King M.R., Mody N.A. Numerical and Statistical Methods for Bioengineering: Applications in MATLAB

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Box 1.6 Monod growth kinetics: using finite difference approximations

Monod, in 1942, devised a mathematical relationship to relate the specific growth rate μ of a micro-

organism to the concentration of a limiting nutrient in the growth medium. If μ

max

is the maximum growth

rate achievable when the substrate (nutrient) concentration s is large enough such that it is no longer

limiting, and K

is the substrate concentration when the specific growth rate is exactly one-half the

maximum value, then the Monod equation

states that

μ ¼

max

þ s

We want to determine the rate of change in specific growth rate with substrate concentration s. Taking

the first derivative of the Monod equation with respect to s, we obtain

max

ðK

þ sÞ

The unit of specific growth rate μ is per hour. The product of μ with the cell mass per unit volume (g/l)

gives the rate of growth of cell mass (g/(l hr)). For μ

max

= 1.0 hr and K

= 0.5 g/l for a fungal strain in a

particular culture medium, determine μ

at s = 1.0 g/l.

The exact value of μ

ðs ¼ 1:0g=lÞ¼0:2222 l =ðghrÞ:

We use the finite difference approximations to determine the first-order derivative. Let the step size

be 0.1 g/l. Then,

μ ðs ¼ 1:0g=lÞ¼0:6666=hr;

μ ðs ¼ 1:1g=lÞ¼0:6875=hr;

μ ðs ¼ 0:9g=lÞ¼0:6428=hr:

Using a forward difference approximation to determine μ

, we obtain

ﬃ

0:6875  0:6666

0:1

¼ 0:209 l=ðghrÞ; relative error E

¼ 5:94%:

Using a backward difference approximation to determine μ

, we obtain

ﬃ

0:6666  0:6428

0:1

¼ 0:238 l=ðghrÞ; relative error E

¼ 7:11%:

Using a central difference approximation to determine μ

, we obtain

ﬃ

0:6875  0:6428

2  0:1

¼ 0:2235 l=ðghrÞ; relative error E

¼ 0:585%:

Note the smaller error in the approximation of μ

when using the central difference formula.

Observe what happens if we reduce the step size by half:

μ ðs ¼ 1:05 g=lÞ¼0:6774=hr;

μ ðs ¼ 0:95 g=lÞ¼0:6552=hr:

Now recalculate the forward, backward, and central differences. Using a forward difference approxima-

tion we obtain

ﬃ

0:6774  0:6666

0:05

¼ 0:216 l=ðghrÞ; relative error E

¼ 2:79%:

This model is mathematically identical to the well-known Michaelis–Menten kinetics for enzyme reactions.

1.6 Taylor series and truncation error

Using a backward difference approximation we obtain

ﬃ

0:6666  0:6552

0:05

¼ 0:228 l=ðghrÞ; relative error E

¼ 2:61%:

Using a central difference approximation we obtain

ﬃ

0:6774  0:6552

2  0:05

¼ 0:222 l=ðghrÞ; relative error E

¼ 0:09%:

By halving the step size, the truncation errors for the forward and backward difference approximations are

reduced by approximately one-half. This is expected since the truncation error is of the same order as

the step size. On the other hand, halving the step size resulted in an approximately four-fold decrease in

the truncation error associated with the central difference approximation. The reason for the pronounced

decrease in the truncation error should be clear to you.

Box 1.7 Monod growth kinetics: truncation error versus round-off error

We again use the forward finite difference approximation (Equation (1.21)) to estimate μ

, the rate of

change in specific growth rate at a substrate concentration s = 1.0 g/l, where

max

ðK

þ sÞ

We prog r essively decrease the step size h by an order of magnitude and accordingly determine the relative

error in the prediction of the first-orderderivative. In other words, h assumes values 0.1, 0.01, 0.001, ...,10

–15

Figure 1.9 shows that as h is decreased from 0.1 to 10

–8

, the relative error is dominated by the

truncation error and decreases with a slope of 1. (We know that the truncation error is dominant for this

range of step-size values because the error reduces in a regular way with decrease in step size.) The

relative error in the finite forward difference approximation is found to be linearly proportional to the step

size. For h <10

–8

, the round-off error dominates the overall error in the estimated value of μ

. The large

round-off error stems from the loss of significant digits (loss of numeric information) in the subtraction

step. Note that the error at h =10

−15

is even greater than the error at h = 0.1. Thus, there is an

optimum step size at which we attain the least error in our estimation. On the right or left side of this

optimum value, the truncation error and round-off error, respectively, constrain the attainable accuracy.

