36. Using the regression equation in Problem 31, if your training program exam score was 84, what does the regression equation predict for your board score? a. 92 b. 82 c. 72 d. 62 ANS: C The standard deviation can be expressed as a percentage of the mean value, which is frequently called the percent standard deviation, or coefficient of variation (CV): In this problem, the score would be substituted for the mean in the equation. REF: p. 23 37. A thyroid uptake dose of 10 Ci is equal to how much radioactivity in SI units? a. 37 kBq b. 370 kBq c. 3.7 MBq d. 37 MBq ANS: B The curie = 3.7  1010 disintegrations per second [dps], and the Bq =1 dps; therefore, 1 Ci = 37 GBq and 1 mCi = 37 MBq and 1 Ci = 37 kBq. REF: p. 5-6 38. What is the proper number of significant figures for the quotient (6.1  10−2)/(0.1232)? a. 0.49513 b. 0.4951 c. 0.495 d. 0.50 ANS: D The exponent on the 10 specifies how many places the decimal point in the number is to be shifted to the left (for negative exponents) or shifted to the right (for positive exponents). In addition, the accuracy of the result in multiplication or division is such that the product or quotient has the number of significant figures equal to that of the term with the smaller number of significant figures. REF: p. 2 39. What is the proper number of decimal places in the sum of (0.3) + (3.5264)? a. 3.8264 b. 3.826 c. 3.83 d. 3.8 ANS: D For addition and subtra ctioTnEthSeTfBinAalNreKsSulEt LhaLsEthRe.saCmOeMnumb er of signifi cant decimal pla ces (rather than significant figures) as the number in the problem with the least number of significant decimal places. REF: p. 9 40. It is known that 75% of a radionuclide decays away in 12 hours. If starting with 10 mCi, how much will be left after 6 hours? a. 7.5 mCi b. 5 mCi c. 2.5 mCi d. 1.25 mCi ANS: B Calculations in nuclear medicine often involve expressing a mathematical concept as a ratio. The equation generally contains several numbers and one unknown value; here the unknown value is (x). The object is to solve for the unknown value by rearranging t he terms in the equation. In this situation, the half -life of the material is t hours. Start with 10 mCi. After t hours, there is 5 mCi left. After another t hours, there is 2.5 mCi left. Starting with 10 mCi, after 2 t hours, there is 2.5 mCi left, which i s only 25% of the initial quantity, and so 75% decayed away. It is given that 75% decays away in 12 hours. So 2 t = 12 hours and t = 6 hours. With a half-life of 6 hours, starting with 10 mCi, there would be 5 mCi left after 6 hours. REF: p. 3 41. Convert 5 mBq to kBq and GBq. a. 50 kBq and 0.5 GBq b. 500 kBq and 0.05 GBq c. 5000 kBq and 0.005 GBq d. 5  104 and 5  10−4 GBq ANS: C The decimal point can be shifted left if the exponent is increased by 1 for each left shift and the decimal can be shifted to the right if the exponent is decreased. REF: p. 3 42. 42. ANS: D Fractions, such as , consist of a numerator (1) that is to be divided by a denominator (3). When using fractions to perform the calculations, a common denominator must be used. The math ematic al manipul ationToEfSfrTaBctAioNnsKrSeqEuLirLesEcRa.reCaOs Mto the number of digits and pla cement of the decimal point. The safest way to handle the mathematical manipulation of fractions is to simply use the power of the pocket calculator to perform calculations such as 1/3 + 9/4 by first converting the fractions to decimals and then carrying out the other arithmetic. REF: p. 3 43. 30% of 7 ml = a. 0.23 ml. b. 2.10 ml. c. 2.30 ml. d. 4.90 ml. ANS: B Percentages are values expressed as a fraction of some whole, entire value: 75% of some number is the same as 0.75 multiplied by that number. REF: p. 3 44. A patient may have had a kidney function te st last month that showed a kidney clearance rate of 45 milliliters per minute (ml/min). The patient returns today and has a kidney clearance rate of 60 ml/min. What is the percent increase in kidney function? a. 7% b. 14% a. b. c. d. c. 25% d. 33% ANS: D Percentages are often used to express percentage change between two values: % change = [(new value – old value)/old value]  100. REF: p. 3 45. An intensity setting of 250 on a display screen may produce a suitable image for a bar phantom with 500K counts in the acquisition. Now it is desired to change to 350K counts in the image. The intensity on the screen is known to change linearly with the in tensity control setting. What intensity setting should be used for the new 350K -count image? a. 100,000 b. 150,000 c. 175,000 d. 200,000 ANS: C Calculations in nuclear medicine often involve expressing a mathematical concept as a ratio. The equation generally contains several numbers and one unknown value; here the new intensity setting is the unknown value (x). The object is to solve for the unknown value by