Calculations

Question 1

BB Guy’s Easy Way to Calculate RhIg vials from KB result

Answer 1

KB% x 5/3 = number of vials

Remember, the 5 is if you are assuming a BV of 5000mL because the test did not give you amount. Otherwise, if they tell you the weight of the mother, then you can calculate actual BV by using 70mL/kg conversion. If it is 6 liters, you would multiply by 6 in the above example.

Rounding rules: If number after decimal is < 5, round up once. If > 5, round up twice. So, “3.4” would mean give 4 vials, while “3.5” would mean to give 5 vials

Question 2

Calculating cryo needed for hypofibrinogenemia

Answer 2

Critical Data Needed: Weight in Kg, Hct, and current fibrinogen level, target fibrinogen level

If they do not give target level, 100mg/dL is “adequate” and 150 mg/dL is “optimal”.

1. Calculate Blood Volume
   Body weight x 70 ml/Kg
2. Calculate Plasma Volume (PV)
   Blood volume x (1 - hematocrit)

Divide by 100 to convert to dL
3. Calculate mg fibrinogen needed
Plasma volume x concentration change desired
*Subtract desired level from current
(ie, 150 mg/dl – 50 mg/dl)
*Multiply level change by PV
(ie, 100 mg/dl x 36 ml)
4. Calculate bags of cryo needed
Fibrinogen needed / 250 mg per bag

Question 3

Calculating amount of FVIII and IX to give

Answer 3

STEPS:

1. Calculate Blood Volume
   Body weight x 70 ml/Kg
2. Calculate Plasma Volume (PV)
   Blood volume x (1 - hematocrit)
3. Calculate FVIII units needed
   Plasma volume x increase desired = FVIII units needed

If not given a target level, use 50% (remember FVIII is measured in percent present) for hemarthrosis, minor bleed or minor surgery and use 100% for major surgery or life threatening hemorrhage but convert to decimal for calculation
4. Only if asked, calculate bags of CRYO needed
FVIII units needed / 80 units per bag

IF ASKED TO CALCULATE FOR FACTOR IX, IT IS THE SAME CALCULATION EXCEPT MULTIPLE BY 2 AT THE END and CRYO amount isn’t issue

Question 4

Correct Count Increment

Answer 4

(Post Plt Count – Pre Plt Count) x BSA / number of plts transfused

(1 apheresis unit = ~3.0 x 10¹¹ plts)

CCI>7,500 /uL is considered adequate response while 5000 is considered refractoriness

Another similar calculation is post-transfusion platelet recovery (PPR):

BV x (post count - pre count) / number platelets transfused

>20% is considered adequate

Question 5

Neonates dose for plts, RBCs and FFP

Answer 5

= 10-15 mL/kg (will raise hemoglobin by 2-3)

Question 6

Anion gap

Answer 6

[Na+] – [Cl-] + [HCO3-] ​

normal is 12 +/- 4

Question 7

Osmolal Gap

Answer 7

Osmolarity measured – (2[Na] + [glucose]/18 + [BUN]/2.8)

normal is 10 mOsm/kg

Question 8

Friedewald Equation

Answer 8

LDL = Total cholesterol – HDL – TG/5

Question 9

Creatinine Clearance

Answer 9

Volume_urine x Urine_creatinine / Plasma_creatinine

Question 10

Fractional excretion of sodium

Answer 10

(urine_Na x plasma_creatinine)/(urine_creatinine x plasma_Na)

FENa 1% is indictative of primary glomerular disease and hepatorenal syndrome while FENa >1% is found in ATN, prerenal azotemia and postrenal azotemia

Question 11

MCV and MCHC

Answer 11

MCV = Hct x 10/RBC

MCHC = (Hgb/Hct) x 100

Question 12

Breakeven point

Answer 12

Z (# of tests to breakeven) = Fixed cost / (Revenue – Variable Cost)

To calculate revenue = total charges- (allowances + bad debt)

To think about it in simple terms, Breakeven is the fixed cost divided by the amount you are profiting from each test

Question 13

Standard deviation index for proficiency testing

Answer 13

Standard Deviation Index for Proficiency Testing =

*SDI > 2 is sometimes used as cutoff to identify labs with unacc perform

(lab result – peer group mean)/peer group standard deviation

Question 14

Relative Risk

Answer 14

(# ppl with condition that develop outcome/# ppl with condition)

(# ppl in population that develop outcome/# ppl in population)

Question 15

Percent carryover

Answer 15

(Results from potentially contaminated low sample – original results from uncontaminated low sample) / Original results from uncontaminated low sample x 100

Question 16

Likliehood ratio of positive test? Negative test?

