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.
Thursday, 26 December 2013
Tuesday, 10 December 2013
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.
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.
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