Euler's Number, e

e is perhaps the most celebrated, most feted number in Maths this side of pi. First described by Euler, it is difficult to define as it's an irrational number, i.e. without an integeric value and unexpressable as a fraction. It is best denoted as the limit of (1+1/n)^n, when n tends to infinity. The mathematical value of e is approximately 2.71828..

You do not begin to appreciate the importance of e until you consider the following problems, provided as examples:

1. A bank offers you 5.6% interest rate, compounded for 5 years. How much would you get back after 5 years?
2. UK has a population of 63 million, Germany has 78 million. UK is growing at the rate of 2 new subjects per 1000 per year, Germany at roughly half that. How many years would it take for UK's population to catch up with that of Germany's?
3. A food item will spoil at a critical bacterial mass of 5 million per gram of foodstaff. If initially, there were only 10,000 bacteria, and if the number of bacteria grow from 10,000 to 20,000 in 1 minute and to 250,000 in the next 10 minutes, how long would it take for the food to spoil?

You need e to solve all the problems. Essentially, if an entity is growing continuously, at a given rate, you can use e to calculate its growth. e is the natural base log (log 10 being the "common" log). OTOH, the natural log of a number when the result is expressed as an exponent of e is called "ln" (l for Lima, n for November, using phonetics.

A simple way to remember this is that e^kt gives you the value of an entity growing at the rate of "k" after a certain time period "t", while ln x gives you the exponent- i.e. the power by which the entity has grown. If the rate of growth is known, ln x will let you find the time taken for the entity to grow from y to x.

Things will become clearer with the above examples.

1. A bank offers you 5.6% interest rate, compounded for 5 years. How much would you get back after 5 years if you invested £5000?

You might think the answer is 5000*(1.056)^5, but you would be wrong. The answer of 6565 worked this way is less than what you'd actually get.

The answer is 5000*e^(0.056*5)= approximately 6616. This is how compound interest is calculated.

On the other hand, you might wonder how long it would take for your money to double at this rate of 5.6%? If time for doubling is denoted as t, you get e^(0.056t)= 2. However, you know that ln 2 = 0.693 from a scientific calculator. Hence ln 2 = 0.693 = 0.056t or t = 12.375 years.

Since interest rates are expressed as percentages, you can simply multiply both numerator and denominator by 100 and get t=69.3/5.6= 12.375 years.

In practice, 69.3 is rounded off to 72 to allow us to take advantage of the fact that 72 is divisible by many small integers (the usual duration of a fixed deposit). However, this gives us an approximate result. Thus 72/5.6 in the previous example would give 12.85 years. This is called the rule of 72.

Similarly, you can calculate the trebling time for your deposit at the same rate of interest. Thus, e^0.056t = 3 . ln 3 = 1.10 approximately. Thus for trebling, you can use the rule of 110. Your money will treble in 110/5.6 = 19.64 years.


2. UK has a population of 63 million, Germany has 78 million. UK is growing at the rate of 2 new subjects per 1000 per year, Germany at roughly half that. How many years would it take for UK's population to catch up with that of Germany's?

In this case, if the projected period for equalisation of numbers is t, you can write the equation as:

63 million*e^(0.002t)= 78 million* (e^0.001t)

Or e^.002t/e^.001t = 78/63.

Here, it is useful to know that since e is a logrithmic function, e^x/e^y = e^(x-y).

Thus, in the above problem, e^(.002t-.001t) = 78/63.

Or e^.001t = 1.238

now, ln 1.238= 0.213 = .001t

or t = .213/.001 = 213 years.


3. A food item will spoil at a critical bacterial mass of 5 million per gram of foodstaff. If initially, there were only 10,000 bacteria, and if the number of bacteria grow from 10,000 to 20,000 in 1 minute and to 250,000 in the next 10 minutes, how long would it take for the food to spoil?

Here, after 1 minute, 10,000* e^(k.1), where k is the rate at which bacteria are multiplying.

