Calculation Session. ACB Trainees Meeting FOCUS 2013. Model Answers
1. Calculate the number of mmols of sodium reabsorbed in 24 hours by the kidneys of a healthy 70 Kg man. Assume a daily urinary sodium excretion of 100 mmols. Clearly state the assumptions you make.
Assuming a glomerular filtration rate (GFR) of 120 mL/minute for an average adult with normal renal function then the number of litres of plasma filtered per 24 hours (1440 minutes) can be approximated as follows: -
Volume Filtered = (mL/minute) x 1440 = 120 x 1440 = 172.8 L
1000 1000
If the average sodium concentration of plasma is 140 mmol/L, then the approximate amount filtered per 24h is given by: -
140 x 172.8 = 24,192 mmol/24 hours
You are given that the daily sodium excretion is 100 mmol therefore the amount reabsorbed =
24,192 - 100 = 24,092 mmol/L
2. A new test has been developed to screen for a disease with a prevalence of 1 in 500,000 of the population. It has a diagnostic sensitivity of 99.8% and a specificity 97%. Calculate the diagnostic efficiency, and the negative and positive predictive values of the test.
Prevalence of the disease is 1 in 500,000 therefore in a population of 50 million you would expect to find 100 cases.
The test has a specificity of 97% which means that 97% of the population without disease (49,999,900) would have negative results and 3% false positive results
The test has a sensitivity of 99.8% and therefore positive in 99.8% of the disease affected population.
Set up a contingency table: -
Diseased / Non-diseased / TotalsPositive test / 99.8 (TP) / 1,499,997 (FP) / 1,500,096.8
Negative Test / 0.2 (FN) / 48,499,903 (TN) / 48,499,903.2
100 / 49,999,900 / 50,000,000
Diagnostic efficiency is a measure of the number of times the correct test result is obtained: -
Efficiency =(TP + TN)/(TP + TN + FN +TN) %=[(99.8 + 48,499,903)/50,000,000)] x100= 97.0%
Positive Predictive Value = TP/(TP + FP) = 99.8/(99.8 + 1,499.997) = 6.6 x 10-3
Negative Predictive value = TN/(TN + FN) = 48,499,903/ (48,499,903 + 0.2) = 0.999
3. A screening test for a lethal, but easily curable, disease with a population prevalence of 0.002 has a diagnostic sensitivity of 95% and specificity of 99%. Calculate the negative and positive predictive values and diagnostic efficiency of the test for a population of 1 million. The cost of this test is £1.00 per patient and the cost of the confirmatory test is £10.00 per patient. What would be the impact on total cost of provision of screening and confirmatory testing if a new screening test is adopted which has a sensitivity of 99% and specificity of 97%, but costs £2.00 per test.
Test +ve / Test -ve / TotalsDiseased / 1900 TP / 100 FN / 2000
No Disease / 9980 FP / 988020 TN / 998000
Totals / 11880 / 988120 / 1000000
TP = True positive, TN = True Negative, FP = False Positive , FN = False Negative
Negative Predictive Value = TN/(TN+FN) = 988020/(988020+100) = 0.9988
Positive Predictive Value = TP/(TP+FP) = 1900/(1900+9980) = 0.1599
Efficiency = (TP+TN)/(TP+TN+FP+FN) = ((1900+988020)/1000000) x 100 = 98.99%
Scenario 1 costs
Cost of testing = cost of screening test Plus cost of confirmatory tests on screen positives.
Cost of testing = (1000000 @ £1.00) + (11,880 @ £10:00) = £1,118,800.00
If the screening test sensitivity changes to 99% with a specificity of 97% then the following contingency table applies: -
Test +ve / Test -ve / TotalsDiseased / 1980 TP / 20 FN / 2000
No Disease / 29940 FP / 968060 TN / 998000
Totals / 31920 / 968080 / 1000000
Scenario 2 costs
Cost of testing = cost of screening test Plus cost of confirmatory tests on screen positives.
Cost of testing = (1000000 @ £2.00) + (319,200 @ £10:00) = £2,319,200
Cost per true positive identified = £588.84 in scenario 1
Cost per true positive identified = £1171.31 in scenario 2
4. You are required to produce a working glucose (MW 180) standard solution of 8.33 mmol/L, from a commercial stock solution of 9.01g/L in saturated benzoic acid (5g/L).
a. What volume of stock solution would you make up to a final volume of 500 mL to produce the required standard?
b. What would be the concentration of benzoic acid in the resulting standard solution?
