ECG Problem

This ECG, from an elderly man with chronic heart failure, is relatively unchanged over at least 6 months. What did the echocardiogram show?

Courtesy: ECG Wave-Maven

http://ecg.bidmc.harvard.edu/maven/mavenmain.asp

The Curious Case of Sickle Cell C

Unusual observations in Medicine sometimes have very simple explanations. Take the case of sickle cell C disease, for example.

Evolutionary pressures have led to the existence of several mutants of the beta chain of haemoglobin. Thus Hb S, Hb C and Hb E all have mutations on the beta chain. For example, in Hb S, glutamic acid is replaced by valine in position 6, while in Hb C, lysine is substituted in the same position. Early on, epidemiologists noticed that these mutated haemoglobins were found in areas with high prevalence of falciparum malaria. For example, Hb C is found in Western Africa, Hb E is present in around 60% of subjects in the Indian subcontinent, and Hb S is widely prevalent in Africa. These variants have evolved because heterozygotes with Hb S, C or E are resistant to severe infestation with P.falciparum, and thus provide a survival advantage in these geographical locations.

In the normal adult, two beta chains combine with two alpha chains to form the complete globin chain (alpha2-beta2) and thus constitute the most abundant form of haemoglobin present in adults, known as Hb A. While the beta chain has only one gene, the alpha chain is coded by two genes. Thus, the alpha chains have 4 different alleles across the two chromosomes.

Heterozygotes with the sickle haemoglobin (sickle cell trait) have one normal allele producing the beta chain, and one mutant allele producing Hb S. Since each allele produces an equal amount of Hb A and Hb S, you'd expect an equal (50% each) proportion of Hb S and Hb A in subjects with sickle cell trait. Yet, this is not so. On haemoglobin electrophoresis, these subjects have 50-60% Hb A, and only 35-45% Hb S [the rest being contributed by Hb A2 (alpha2-delta2) and Hb F (alpha2-gamma2)]. Why does this happen?

As it happens, the reason beta chains and alpha chains join so harmoniously is because they carry an almost equal, and importantly, opposite electrical charge. Beta chains carry a negative charge of -2.5 coulomb (C), while alpha chains carry a positive charge of +2.4 C, thus ensuring electroneutrality (almost) when they combine.

However, the beta chain mutants are less negatively charged than the native beta chains. Thus, they combine less effectively with the alpha chain to form Hb S, C or E. This is why, in heterozygotes, instead of a 50-50 split, Hb A produced by the normal allele predominates over the variant haemoglobin. Thus, subjects with sickle cell trait have ~55% Hb A, and 40% Hb S, while heterozygotes for Hb E, have~ 70% Hb A and only 30% Hb E. This also explains why such heterozygotes are not anaemic. Subjects with sickle cell trait can only be picked up on electrophoresis, while heterozygotes with Hb E are only revealed by microcytosis with an absence of iron deficiency.

The principle is further illustrated in subjects with Hb SC disease. Here both alleles of the beta chain are mutant- one is producing Hb S, the other Hb C. As these two beta chain mutants have roughly equal charge (and thus affinity for the alpha chain), they are present in roughly equal concentration on electrophoresis~45-50% each. There is no normal beta chain to compete with.

A similar phenomenon occurs in sickle cell beta thalassaemia. As you may know, the defect in beta chain production in beta thalassaemia may be only partial (denoted as beta thal+) or severe (denoted as beta thal 0). Despite the deficit in production of normal beta chains, subjects with sickle cell beta(+) thalassaemia still have Hb A comprising around 30% of the total Hb in RBC, the other 70% being Hb S, as even in diminished quantities, the available normal beta chains combine more efficiently with alpha chains than the mutated beta chain found in Hb S. Thus, these subjects have a less severe phenotype than those with sickle cell beta (0) thalassaemia, who can't produce any Hb A.

This principle can be put to good use in the diagnosis of newborn subjects (with carrier parents) with one of the mutated beta chains. While Hb F is the predominant haemoglobin in newborns, the proportion of Hb A and Hb S will vary depending on homozygosity, heterozygosity and the co-existence of beta thal (+) trait. Thus, newborn with sickle cell disease will have a FS (F>S) pattern at birth, subjects with sickle cell trait will have a FAS (F>A>S)pattern, while a FSA (F>S>A)pattern at birth is diagnostic of sickle cell beta (+) thalassaemia.

Finally, a correction. In my post on hereditary spherocytosis, I had said that I did not know of any other condition that caused a high MCHC. This is incorrect. Subjects with Hb AC or Hb SC have RBC that are prone to dehydration due to a chloride channel defect, a condition known as xerocytosis. Due to loss of water, the RBC have a high MCHC, which might be the only clue to diagnosis in subjects with Hb AC.