Or 10,000*e^k= 20,000- This is equation 1.

After a further 10 minutes, 20,000*e^[k.(1+10)]= 250,000- This is equation 2.

Using logrithmic notation, e^[k(1+10)] can also be written as e^k * e^10k

Dividing equation 2 by equation 1, we get:

(20,000*e^k*e^10k = 250,000)/(10,000*e^k = 20,000)

You get 2e^10k = 12.5

or, e^10k = 6.25.

ln 6.25 = 1.83 = 10k, or k = .183

How long will it take to attain the critical mass of 5 million bacteria per gramme? (call this T)

10000*e^(.183T) = 5 million

or e^.183T = 500

ln 500 = 6.21 = .183 T,

or T= 6.21/.183 = 33.93 minutes.

The Nyquist Limit

Doppler echocardiography is often used to assess stenotic or regurgitant valves. Two types of doppler signals are used- pulsed wave doppler (PWD) and continuous wave doppler (CWD). Both serve different purposes and can be viewed as complementary.

PWD is used as a localising tool. It accurately detects that a systolic murmur is, for example, a consequence of aortic stenosis rather than mitral regurgitation. By producing a spectral image, PWD demonstrates a direction of flow towards or away from the tranducer. A spectral wave of aortic stenosis would, for example, be directed away from a transducer placed at the apex. In the case of PWD, a single transducer does both the sending and receiving.

Echocardiography relies on the shift in ultrasound frequency caused by red cells flowing towards or away from the transducer. This is called doppler shift and is given by F= 2Fo.v.cos theta/c, where Fo is the transmitted frequency, v denotes velocity of blood flow, theta is the angle between the transducer and plane of flow and c is the velocity of ultrasound waves in the medium in use, in this case, blood. When the transducer is parallel to the direction of flow, theta is 0, and cos theta is 1. Thus F= 2Fo. v/c.

Note that the doppler shift, i.e, the detected change in frequency is proportional to twice the emitted frequency. This illustrates an important limitation of PWD called "Nyquist limit". The Nyquist limit is always half the sampling frequency. That is to say that the maximum frequency accurately detectable with a sampling frequency of f is f/2. If emitted frequency is more than the Nyquist limit for the sampling frequency, than a phenomenon called "aliasing" occurs, where the recorded spectral wave is cut off at its peak and appears on the other side of the baseline (mimicking combined stenosis and regurgitation in the case of pure stenosis, for example), thus giving a distorted image. One way of reducing aliasing is by reducing the "sample volume", i.e. by placing the transducer as close to the valve being examined as possible. Thus, the ultrasound waves have to travel a shorter distance, thus raising the frequency at which sampling occurs, and thus the Nyquist limit.

CWD overcomes this shortcoming by using 2 transducers- one to transmit, and one to receive. There is thus no Nyquist limit. CWD is thus used to measure high velocity flows, such as through a severely stenotic valve (velocity being a function of Doppler shift in the above equation). Using the modified Bernoulli equation, one can estimate the pressure change across a defective heart valve. Thus Delta P (change in pressure)= 4 V^2. For example, if blood is flowing through a stenotic aortic valve at 4m/s, the pressure differential across the valve is 64 mm Hg.

The limitation of CWD is that while it can measure, it cannot localise. Thus, it is likely to confuse AS with MR if the jets happen to be in range. This distinction is only achievable by PWD, which samples a limited frame. In practice therefore, one should localise the jet with PWD, taking care to avoid aliasing and then measure the velocity and thus delta P with CWD.

Immunohistochemistry as a Diagnostic Tool in Cancer

Unfortunately, cancer has often spread before it is diagnosed. In such cases, it can be a diagnostic challenge to determine the site of the primary neoplasm. In some cases, an extensive search for the primary may prove fruitless, leading to a diagnosis of carcinoma with unknown primary or CUP. It is in such cases that immunohistochemistry comes into its own.