The MATLAB code used to generate Figure 1.9 is available on the Cambridge University Press website.

Figure 1.9

An optimum step size h exists that yields minimum relative error in the estimation of a first-order derivative using a

forward finite difference approximation.

−15

–8

–6

–4

–2

Step size, h

Relative error

−10

−5

Types and sources of numerical error

1.7 Criteria for convergence

In Sectio ns 1.1–1.6 we discussed the types of errors generated as by-products of

numerical computations and the different ways to assess the magnitude of the error.

In this section, we introduce different criteria for determining whether the error in

the solution has been reduced to a tolerable limit, or, in other words, whether a

numerical algorithm has converged upon a solution. There are several convergence

criteria that may be used, and you have the option of employing one or multiple

criteria as you deem appropriate.

One condition that speciﬁes whether convergence has been realized involves

computing the change in the magnitude of the solution with every time step.

The convergence condition may be based on the absolute difference of two

successive iterative values of the numerical solution of x or their relative

difference:

(a) x

iþ1

 x

 E

(b)

iþ1

 x

iþ1



 E,

where E is the acceptable tolerance. In condition (a) E is called absolut e t ol er -

ance, while in condition (b) it is called relative tolerance. The stringency of

condition (a) in comparison to condition (b) depends on the magnitude of x

i+1

if E is the same in both cases. The ﬁrst condition will allow termination of the

iteration procedure only after x

 E  x

iþ1

 x

þ E. In this case, the i mposed

tolerance governs the precision of the solution by specifying the rightmost digit

of the number that must be c orrect or signiﬁcant. C ondition (b) will direct the

algorithm to exit the iterative procedure when x

=ð1 þ EÞx

iþ1

 x

=ð1  EÞ.

The imposed tolerance governs the t otal number of signiﬁcant ﬁgures in the

solution. If x

i+1

is large, i.e. much greater than unity, then condition (a) is

more stringent. If x

i+1

is small, i.e. 1, then condition (b) is a stricter

convergence criterion than condition (a).

Another method to determine if convergence has been reached involves periodi-

cally monitoring the value of a function of the iteration variable(s). If a zero of the

function is being sought, then the approach of the function towards zero can be the

test for termination of the iterative method, such as

ðÞ

where E is the tolerance limit close to zero.

Finally, other checks can be used to ensure that the algorithm is functioning

properly and whether, in certain situations, convergence is not reached in an accept-

able time limit, in which case the algorithm can be terminated. Iteration checks that

are inserted into computational procedures to prevent an endless iterative loop

include capping the total number of iterations allowed and placing bounds on

the magnitude of the iteration variables or the function values being calculated. In

the next few chapters, such convergence criteria will be suitably incorporated

into the numerical algorithms introduced. It will become clear from the speciﬁc

examples in subsequent chapters how the appropriate condition(s) may be chosen

for terminat ing iterative procedures.

1.7 Criteria for convergence

1.8 End of Chapter 1: key points to consider

(1) Numerical methods of solution usually possess two inherent sources of numerical

error: round-off errors and truncation errors.

(2) Computers store numbers using binary ﬂoating-point representation and have ﬁnite

memory constraints. As a result:

(a) the range of numbers that a computer can represent is ﬁnite;

(b) the precision by which real numbers can be repres ented by a computer is limited –

many real numbers cannot be stored with exact precision;

decimal fractions such as 0.1, 0.2, or 0.7 with a ﬁnite number of binary

digits.

This leads to discontinuity of the ﬂoating-point number line and the generation of

round-off errors.

(3) The magnitude of round-off error can be calculated by determining the absolute

error and relative error of the calculated resul t with respect to the true v alue, if

known.

(4) In most cases, round-off error generated when using double-precision format is

minimal and can be neglected. However, the magnitude of round-off error can

become signiﬁcant when performing arithmetic operations on ﬂoating-point num-

bers such as

(a) add ition of two numbers of disparate magnitudes;

(b) subtra ction of two numbers close in value that results in loss of signiﬁcance;

times magni fy a round-off error.