Chronic Atrophic Gastritis- Lessons In Heterogeneity

Like I, you probably learnt in medical school that Chronic Atrophic Gastritis is associated with pernicious anaemia and possibly carries an increased risk of gastric cancer. This dogma is in fact true, but it leaves out more than it reveals. What about H.pylori? Which type of gastric cancers are increased? Is there a role for surveillance? It's time to take a look.

The stomach is probably best viewed as a viscus of two parts- the first- the acid and pepsin secreting part is called the oxyntic mucosa, and comprises the cardia, fundus and body of the stomach, the second- the non-acid secreting part, is the antrum, which of course leads to the pylorus.

The oxyntic mucosa contains two principal types of cells- the parietal cells, which secrete hydrochloric acid, and the chief or peptic cells, which secrete pepsin. The antrum does not contain either of these specialised cells, but has epithelial cells that can secrete mucin, like similar cells in the oxyntic mucosa. Gastric mucin differs from intestinal mucin in having neutral pH. On the other hand, intestinal mucin has acidic pH and can be sialo-mucin (containing N-acetyl muramic acid) or sulfo-mucin, depending on the negatively charged components that it comprises.

The stomach does not contain goblet cells, unlike the intestinal epithelium. Goblet cells are mucus secreting cells found in the intestine.

Pepsin is derived from a zymogen called pepsinogen. There are two isoenzymes of pepsinogen- types I & II. The oxyntic mucosa secretes both isoenzymes of pepsinogen, while the antral mucosa only secretes pepsinogen II.

It is widely believed that the sequence of change in the gastric mucosa in response to inflammation proceeds thus: gastritis-->atrophy-->metaplasia-->dysplasia-->cancer.

The three processes of gastritis, atrophy and metaplasia form a continuum and have been subsumed into one term- metaplastic atrophic gastritis or MAG. There are two principal triggers that drive MAG- autoimmunity (A) and environmental (E) factors- thus the two subtypes of MAG are described as AMAG and EMAG.

Metaplasia connotes a change in the type of gastric mucosal epithelium. The normal gastric mucosa may change into a pseudopyloric or an intestinal phenotype.

AMAG is due to an autoimmune attack on the resident cells of the oxyntic mucosa. This is typically accompanied by the presence of anti-parietal cell and anti-intrinsic factor antibodies in the serum. Autoimmune gastritis slowly destroys the parietal and chief cells, in a patchy manner at first, and more extensively as time wears on. There is therefore a lack of intrinsic factor, leading to pernicious anaemia, and if the process carries on, achlorhydria ensues. Achlorhydria leads to hypertrophy of G cells or gastrin secreting cells present in the antrum. Thus, hypergastrinemia is one of the key features of AMAG.

H.pylori infection of the oxyntic mucosa is uncommon in AMAG. This may be because the atrophic mucosa of AMAG may not form a good substrate for H.pylori, or because of colonisation by other bacteria.

On the other hand, the principal trigger of EMAG is H.pylori infection of the gastric mucosa. While AMAG involves the oxyntic mucosa, EMAG favours the antral mucosa. Diet is thought to be involved in some cases, particularly a high salt intake and a group of chemicals called nitrosoamines, that are produced in the stomach from dietary nitrates. Unlike in AMAG, complete acholrhydria rarely occurs in EMAG, pernicious anaemia does not occur, and hypergastrinemia is not a feature, as the G cells of the antrum are lost to the inflammatory process.

An useful differentiating feature between AMAG and EMAG is the ratio between serum Pepsinogen I & II. As Pepsinogen I is only secreted by the oxyntic mucosa, and and Pepsinogen II by both oxyntic and antral mucosa, a low Pepsinogen I : Pepsinogen II ratio is found in AMAG and in patients with pernicious anaemia. It can be used as a risk marker for the development of AMAG, pernicious anaemia and gastric adenocarcinoma in relatives of affected subjects.

In time, in some subjects, the mucosa becomes dysplastic, a precursor to development of gastric cancer. Gastric adenocarcinoma is the commonest malignancy and occurs more commonly in the antral mucosa than in the oxyntic mucosa. The principal risk factor for gastric adenocarcinoma is untreated H.pylori infection leading to EMAG. Gastric cancer can also arise post-pernicious anaemia in subjects with AMAG.

In subjects with AMAG, a second type of cancer- carcinoid tumour- may arise in the oxyntic mucosa. Hypergastrinemia in AMAG is a powerful trigger for hypertrophy of enterochromaffin type cells (ECL)present in the oxyntic mucosa. These cells are normally responsible for secreting histamine, an important secretagogue for acid (hence the role of H2 blockers in treating peptic ulcer disease). With continued stimulation from gastrin in AMAG, the ECL cells first form polyps, which may, in time, turn into carcinoid tumour. Thus, antrectomy is sometimes employed in treating gastric carcinoid to remove gastrin as a driver.