Immunohistochemistry (IHC) is the application of a large panel of monoclonal antibodies directed to a plethora of tissue specific antigens to the putative cancer tissue, followed by the use of immunofluorescence to identify stained areas.

Every tissue has its own signature antigens. Because such antigens will often be present in other tissues, they can be rather non-specific and no one antigen should be relied upon while looking at a biopsy sample. Rather, pathologists rely on a panel of antigens while performing IHC.

Carcinomas, i.e cancers of epithelial origin will demonstrate cytokeratins (CK). CK 8 and 18, and CK 1-3 are universally present in all carcinomas and are initially looked for to differentiate carcinomas from tumours of mesenchymal origin, i.e. sarcomas.

Sarcomas themselves will always have a protein called vimentin. Unfortunately, vimentin can also be produced by the occasional carcinoma.

The most useful cytokeratins are CK7 and CK20. Presence or absence of one or both of these cytokeratins is initially employed to narrow down the possible range of neoplasms.

Thus CK7+ and CK20+ will include transitional cell carcinoma, ovarian mucinous carcinoma and pancreatic carcinoma. CK7+ but CK20- will identify breast, lung, thyroid, gallbladder and cholangiocarcinoma. CK20+ and CK7- comprises gastrointestinal, particularly colon carcinoma and Merkel cell carcinoma. CK7- and CK20- narrows the field down to prostate, renal, adrenal and hepatocellular carcinomas.

Here I will give examples of some of the commonest clinical challenges faced by pathologists and how they are resolved.

1.Tumour deposits in pleura. Primary may be lung, breast, thyroid or pleura itself. Stain for TTF1, CK5/6, WT-1, calretinin, CK7, CK20, CEA, PAX 8, gross cystic disease fluid protein 15 (GCDFP15), mammaglobin, ER, PR.

Thus, TTF-1 positive- lung or thyroid. Thyroid will demonstrate thyroglobulin and PAX 8, lung will not. If lung, stain for BER-EP4, BG8 and MOC-31. All three are seen only with adenocarcinomas.

TTF1 negative- GCDFP15+, mammaglobin+, CEA positive, ER+, PR+ means breast cancer.

2. Deposits in axillary nodes. Cytokeratin negative, but positive for Melan A, HMB 45, S100. Diagnosis is metastatic melanoma.

3. Liver mass- primary or secondary? CK7-, CK20-, Hep-par 1+, alpha-fetoprotein positive, CEA (p), in situ hybridisation positive for albumin-primary hepatocellular Ca.

4. Liver deposit- CK20+, CK7-, CEA+, CDX2+, villin+ denotes metastatic colon cancer.

5. Peritoneal deposits- CD10+, RCA+, CK7-, CK20-, PAX2+ will mean this is a Clear cell carcinoma of the Kidney.

6. Lymphadenopathy- CD5+, Cyclin D1 positive, CD10-, CD23-, CD19, 20, 22 and 79A+ (pan B cell antigen) denotes Mantle cell lymphoma.

7. Adrenal deposits- CK7-, CK20-, simple CK (1, 8, 18+), EMA-, Melan A+, S100+, AdCB4P means adrenocortical carcinoma

8. Liver deposits- CK5/6+, CK7+, CK20+, uroplakin+, thrombomodulin+ denotes TCC of urothelial tract.

9. Pleural tumour- CK5/6+, CK7+, CK20-, TTF1-, BER-EP4-,WT-1+, calretinin+ denotes mesothelioma.

10. Vulval tumour- CK7+, CK20-, GCDFP15+, CEA+, ER+, HER2- is consistent with Pagets disease of vulval appendage- usually sweat duct.

The Pathobiology of "Autosomal Dominant"

Have you ever wondered why certain traits are autosomal dominant and others are autosomal recessive?

In most cases, the explanation is fairly straightforward. Diseases, such as Huntington's or myotonic dustrophy, which are transmitted in an autosomal dominant fashion are mostly due to the toxic effect of the mutant protein encoded by the abnormal gene. This protein builds up inside the endoplasmic reticulum and interferes with the synthesis of other vital proteins, which then leads to the abnormal phenotype.