Care should be taken to formulate numerical algorithms such that disruptive or fatal

numerical errors, such as loss of signiﬁcance due to subtractive cancellation, increase

in round-off error from addition of disparate numbers, or generation of ﬂoating-

point exceptions due to overﬂow error, are avoided.

(5) The number of signiﬁcant digits present in a number determines the accuracy of the

numeric quantity in question.

(6) Many numerical methods employ the Taylor series expansion for approximating a

function f(x). The inclusion of only a ﬁnite number of terms in the series leads to the

origin of truncation error. The order of magnitude of the truncation error can be

easily estimated from the Taylor series.

(7) While round-off error and truncation error arise independently of each other, there

exists a trade-off between reducing truncation error and keeping the round-

off error small. When using large step sizes or when retaining only a few terms of

the inﬁnite series to approximate the function of interest, t he truncation error

dominates. On the other hand, when the step size is very small or when many

terms of the inﬁnite series th at represents the function are retained, round-off error

can dominate.

1.9 Problems

1.1. Write a MATLAB program to convert a decimal fraction into its equivalent binary

fraction. Run the program to determine the binary fraction representation for the

decimal fract ions 0.7 and 0.2. Verify the correctness of your algorithm from the

answers in the chapter.

Types and sources of numerical error

1.2. (i) Determine the 4-bit signiﬁcand (binary fraction) for the following decimal

fractions:

(a) 1=9,

(b) 2 = 3.

You can use the program you de veloped in Problem 1.1.

(ii) Add the 4-bit binary signiﬁcands determined in (i) to ﬁnd the binary equivalent

for ð1=9Þþð2=3Þ. When performing binary addition, note these simple rules:

0 + 1 = 1 and 1 + 1 = 10. As an example, the addition of 1.01 and 0.11 would

proceed as follows:

1.01

+ 0.11

————————

10.00

(iii) Convert the 4-bit binary signiﬁcand determined in (ii) to its decimal equivalent.

(iv) Determine the relat ive error in the binary addition of two decimal fractions, i.e.

compare your answer in (iii) to the exact solution of the addition of the two

decimal fractions, which is 7=9.

1.3. Write the binary ﬂoating-point representations (1.b

... b

p-1

 2

)of

the following numbers (you can use the program you developed in Problem 1.1 to

extract the binary signiﬁcand ):

(a) 100,

(b) 80.042 ,

“1” appears to the left of the decimal point.)

1.4. Evaluate the function fxðÞ¼

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

þ 1



ﬃﬃﬃﬃﬃ

for x = 300 using only six-digit arith-

metic. Calculate the relative error in your result. Find another way to evaluate f (x)

again using only six-digit arithmetic in order to obtain a better approximation of the

value of the function at x = 300. (Hint: Algebraic rearrangement of f (x).)

1.5. The equation fxðÞ¼x

þ 5x

 2x þ 1 ¼ 0 can be rearranged to obtain an alternate

nested form gxðÞ¼ x þ 5ðÞx  2ðÞx þ 1 ¼ 0, both expressions being equivalent.

Using four-digit rounding arithmetic, determine the value of f (x) and g(x) for

x = 1.17 and check your results against the exact result of 7.106 113. Which expres-

sion produces a more accurate result, and can you explain why?

1.6. Using three-digit arithmetic, add the following numbers from left to right:

1:23 þ 0: 004 þ 0:0036 þ 5:6:

With every addition step, the intermediate result can only have three digits after

rounding.

Check the precision of your ﬁnal result by comparing with the exact answer. What

is the absolute error?

Now rearrange the numbers so that the smallest numbers are summed ﬁrst and

this is followed by adding the larger numbers. Again, use three-digit arithmetic.

Check the precision of your ﬁnal result by comparing with the exact answer. What is

the absolute error now?

Determine the number of signiﬁcant digits in both calculated results. How has

changing the order in which the numbers are added affected the precision of your

result?