Standard glucose concentration required = 8.33 mmol/L
Stock standard = 9.01 g/L glucose in benzoic acid 5.0 g/L
Molar concentration of glucose stock = Mass in g/Molecular Wt glucose = 9.01/180 = 0.050 M
= 50.0 mmol/L
To work out the volume of stock solution required to make up 1L of 8.33 mmol/L glucose the amount by which the stock must be diluted may be calculated or the volume of stock solution containing 8.33 mmol of glucose calculated. The former method enables a very quick solution to part 2 of the question: -
Required dilution factor = Actual Concentration/ Required concentration = 50/8.33 = 6.002
Therfore stock must be diluted 1 part in 6.002 to produce the required concentration.
A) Volume required to produce 1L of stock = 500 mL/ Dilution factor = 500/ 6.002 = 83.3 mL
B) The concentration of benzoic acid in the final solution can be determined very simply =
Benzoic Acid Concentration g/L/ Dilution factor = 5.0/6.002 = 0.833 g/L
5. Injection loops for high performance liquid chromatography systems are made out of coiled lengths of stainless steel tubing of known internal diameter. When filled the loops enable the injection of a fixed known volume of sample reproducibly into the system. If an injection loop is constructed from tubing with an internal diameter 0.25 mm, and is 0.306 M long, how many µL of sample in would be injected onto the system?
This problem can be solved using general knowledge: -
Volume of a cylinder (V) = cross sectional area (A) x length (L) = π r2 L
In order to determine the the volume contained in the tubing, we know that 1mL = 1 cm3 or 1000 mm3. Which means that 1 µL must = 1 mm3.
we know the length of the tubing is 0.306 m = 30.6 cm = 306 mm
The Internal volume is thus easily established as indicated above: -
V=A x L = π r2 L
therefore
= (3.147 x (0.25/2)2) x 306
V= 0.049 x 306 = 14.99 mm3 = 14.99 µL
6. A reference interval has been constructed for a new analyte. 1000 patient samples were analysed and the values were found to be normally distributed following application of appropriate statistical tests. Calculate the theoretical number of patients that may be expected to have values that fall within the following ranges within the population tested: -
a. -2 and +2 standard deviations
b. over +3 standard deviations
c. below -1 standard deviations
d. within the coefficient of variation
This question can be fairly easily be answered by reference to the above diagram. A population that follows a normal or Gausian distribution has a characteristic proportion of the population contained within the boundaries described by the standard deviations. Therefore the % of the population having values equal to the mean plus 1 standard deviation would be 34.1%, and those with mean plus 2 standard deviations would be 34.1 + 13.6% = 47.7%.
Alternatively:-
Mean ± 1SD = 68 % of the population
Mean ± 2SD = 95% of the population
Mean ± 3SD = 99.7% of the population
Therefore.
a) % of population with results between ± 2SD = (34.1 + 13.6) * 2 = 95.4
therefore number of subjects = 1000/100 *95.4 = 954
b) % population outside 3SDs =100 –( 34.1 + 13.6 + 2.1)x 2) =100 - 99.6% = 0.4%
Therefore percentage over +3 standard deviations = 0.4/2 = 0.2%
Therefore number of subjects = (1000/100) x 0.2 = 2
c) Below -1 sd = 13.6% + 2.1% + 0.2% = 15.9% of total number = 159.
Another approach would be that given 50% of the population must be below the mean then the percentage below 1SD must = 50 -34.1 = 15.9% = 159 subjects
d) The CV is the SD x 100/mean. As we know 1 SD contains 34.1% 0f the population ± 1SD would contain 68.2% or 682 subjects
7. You are given below copies of HPLC chromatograms of a catecholamine standard solution and extract of a urine from a patient with phaeochromocytoma. The standard concentrations and retention times are given in the Table. The method uses an internal standard (DHBA) to correct for losses incurred during the extraction. If the patient had a urinary output of 2400 mL in 24 hours, calculate the 24 hour outputs of noradrenaline, adrenaline and dopamine in this case.
24 Urine volume = 2400 mL/24 hour = 2.4 L
This problem is solved using the peak height ratios approach. The data may be used as follows: -
From the standard chromatogram A measure the height of the noradrenaline, adrenaline, dopamine and DHBA (internal standard) peaks after constructing a suitable baseline. The peaks are identified by their retention times given in the question. The peak height ratios are determined using the following formula: -
Peak Height Ratio (PHR) = Peak height of the catecholamine peak in mm
Peak height of the DHBA (Internal Standard)
The same process is undertaken for the urinary extract chromatogram B.