With autosomal recessive inheritance, it is usually an useful function that is lost. Both alleles encoding an essential protein are lost, and thus the product can no longer be synthesised, for e.g. the CFTR protein in cystic fibrosis.

There are notable exceptions to this general principle. Sometimes, the product of one normal allele is insufficient to perform the requisite function of the putative gene. This is called haploinsufficiency. Thus, the condition is transmitted as autosomal dominant, with a dose-response gradient between those heterozygous and homozygous for the abnormal allele. Examples of haploinsufficiency contributing to autosomal dominant transmission are dyskeratosis congenita, Williams syndrome and autosomal dominant retinitis pigmentosa. Marfans and Ehlers Danlos syndromes are also inherited in a similar way.

A different, but nonetheless fascinating mechanism of autosomal dominant inheritance is illustrated by an autoinflammatory condition called TRAPS (TNF Receptor Associated Periodic Syndrome). In this syndrome, the 55 kDa TNF receptor on the cell surface is abnormal, and does not shed on binding to the pro-inflammatory cytokine TNF alpha. Thus prolonged signalling through TNF alpha leads to an augmented inflammatory response, with excessive production of NF-kappa B. Further, it is thought that the usual neutralising effect of circulating "soluble" (non-membrane bound) TNF receptor on TNF alpha is reduced.

The TNF alpha receptor is homotrimeric, i.e. it is composed of three similar units. Most subjects with TRAPS have only 1 abnormal allele, usually due to a missense mutation, while the other allele is normal. However, assuming that even one mutated component of the trimer would lead to abnormal receptor function, one can see that even with one normal allele, 7 out of 8 receptors would be abnormal (the odds are only 1 in 2^3 that in a given trimeric receptor, all 3 components would be from the normal allele). Thus, the condition would transmit as autosomal dominant.

The Problem of Isolated Uveitis

One of the most common reasons for referral to the Rheumatologist from the Ophthalmologist is the young subject with recurrent or troublesome episodes of uveitis, often in association with a positive ANA or other features suggestive of an autoimmune aetiology such as a raised ACE. The query is whether such subjects have an underlying systemic disease contributing to their uveitis. Certain features can help narrow down the D/D.

1. HLA B27 associated spondarthritis presents with acute unilateral anterior uveitis that improves within 3 months but often recurs in the other eye. Therefore simultaneous or closely spaced occurrence of uveitis in both eyes is not characteristic of this condition, even if arthralgias are present. Prognosis is excellent.
2. The uveitis associated with IBD is often more chronic, posterior to the lens, and bilateral and more common in women. In subjects with IBD, the uveitis often presents prior to bowel symptoms (10/17 in one series).
3. Anterior uveitis is associated with the presence of deposits on the back of the cornea, called keratic precipitates (KP). In the case of sarcoidosis, these KPs are often large and greasy, therefore fine keratic precipitates make sarcoid unlikely.
4. The uveitis of sarcoid is often chronic, bilateral, posterior as well as anterior and unaccompanied by systemic features.
5. Syphilitic uveitis may not be accompanied by systemic features, In Asians, always rule out tuberculosis.
6. In subjects above the age of 45 with chronic posterior uveitis, rule out lymphoma, particularly DLBCL type of NHL. Higher risk in HIV positive cases. Thus, lymphoma is an uveitis mimic.
7. In children with JIA, uveitis is more common in those with oligoarticular arthritis and positive ANA. Polyarticular involvement and absence of ANA are less commonly associated with arthritis.
8. Behcet's disease often causes a panuveitis, is often associated with retinal vasculitis, is silent, and thus can lead to blindness. It's important to be aware that while anterior uveitis presents with pain, redness, photophobia, headache or brow pain and constricted pupils, posterior uveitis presents with a white eye, is silent, and can only be picked up initially by the presence of vitreous cells on slit lamp examination. Being silent, it can lead to blindness, and therefore needs a high index of suspicion. This is also true for children with JIA.
9. Multiple sclerosis can be associated with pars planitis (intermediate uveitis).
10. TINU or Tubulointerstitial Nephritis with Uveitis is a rare condition that combines uveitis with interstitial nephritis. It can be sen in subjects with Sjogren's syndrome or sarcoid.
11. Other autoimmune conditions that present less commonly with uveitis are SLE and GPA (Wegener's).