rearranging the terms in the equation. Cross -multiplying is a technique used to solve a proportion . REF: p. 3 46. The morning elution of the 99Mo-99mTc generator yields 850 mCi of 99mTc radioactivity in 20 ml of saline eluate. What volume should be withdrawn from the eluate vial into a patient syringe to immediately perform a 25 -mCi patient scan? a. 0.2 ml b. 0.30 ml c. 0.59 ml d. 1.7 ml ANS: C Calculations in nuclear medicine often involve expressing a mathematical concept as a ratio. The equation generally contains several numbers and one unknown value; here the unknown value (x). The object is to solve for the unknown value by rearranging the terms in the equation. Cross -multiplying is a technique used to solve a proportion . REF: p. 3 Chapter 02: Cell and Molecular Biology Waterstram -Rich: Nuclear Medicine and PET/CT: Technology and Techniques, 8th Edition MULTIPLE CHOICE 1. Which of the following is an organelle of a eukaryotic cell? a. Photolipid b. Cisternae c. Centrosome d. Endocyte ANS: C The centrosome is the main site of microtubule assembly in the cell and is therefore referred to as the microtubule organizing center. REF: p. 43 2. The large organelle often referred to as “the control center” of the cell is the a. lysosome. b. nucleus. c. centrosome. d. Golgi body. ANS: B The nucleus is the large organelle that is often referred to as “the control center” of the cell. REF: p. 43 3. An organelle composed of a network of interconnected, closed, membrane -bounded vesicles that extend from the nucleus to the cell membrane is the a. Golgi body. b. endoplasmic reticulum. c. centrosome. d. lysosome. ANS: B The endoplasmic reticulum (ER) is an organelle composed of a network of interconnected, closed, membrane -bounded vesicles that extend from the nucleus to the cell membrane. REF: p. 42 4. Phagocytosis, endocytosis, and autophagy are processes that are performed by the a. Golgi bodies. b. ribosomes. c. nucleus. d. lysosomes. ANS: D Lysosomes degrade extracellular materials taken up by the cell through three main processes; phagocytosis, endocytosis, and autophagy. REF: p. 42 -43 5. Messenger RNA (mRNA) is converted into a protein within the of eukaryotic cell. a. nucleus b. mitochondria c. ribosome d. Golgi body ANS: C The process of converting (translating) mRNA into protein is carried out by ribosomes. REF: p. 42 6. Which term denotes the process by which a gene is transformed into messenger RNA (mRNA)? a. Transcription b. Initiation c. Translation d. Synthesis ANS: A Transcription is the process by which a gene is transcribed into mRNA. The three steps of this process are initiation, elongation, and termination. REF: p. 45 7. The three steps of the transcription process are a. initiation, elongation, and termination. b. interphase, prophase, and telophase. c. expression, translation, and suppression. d. synthesis, regulation, and termination. ANS: A Transcription is the process by which a gene is transcribed into mRNA. The three steps of this process are initiation, elongation, and termination. REF: p. 45 8. The process by which a gene is used as a template to transcribe messenger RNA (mRNA) is termed: a. gene mutation. b. gene expression. c. regulation. d. post-translation modification. ANS: B Gene expression is often described as the process by which a gene is used as a template to produce (transcribe) mRNA. REF: p. 45 9. All of the following are steps in gene expression, EXCEPT a. transcription. b. translation. c. termination. d. post-transcriptional modifications. ANS: C Gene expression is an extremely complex, multistep, tightly regulated process that involves hundreds of proteins and other molecules. Steps include the mechanisms of transcription, post-transcriptional modifications, translation, and post-translational modification of proteins. REF: p. 45 10. The phases of mitosis are a. interphase, metaphase, synthesis, and replication. b. telophase, mitosis, interphase, and S-phase. c. G0, G1, interphase, and S-phase. d. prophase, metaphase, anaphase, and telophase. ANS: D Mitosis is divided into four separate steps, prophase, metaphase, anaphase, and telophase, that are summarized in Table 2.1. REF: p. 48 11. CDK2 is the partner for both cyclins: a. A and E. b. D and E. c. A and B. d. D and B. ANS: A Cyclin E binds to CDK2 at the end of G 1. This interaction is required for the cell to commit to DNA replication and advances the cell into S -phase. Cyclin A binds to CDK2 during S -phase and this interaction is requ ired to initiate the G 2 to M transition. See Table 2 -2. REF: p. 49 12. Which are the three main cell cycle checkpoints? a. G1, restriction, and metaphase/anaphase b. G1, G2/M, and mitotic spindle c. G1, G2/M, and anaphase/telophase d. Restriction, DMA damage, and anaphase/telophase ANS: B In addition to