Because the urinary extract and the standard have the same concentration of DHBA added as internal standard it is possible to quantify the concentration of the endogenous catecholamine in the urine using the following formula: -
Concentration = Peak Height Ratio of the extract x Standard Concentration
Peak Height Ratio Standard
e.g. See table
Concentration Noradrenaline = 1.607 x 500
0.586
Urinary output = concentration x urine volume in Litres
e.g. Urinary output Noradrenaline = 1371 * 2.4 = 3290 nmol/24h
Catecholamine / RetentionTime (Min) / Peak Height (mm) / Peak Height Ratio / Concentration
nmol/L / Urinary
Output
nmol/24h
Standard
Noradrenaline / 1.9 / 8.5 / 0.586 / 500Adrenaline / 3.2 / 3.5 / 0.241 / 150
DHBA / 3.9 / 14.5
Dopamine / 7.1 / 9.5 / 0.655 / 1500
Extract
Noradrenaline / 1.95 / 22.5 / 1.607 / 1371 / 3290Adrenaline / 3.15 / 5.0 / 0.357 / 222.3 / 533.5
DGBA / 4.0 / 14.0
Dopamine / 7.1 / 27.0 / 1.862 / 4264 / 10,233
8. Specimens were taken from a patient for measurement of a drug that has a half life of 56 hours. Three specimens were taken for assay on cessation of the drug and received in the lab labelled time 0, 74.1, and 97.3 hours. The assay produced gave results of 65, 20, and 15 mg/L respectively for the sequential samples. The requesting physician contacts the lab indicating that the patient was a subject in clinical trial and that the elimination profile looked odd. Is he correct in his assessment? The critical point for his data set is the result at 74 hours. The samples are not available for re-assay as they were accidentally thrown away post assay and he is not very happy. Is the situation salvageable?
The question simply asks the question as to whether the results are likely to be correct, and if not where does the problem lie, and can we retrieve anything form the situation.
You are given the following information: -
The usual half life (t) is 56 hours
Measured concnentraions of the drug: -
1. Hours = 65 mg/L (C0)
74.1 hours= 20 mg/L (C74.1)
97.3 hours = 15 mg/L (C97.3)
The first thing that could be done is to check whether the results fit mathematically.
We know that Ct = C0 e-kt …………. (1)
Therefore the elimination (k) constant for the drug can be determined as we know that at one half life t = 56 and the ratio of Ct to C0 will = 0.5: -
Ct/C0 = e-kt …………………….. (2)
or
1/( Ct/C0 )= kt ……………………..(3)
Therefore; -
ln 1/( Ct/C0 )= ln 1/(0.5) = ln2.0 = k x 56
therefore
0.693/56 = k = 0.01237
Knowing k and t the measured concentrations can be compared with the theoretical: -
If it is assumed that the measured concentration at time 0 is correct then C0 = 65 and t = 74.1
Substituting into equation 1: -
Ct = Co e-0.01237 x 74.1
Ct = 26 mg/L
Theoretical concentration at 74.1 hours assuming measured concentration at time 0 is correct = 26mg/L
Similarly assuming measured concentration at time 0 is correct and t = 97.3
Substituting into equation 1: -
Ct = 65 e-0.01237 x 97.3
Ct = 19.5 mg/L
Theoretical concentration at 74.1 hours assuming measured concentration at time 0 is correct = 19.5 mg/L
These values are discrepant from the measured concentrations at the two later time points. The possibility of analytical error arises or the possibility of biological variability in the drug handling. Of course the assumption has been made that the measurement at time t is correct and this could be cross checked by calculating the theoretical concentration at t = 0 assuming the 97.3 or 74.1 hour assay value was correct.
Assume the measured concentration at t= 97.3 is correct and k = 0.01237
Substitute into equation (1): -
15= C0 e-0.01237 x 97.3
C0 = 49.9 mg/L
9. A diabetic with a 24 hour creatinine clearance of 25 mL/minute, and a serum creatinine of 200 µmol/L, was found to have an albumin excretion rate of 25µg/minute. If the 24 hour urine volume was 2000 mL calculate the albumin/creatinine ratio in mg/mmol of creatinine.
Creatinine clearance = 25 mL/minute
Serum creatinine = 200 µmol/L
Albumin excretion = 25 µg/minute
Urine volume = 2000 mL
Amount of albumin in 2000 mL = Albumin excretion x Number of minutes/day
= 25 x 1440 = 36000 µg = 36 mg
Concentration of albumin/L =36/2 = 18 mg/L
To calculate concentration of creatinine in urine in mmol/L:-