1.7. If b in the quadratic equation ax

þ bx þ c ¼ 0 is negative, explain why the choice of

formulas to obtain the solutions x

and x

should be chosen as

1.9 Problems

b þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac

(1:6)

and

2c

b 

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac

(P1:1)

in order to minimize round-off error. Show how the expression for x

is obtained.

Choose appropriately two out of the four Equations (1.6), (1.7), (1.9) and (P1.1)

and solve the equations below. Justify your choice of equations each time.

(1) x

 75:01x þ 1 ¼ 0,

(2) x

þ 100:01 þ 0:1 ¼ 0.

1.8. Write an algorithm that accurately computes the roots of a quadratic equation even

in situations when subtractive cancellation may severely corrupt the results, i.e.



ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

 4ac

. Convert the algorithm into a MATLAB program and solve the

quadratic equations given in Problem 1.7.

1.9. Use Equation (1.11) to derive the Taylor series expansion for fxðÞ¼lnð1 þ xÞ. Use

= 0. Show that

ln 1 þ xðÞ¼x 



þ:

Calculate ln(1 + x) for x = 0.5 and 2. Estimate the relative error based on the

exact solution if you include either one, two, three, or four terms. Construct and

complete Table P1.1 to compare your answers. What conclusions can your draw

from your solution?

1.10. Write a MATLAB program to determine the value of sin x for values of x such that

0<x < π, using the sine series (Equation (1.13)). Find the answer to within a

relative accuracy of 0.001 of the exact value of sin x, the criterion for convergence

being

abs((sum – sin(x))/sin(x)) < 0.0001,

where sum is the summation of terms in the sin x series. Plot the number of terms

that must be included in the sum to achieve this relative accuracy for each value of x.

Redetermine the value of sin x using the sine series for ﬁve values of x such that

0<x < π; however, this time use the following tolerance speciﬁcation:

absolute value of ðlast term=summation of termsÞ

0:0001

as the convergence criterion. Plot the number of terms that must be included in the

sum to achieve this tolerance for each value of x.

How well does the tolerance speciﬁcation for the second criterion match the ﬁrst

criterion of convergence?

Table P1.1. Relati ve error in approximating ln(1 + x)

Terms included x = 0.5 x =2

Types and sources of numerical error

1.11. Determine the ﬁrst-order derivative of the function

fxðÞ¼4x

 9x

þ 1:5x

þ 7x  2

at x = 2, using forward difference, backward difference, and central differen ce

(Equations (1.20),(1.22) and (1.23)). Use a step size of 0.1. Calculate the absolute

and relative error for each case based on the exact derivative f

2ðÞ.

1.12. Novel hematopoietic stem cell separating device Each year, more than 35 000

people in the United States are diagnosed with leukemia, lymphoma, or other blood,

metabolic, or immune system disorders for which a bone marrow or adult blood

stem cell transplant could be a cure. However, 70% of these patients do not receive a

transplant due to both a lack of matched donors and the high, non-reimbursed costs

incurred by treatment centers. For many patients, bone marrow or stem cell trans-

plants provide a cure to an otherwise fatal or greatly debilitating condition. One of

the most important steps in processing a bone marrow or peripheral blood sample

for transplantation is separating out the hematopoietic stem cells. These cells are

present in very low numbers and are responsible for restoring function to the

patient’s damaged bone marrow. Current separation technology is a labor-intensive

and inefﬁcient process. There is often a low yield of stem cells, and high expenses

must be incurred by transplant centers.

An extracorporeal device to capture hematopoietic stem cells (HSCs) from blood

quickly and efﬁciently is currently under design. This device is made of tubes that are

coated internally with speciﬁc sticky molecules called selectins that preferentia lly

bind HSCs (Wojciechowski et al., 2008). Stem cells have a surface marker called

CD34 that is absent on mature blood cells. Selectins can also bind white blood cells

or leukocytes. If blood or bone marrow cells are perfused through such tubing, stem

cells and some amount of white blood cells are selectively removed from the bulk of

the ﬂuid due to adhesion of these cells to the wall.

Rat experiments were performed to assess the feasibilty of purifying stem cells from

blood by using this capture device. The inner surface area (SA) of a single tube is

35.34 mm

. It was found that, on average, 1770 cells/mm

were found to be attached to

the inner tubing surface following perfusion of blood for 60 min. The total number of

stuck cells is calculated to be 1770 × SA = 62 550 (rounded off to the nearest integer).