the complicated series of steps and layers of regulation discussed above, there are three main cell cycle checkpoints: the G1 (or restriction checkpoint), the G2/M checkpoint (or the DNA damage checkpoint), and the mitotic spindle checkpoint (also called the metaphase/anaphase checkpoint). These control mechanisms can detect errors that may have occurred (e.g., DNA replication errors or chromosomes that did not segregate proper ly) and halt progression of the cell cycle at specific points, giving the cells an opportunity to correct any errors before they are passed on to the daughter cells. REF: p. 49 13. All of the following are the groups of genes associated with cancer, EXCEPT a. proto-oncogenes. b. angiosynthesis genes. c. tumor suppressor genes. d. DNA repair genes. ANS: B The genes associated with cancer can be classified into three groups. (1) Proto -oncogenes are genes that produce proteins that promote cell division (e.g., growth factors) or inhibit apoptosis. The mutated forms of these genes are called oncogenes and will promote unregulated cell division. (2) Tumor suppressor genes produce proteins that normally prevent cell division or promote cell death. Mutations that destroy production of tumor suppressor proteins therefore promote cell proliferation. (3) DNA repair g enes produce proteins that prevent or correct mutations in DNA. REF: p. 50 -51 14. Which t erm describes a groTuEpSoTf dBiAseNasKeSs iEnLwLhEicRh.ceCllOsMno long er respond to the norm al constraints placed on their proliferation and instead grow indefinitely? a. Metastasis b. Angiogenesis c. Cancer d. Mutation ANS: C Cancer is a term used to describe a group of diseases in which cells no longer respond to the normal constraints placed on their proliferation and instead grow inde finitely. REF: p. 50 15. All of the following are viruses linked to cancer, EXCEPT a. influenza A. b. hepatitis B. c. human papilloma virus (HPV). d. Epstein -Barr. ANS: A Influenza A is not linked as being a cancer -causing virus. REF: p. 52 -53 Chapter 03: Chemistry/Biochemistry Waterstram -Rich: Nuclear Medicine and PET/CT: Technology and Techniques, 8th Edition MULTIPLE CHOICE 1. Titrations are procedures in which a. an accurately measured but unknown volume of solution must be determined. b. an accurate volume of a known substance must be determined. c. an accurate volume of a ligand must be determined and measured. d. an accurate volume is used to determine an unknown redox reaction. ANS: A Procedures such as titrations, in which an accurately measured but unknown volume of solution must be determined, are conveniently performed by use of a buret . Titrations are performed by taking an aliquot (V2, a precisely measured volume) of the solution containing an unknown quantity ( N2) of acid or base and placing it in a titration vessel, usually an Erlenmeyer flask. A standard solution of acid or base, whose concentration is precisely known, is then carefully delivered into the titration flask by use of a buret. REF: p. 69 2. The physical properties of a substance may be affected by a change in conditions. If this occurs, when the substance returns to its original condition, the physical properties a. result in the formation of one or more “new” substances. b. will be the same as theTinEitSiaTl BprAoNpeKrtSieEs.LLER.COM c. will still be different because they cannot have their change reversed. d. result in a change of physical state. ANS: B A substance that has undergone a change in physical properties because of a change in conditions will exhibit the initial properties when the substance is returned to its original condition. REF: p. 57 3. The chemical properties of a substance may be affected by a change in conditions. If this occurs, the change will a. result in the formation of one or more “new” substances. b. exhibit the initial properties of the substance. c. always be irreversible. d. result in a change of the physical state of the substance. ANS: A A chemical change always results in the formation of one or more “new” substances, each of which has unique chemical and physical properties. REF: p. 57 4. The atomic number of an element is defined as the a. number of neutrons in an atom. b. atomic mass of an atom. c. number of protons in an atom. d. the number of protons minus the number of neutrons. ANS: C The number above a chemical symbol is the atomic number (Z number) of the element, and the number below a chemical symbol is the average atomic mass ( A number) of the element. Z number indicates the number of protons. REF: p. 58 5. The Pauli exclusion principle states that a. no electrons can be in the nucleus. b. no two electrons can have the same set of quantum numbers. c. only one electron can occupy the same orbital. d. no electron can be at a subenergy level. ANS: B