A third type of cancer can rarely arise from the gastric mucosa from mucosa associated lymphoid tissue (MALT). These lymphomas typically arise in subjects with Sjogren's syndrome and other allied conditions such as Rheumatoid arthritis, leading to a histological subtype of lymphoma called "Extranodal marginal zone lymphoma". The main driver for MALT associated lymphoma is again H.pylori infection. While extranodal marginal zone lymphoma may arise elsewhere, such as in the parotid glands, in a majority of cases, the stomach is also involved and should be examined through endoscopy and biopsy.

The role of surveillance in early diagnosis of gastric cancer is clearer in high risk subjects such as those from Far Eastern countries, and those with a family history of gastric cancer, but is less clear in Western subjects who generally have a lower risk of progression from MAG to gastric cancer. In general subjects with pernicious anaemia should have one endoscopic examination to look for AMAG and gastric cancer but repeat endoscopies are not advised. In contrast, subjects in or from high risk countries such as Japan, or those with a family history of cancer should have OGD every 2-3 years. Apart from biopsying abnormal lesions, non-targeted biopsies should be taken from the fundus, antrum and incisura- at least two each from the fundus and antrum and one from the incisura. These should be labelled in different containers and the pathologist should report the biopsy by area examined in his/her report. The incisura is usually involved in extensive EMAG.

Finally, some commonly used terms. The term Type I or "complete" gastric metaplasia is used to describe replacement of the gastric mucosa by small intestinal mucosa (containing goblet cells and brush border). Type III or "incomplete" gastric metaplasia describes replacement of the gastric mucosa by colonic mucosa- large droplets of mucin but no brush border. This classification is of some importance as Type III or "incomplete" metaplasia is associated with a higher risk of gastric adenocarcinoma than Type I or complete metaplasia. The pathologist can identify intestinal metaplasia by the presence of acid sialo-mucin or acid sulpho-mucin, as opposed to the presence of neutral mucin in normal gastric mucosa.

The terms "complete" and "incomplete" metaplasia do not desribe the extent of involvement. If only one area of the stomach is involved, say the antrum or fundus, this is described as "limited" involvement. With metaplasia of more than one area, say antrum, fundus and body of stomach, involvement is termed "extensive" and carries a higher risk of cancer.

What's the diagnosis?

Young man with new onset abdominal distension and right hypochondrial pain. CT scan of abdomen is shown.

(courtesy UpToDate)

Behaving Oddly- Hereditary Spherocytosis

Hereditary Spherocytosis (HS) is the commonest disorder of red cell membrane leading to haemolysis (incidence~500/million). It is inherited in an autosomal dominant manner in 75% of cases and recessively in the remaining quarter. Subjects with recessive inheritance tend to have more aggressive disease and present earlier in life, as with many other disorders with recessive inheritance.

HS has many odd, non-intuitive features. Newborn with HS do not have anaemia, but are diagnosed due to prolonged jaundice. Adults present with anaemia, jaundice and with bilirubin containing gallstones. Thus, recurrent gallstones with jaundice may not necessarily presage primary gallstone disease, but may indicate underlying HS. Anaemia may be mild or absent because of compensation by the marrow, thus leading to difficulty in diagnosis. The diagnosis is usually made by noticing spherocytes on the blood film. It is thought that because of defective anchoring of the red cell membrane to the underlying cytoskeleton, bits of the membrane are gradually lost, reducing the surface area to volume ratio, causing spherocytosis.

Oddly, subjects with HS may develop obstructive jaundice due to CBD stones in addition to underlying indirect hyperbilirubinaemia. When this happens, red cell survival is paradoxically increased. This is thought to be due to increase in the cholesterol content of the red cell membrane which increases its distensibility and thus reduces the propensity for lysis while passing through the splenic sinusoids.

As stated above, mild HS may not be associated with anaemia due to compensation by the bone marrow. The reticulocyte count will be raised in such subjects, but may return towards normal after splenectomy. An important clue to the diagnosis of HS, particularly in infants, is the presence of a high mean corpuscler haemoglobin concentration (MCHC). Normally, MCHC would be expected to be around 31. In HS, MCHC is 36 or higher. A combination of high MCHC (>35) and a high RDW (15 or higher), is virtually diagnostic of HS. I know of no other condition that lends itself to diagnosis by looking at the MCHC alone.

Subjects with HS have more porous red cell membrane. While this allows Na to enter freely, thus increasing the risk of osmotic haemolysis (exploited in the osmotic fragility test, now supplanted by more sensitive and specific tests such as EMA), it also leads to a leakage of K ions when blood is cooled after being drawn. This often leads to the finding of pseudohyperkalaemia in such subjects.

Other odd things happen in these subjects. While splenectomy imparts a lifelong higher risk of arterial and venous clots in HS, due to higher leucocyte, red cell, platelet and fibrinogen levels, non-splenectomised subjects with HS actually have a lower cardiovascular risk than their relatives. It is thought that higher serum bilirubin and lower serum cholesterol protects such subjects against cardiovascular events.

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.