All In The Family

A 29 year old man presents with recurrent gout, in association with a very high uric acid level at over 800 umol/l. His blood tests show impaired renal function, with a creatinine of 180umol/l and an e-gfr of 28 ml/min. His father has also had gout from an early age with similarly high serum uric acid levels, but normal renal function. However, a brother had to have dialysis starting at age 36, leading to transplantation. This brother also suffered from gout.

The patient's only dietary risk factor for gout was a fondness for Coca-cola. He did not drink much alcohol and consumed little red meat, shellfish, prawn or offal.

What is the diagnosis?

The Genius of Adolf Eugen Fick

Fick was a 19th century German physiologist. In 1870, he worked out that cardiac output could be measured from a subject's oxygen consumption, if the concentration of oxygen entering and leaving a given organ were known.

Thus, cardiac output= Oxygen consumption/Arterial O2 content-Venous O2 content

It is easy to calculate the arterial or venous oxygen content if you know the O2 saturation (SaO2) in arterial or venous blood, as almost all the O2 is carried by haemoglobin. Each gram of Hb carries 1.34 ml of O2.

Thus, Arterial oxygen content= 1.34*Hb concentration*SaO2

As cardiac output is expressed in l/min and Hb in g/dl, to make matters uniform, you have to multiply the denominator by 10.

Thus, cardiac output (l/min)= O2 consumption (l/min)/[10*1.34*Hb(g/dl)*SaO2]-[10*1.34*Hb(g/dl)*SvO2]

In practice, it is rarely necessary to determine the absolute cardiac output, but Fick's principle has been put to good use to determine the degree of left to right shunt by calculating the respective blood flow to pulmonary (Qp) and systemic circulation (Qs)in conditions such as ASD & VSD. In significant left to right shunt, Qp>Qs. If Qp is more than twice Qs, most authorities would recommend percutaneous closure of ASD or VSD with an Amplatzer device or a surgical closure if the shunt is very large, as with some VSDs.

It is easy to determine O2 saturation during cardiac catheterisation consecutively in the superior venna cava, right atrium, inferior vena cava, right ventricle and pulmonary artery. Pulmonary venous oxygen content is considered to be the same as the oxygen saturation of arterial blood measured during pulse oximetry, provided there is no right to left shunt. The latter can be detected by a drop off in O2 saturation while moving into the putative chamber, e.g decline in SO2 while moving from right atrium to right ventricle, suggests a right to left shunt from a VSD, while the same thing happening while moving down from the SVC to the RA indicates a right to left shunt via an ASD.

As the SVC normally has 3 times the flow as in the IVC, the systemic mixed venous O2 content is calculated as (3*SVC SO2+IVC SO2)/4.

Oxygen consumption can be be calculated by using a special spirometer with the subject rebreathing air and using a CO2 absorber.

Using the Fick principle, Qp/Qs can be written thus:

Qp/Qs= [Oxygen consumption/ 10*1.34*Hb*(Pulmonary venous SO2-Pulmonary arterial SO2)] divided by,
[Oxygen consumption/10*1.34*Hb*(Systemic arterial SO2-Systemic mixed venous SO2)]

or simply, Qp/Qs= Systemic arterial SO2-Systemic mixed venous SO2/Pulmonary venous SO2-Pulmonary arterial SO2

In practice, an alternative and non-invasive way to measure the pulmonary:systemic flow ratio in left to right shunts is to use Doppler Echo to find the area of the aortic and pulmonary outflow tracts and multiply them with their respective velocity time integral.