However, the number of stuck cells per mm

was found to vary between 1605 and 1900

for different experiments (or 0.9068 to 1.073 times the average). Determine the range

of the total number of stuck cells that are captured by a single tube in both absolute

and relative terms. How many signiﬁcant digits can you include in your answer?

1.13. In the experiment discusseed in Problem 1.12, the purity of the captured stem

cells was found to be 18%; in other words, the fraction of captured cells that were

CD34+ stem cells was 18%. Based on the range of the total number of stuck cells

that are captured by a single tube, calculate the range of the number of captured stem

cells by the coated tube. If the purity ranges from 12 to 25%, then determine the

absolute lower and upper bounds of captured stem cells. (Take into account the

variability in the total number of captured cells.)

1.14. Separation of different populations of cancer cells Some leukemic cancer cells

exhibit stem cell-like properties and express the CD34 marker on their surface, a

surface marker exhibited by immature stem cells. In this way, leukemic cancer cells

can be loosely classiﬁed into two categories: CD34+ cancer cells and CD34– cancer

cells. Current cancer therapies are not as effective in eradicating the cancer cells

displaying CD34 on their surfaces. These CD34+ cancer cells are believed to be

responsible for inducing relapse of the disease following treatment.

1.9 Problems

A device for separating CD34+ cancer cells from CD34– cancer cells is currently

being investigated. This device exploits the differences in the mechanical properties

of cell rolling on a surfa ce-bound glycoprotein called selectin. Cell rolling manifests

from repeat ed bond formation and breakage between selectin molecules bound to a

stationary wall and CD34 molecules bound to the cells’ surface, when cells are

ﬂowing over a surface (see Figure P1.1 ).

A series of in vitro experiments were performed to measure the rolling velocities of

primary (directly derived from patients) cancer cells (acute myeloid leukemia

(AML)) to ascertain the differences in their rolling properties. Table P1.2 lists the

observations for both CD34+ and CD34− cells.

Calculate the rolling velocities (in μm/s) using the appropriate number of signiﬁ-

cant ﬁgures. Determine the average rolling velocity and the mini mum and maximum

rolling velocities observed for CD34+ and CD34− cells. What are the maximum

absolute and relative deviations from the average deviations in the measured rolling

velocities for both CD34+ and CD34− cells?

Compare the average rolling velocites of CD34+ and CD34− cancer cells in

absolute terms. How do you think they compare? Do you suggest the need for

more data (observations)? The method used to determine whether the two rolling

populations are signiﬁcantly different using statistical tests is discussed in Chapter 4.

1.15. Treatment of prostate cancer cells with vitamin D A tumor (neoplasm) is a cluster

of abnormally proliferating cells. As long as the tumor cells remain within the

clustered mass and do not detach from it, the tumor is considered benign. A tumor

is deﬁned as malignant or a cancer if the abnormally growing and dividing (neo-

plastic) cells are capable of separating from the tumor of origin and invading other

tissues. The process by which cancerous cells break away from the tumor grow th and

invade other surrounding normal tissues is called metastasis. During metastasis,

these abnormally proliferating cells may even enter into the body’s circulation

(blood or lymphatic system), where they are transpo rted to different regions of the

body. It is important to note that a neoplastic cell cannot metastasize and colonize in

Table P1.2.

CD34+ cancer cells CD34– cancer cells

Observation

Distance

travelled (μ m) Time (s)

Distance

travelled (μm) Time (s)

1 147.27 61.40 124.77 58.47

2 129.93 46.12 162.74 60.90

3 42.29 51.87 173.99 39.67

4 142.54 61.40 163.20 60.90

5 59.70 61.40 167.78 60.90

6 163.94 61.40 127.36 60.90

7 83.99 61.40 112.08 43.67

8 121.9 61.40 159.32 60.90

9 89.85 27.43 123.76 60.90

10 83.35 61.40 176.41 60.90

Types and sources of numerical error

Table P1.3.