Only two electrons may occupy the same orbital and each must have opposite spins; no two electrons can have the same set of quantum numbers. REF: p. 58 6. In the chemical symbol ZXA , the A number is a. the number of protons and neutrons. b. the number of protons and electrons. c. the numb er of protons TmEinSuTs tBhAe NnuKmSbEerLoLfEelRec.tCroOnsM. d. the number of neutrons minus the number of protons. ANS: B The atomic mass minus the atomic number gives the number of neutrons present in the nucleus. The atomic mass is the sum of the protons and neutrons. REF: p. 60 7. Atoms that differ only in the number of neutrons present in the nucleus are called a. isomers. b. isotones. c. isotopes. d. isomorphs. ANS: C Atoms that differ only in the number of neutrons present in the nucleus are called isotopes. REF: p. 61 8. On the periodic table, there is a family of elements that are not very reactive but do form compounds under special conditions. The elements in this family are referred to as a. noble gases. b. transition metals. c. alkaline metals. d. radioactive. ANS: A Certain elements are unreactive and will not combine with other elements to give compounds; these elements are all members of the eighth group shown in the periodic table and constitute a family. They all exist in the gaseous state under normal conditions and, being inert, are commonly referred to as the noble gases. REF: p. 61 9. A compound, regardless of its origin or method of preparation, always contains the same elements in the same proportions by weight; this is called a. the law of constant composition. b. the law of multiple proportions. c. a complex ion. d. a ligand. ANS: A The fact that elements generally react with other elements to attain a noble gas electron configuration leads to the conclusion that any two elements that undergo reaction must do so in a definite ratio of atoms, that is, one Na to one F and one C to four Cl. It then follows that the elements must also react to definite ratios by weight. This is stated by the law of constant composition: a compound, regardless of its origin or method of preparation, always contains the same elements in the same proportions by weight. REF: p. 63 10. Avogadro’s number is a. the ratio of the weight of each element in a compound. b. equal to the atomic mass of an element. c. the number of atoms or molecules in 1 gaw or gmw. d. the electron configuration of an element. ANS: C One gram molecular weight (1 gmw) of any two substances will contain the same number of atoms or molecules. The number of atoms or molecules in 1 gram atom weight (1 gaw) or gmw has been shown by experiment to b e 6.02  1023. This number is called Avogadro’s number, and 1 gaw or gmw of any substance is commonly referred to as 1 mole. REF: p. 64 11. A solution is a homogeneous mixture a. of two substances. b. of particles and a solute. c. of particles and a solvent. d. of a substance and a ligand. ANS: A A solution by definition is a homogeneous mixture of two or more substances. The solute can be a gas, liquid, or solid, and the solvent is generally a liquid. REF: p. 64 -65 12. A colloid is a type of mixture in which a. the particles are aerosolized. b. the particles settle and result in a layered mixture. c. the particles remain suspended in the dispersing medium indefinitely. d. all of the particles are of equal size and shape. ANS: C Dispersed colloid particles remain suspended in the dispersing medium indefinitely; the behavior is attributed to a constant bombardment of the dispersed particles by the molecules of the dispersing medium; thus, the colloid particles are in constant motio n. The phenomenon is called the Brownian movement . REF: p. 66 13. The products of a neutralization reaction are a. an acid and a base. b. salt and water. c. a chelate and a ligand. d. a standard solution. ANS: B The process of neutralization involves the reaction of acids with bases as defined by the Arrhenius concept: H+ or and OH–. The products of neutralization are invariably a salt and water. If equal quantities (as defined by normality of an acid and base) are mixed, the resulting solution is neither acidic nor basic. REF: p. 69 14. The equation used for calculation of the quantity of an acid or base in a given sample is a. C1V1 = C2V2. b. V1N1 = V2N2. c. C V 2 = C V 2. 1 1 2 2 d. V N 2 = V N 2. 1 1 2 2 ANS: B This is a useful relationship for calculation of the quantity of an acid or base: V1 = Volume of the standard solution (prepared in the laboratory) N1 = Normality of the standard solution V2 = Volume of the unknown solution N2 = Normality of the unknown solution REF: p. 69 15. A characterization of a weak acid is that it a. completely dissociates in water. b. does not completely dissociate in water. c. needs a buffer water. d. neutralizes water. ANS: B