Answer 16

Likelihood ratio of POSITIVE test =

Sensitivity/(1-specificity)

Likelihood ratio of NEGATIVE test =

(1-specificity)/sensitivity

Question 17

Post test probability

Answer 17

Post-test odds/(1 + post-test odds)

*to calculate post-test odds = disease prevalence x likelihood ratio

Question 18

Renal Failure Index

Answer 18

Urine sodium/(Urine creatinine/Plasma creatinine)
<1 is indicative of prerenal failure or acute glomerulonephritis.
>2 is indicative of postrenal failure or tubular injury.

Question 19

Odds ratio:

Answer 19

Question 20

EXAMPLE PROBLEM:

A 75kg female has fibrinogen level of 45 mg/dL. Her Hct is 31%. How many bags of CRYO are needed to raise her fibrinogen to an acceptable level for hemostasis?

Answer 20

Acceptable level = 100 mg/dL

Optimal level = 150 mg/dL

ANSWER 7.97 bags

Question 21

EXAMPLE PROBLEM:

A 15 year old, 45 kg boy with hemophilia A presents to the ED with a knee joint that is swollen and discolored. He is diagnosed with hemarthrosis. His Hct is 33% and his FVIII level is 3%. The ED doctor asks for your help in calculating the dose of FVIII concentrate this boy should receive.

How many bags of CRYO would he need?

Answer 21

Targets: minor bleeds, hemarthrosis or minor surgery 50%. Major surgery, life threatening bleeds or CNS hemorrhage 100%.

Answer 992 units of FVIII, 12.4 bags of cryo

*calculation for FIX is the same only you double the amount at the end and you CAN’T USE CRYO

Question 22

Reticulocyte index

Answer 22

RI = Retic count x Hct/normal HCt (which is 45%)

Question 23

Given the frequency of homozygous recessive, homozygous dominant and heterozygous genes, be able to calculate allele frequency.

Hardy-Weinburg Equation

Answer 23

Remember the basic formulas:

p² + 2pq + q² = 1 and p + q = 1

p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p² = percentage of homozygous dominant individuals
q² = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals

EXAMPLE:

You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:

• The frequency of the “aa” genotype. Answer: 36%, as given in the problem itself.
• The frequency of the “a” allele. Answer: The frequency of aa is 36%, which means that q² = 0.36, by definition. If q² = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequency is 60%.
• The frequency of the “A” allele. Answer: Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%.
• The frequencies of the genotypes “AA” and “Aa.” Answer: The frequency of AA is equal to p², and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 16% (i.e. p² is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).
• The frequencies of the two possible phenotypes if “A” is completely dominant over “a.” Answers: Because “A” is totally dominate over “a”, the dominant phenotype will show if either the homozygous “AA” or heterozygous “Aa” genotypes occur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive phenotype is simply the frequency of aa. Therefore, the dominant frequency is 64% and, in the first part of this question above, you have already shown that the recessive frequency is 36%.

Question 24

Determine how many tests must be performed in a year to justify the cost of a new instrument if the price of the instrument is $70,000 and it should last 5 years. It can do 60 tests per hour. A tech can do 4 tests per hour and is paid $6 per hour.

Answer 24

10,000 tests per year

Equipment Evaluation (Straight from Osler Notes)

You have purchased a $70K machine which has a 5 year lifetime. The machine can perform 60 tests/hr. A tech can run the machine at a salary of $6/hr. Currently, with the manual method, a tech can do 4 tests/hr. How many tests need to be done each year to justify the cost of the machine?

$70K/5 = $14K

$6/4 = $1.50 per test (manual method)

$6/60 = $0.10 per test (new method)

Let x = the number of tests.

0.10x + $14,000 = 1.50x

x = 10,000/yr