Vitamin D-treated cells Untreated cells

Observation

Distance

travelled (μ m) Time (s)

Distance

travelled (μm) Time (s)

1 114.865 11.50 87.838 19.70

2 121.622 18.18 165.541 44.53

3 114.865 46.73 351.416 57.93

4 108.108 25.17 442.568 41.27

5 334.528 57.93 362.259 57.90

6 233.108 27.60 375.380 57.90

7 212.865 57.93 432.485 57.90

8 209.568 57.93 523.649 56.73

9 331.512 57.93 496.725 57.90

10 479.777 57.93 317.639 57.90

11 446.266 57.93 375.243 57.90

12 321.017 57.93 351.368 57.90

13 127.297 57.93 425.796 57.90

14 320.964 57.93 270.355 57.90

15 321.106 57.93 459.472 57.90

16 517.289 57.93 341.216 57.90

17 391.906 53.47 365.256 57.90

18 328.138 57.93 405.631 57.90

This data set is available at the book’s website www.cambridge.org/* as

prostateCA.dat.

From K. P. Rana and M. R. King (unpublished data).

Figure P1.1

A CD34+ cancer cell rolling on selectin molecules bound to a wall while flowing through a flow chamber.

wall

CD34 surface

marker

CD34 + cell

surface-bound

selectins

1.9 Problems

foreign tissues (tissues other than that of its origin) unless the cell has acquired

certain properties that allow it to survive in an alien environment.

Prostate cancer cells are prone to metastasize to the bone and cause bone cancer. It

has been shown that treatment of prostate cancer cells with vitamin D inhibits meta-

stasis to the bone (Ali and Vaidya, 2007). In order for cancer cells to metastasize

successfully, they must be able to contact and adhere to the vascular endothelium after

they have entered the blood stream. Once adhered, the cancer cells must then ﬁnd a way

to squeeze through the vascular wall, escape from the blood stream, and enter new

(foreign) tissues. In order for cells to adhere ﬁrmly to the endothelial lining of the vessel

wall, they ﬁrst contact the wall via weak non-covalent bonds that form and break under

ﬂow. The repeated formation of new cell–surface bonds and breakage translates

macroscopically into cell rolling on the endothelial wall.

It is hypothesized that vitamin D may inﬂuence the rolling properties of cells on a

selectin-coated surface. Selectins are glycoprotein molecules that are expressed by

endothelial cells. Selectins reversibly bind the surface receptors of white blood cells

and certain cancer cells and stem cells, the consequence of which is cell rolling on the

surface. To test this hypothesis, the rolling velocities of vitamin D-treated prostate

cancer cells are compared with the rolling velocities of non-treated prostate cancer

cells (see Table P1.3).

Calculate the rolling velocities (in μm/s) using the appropriate number of signiﬁ-

cant ﬁgures. Determine the average rolling velocity and the mini mum and maximum

rolling velocities observed for vitamin D-treated and untreat ed cells. What are the

maximum absolute and relative deviations from the average value in the measured

rolling veloci ties for both cases? Compare the average rolling velocites of vitamin

D-treated and untreated cancer cell s in absolute terms. How do you think they

compare?

References

Ali, M. M. and Vaidya, V. (2007) Vitamin D and Cancer. J. Cancer. Res. Ther., 3, 225–30.

Fletcher, J. E. (1980) On Facilitated Oxygen Diffusion in Muscle Tissues. Biophys. J., 29,

437–58.

Liguzinski, P. and Korzeniewski, B. (2006) Metabolic Control over the Oxygen Consumption

Flux in Intact Skeletal Muscle: In Silico Studies. Am. J. Physiol. Cell Physiol., 291,

C1213–24.

Mathews, J. H. and Fink, K. D. (2004) Numerical Methods Using Matlab (Upper Saddle

River, NJ: Pearson Prentice Hall).

Scarborough, J. B. (1966) Numerical Mathematical Analysis (Baltimore, MD: The John

Hopkins Press).

Tanenbaum, A. S. (1999) Structured Computer Organization (Upper Saddle River, NJ:

Prentice Hall).

Wojciechowski, J. C., Narasipura, S. D., Charles, N., Mickelsen, D., Rana, K., Blair, M. L.,

and King, M. R. (2008) Capture and Enrichment of CD34-Positive Haematopoietic Stem

and Progenitor Cells from Blood Circulation Using P-Selectin in an Implantable device.

Br. J. Haematol., 140, 673–81.

Types and sources of numerical error