How much information opens a door? In
figure 4.1, the combination on the right will open any of the three doors. The parentheses
indicate that all the amino acids inside the parentheses will open the door. To compute
information, it is first necessary to establish the number of possible outcomes and the
likelihood of each.
Table 4.1 - Odds of Each Amino Acid Arising by Chance
Number of Codons |
Odds |
Amino Acids |
6 |
6 in 64 |
ser, arg, leu |
4 |
4 in 64 |
val, pro, thr, ala, gly |
3 |
3 in 64 |
ile |
2 |
2 in 64 |
phe, tyr, his, gln, asn, lys, asp, glu, cys |
1 |
1 in 64 |
met, trp |
The odds of each amino acid arising by chance are listed in table 4.1.
For example, the amino acid, serine (ser), will arise by chance 6 times for every 64
tries. The same is true for leucine(leu) and arginine(arg). The amino acid, methionine
(met), will only arise by chance 1 time in 64 tries. Table 4.1 was created from table 3.2
by assuming that all changes to existing DNA (mutations) are random.
Table 4.1 will now be used to calculate the information in figure 4.1.
In figure 4.1, the required combination is as follows:
pos1 pos2 pos3 pos4
pos5 pos6 pos7
asn gln
his leu
cys gly ser
ala arg
Position 1: asn (odds = 2 in 64 ) and ala (odds = 4 in 64) are both allowed. The odds of
seeing either an asn or ala are found by simple addition. 2/64+4/64=6/64. In other words,
the odds of an ala or asn arising by chance at position one are 6 times in 64 tries. Using
the second equation presented in chapter 1, Information = 3.32xlog(64/6) = 3.4 bits.
Position 1 contains 3.4 bits of information.
Position 2: gln has a 2 in 64 chance of arising by chance. This is equivalent to 1 chance
in 32 tries. So the information is easy to find: 2(information) = 32/1. Since 25
=32, position 2 must contain 5 bits of information. Notice that logarithms can also be
used. Information = 3.32 x log(32/1) = 5 bits. Both equations give the same result.
Position 3: The odds for his are 2 in 64. The odds for arg are 6 in 64. So the sum yields
the odds that one of these two will arise by chance = 2/64+6/64=8/64. Thus, one of
these will arise 8 times in 64 tries. This is equivalent to 1 in 8. The information is 2(information)
= 8. Since 23=8, position three contributes 3 bits of information.
Position 4: Leucine has a 6 in 64 chance of arising by chance. These
odds are the same odds that were found for position 1. So position 4 also contributes 3.4
bits.
Position 5:cysteine (cys) has a 2 in 64 chance of arising by chance. These odds are the
same as those calculated for position 2. So position 5 contributes 5 bits.
Position 6: glycine (gly) has a 4 in 64 chance of arising by chance. This is equivalent to
1 in 16. So the information at position 6 is 2(information) = 16/1. Since 24=16,
position 6 contributes 4 bits of information.
Position 7: serine has a 6 in 64 chance of arising by chance. These odds are the same as
those for position 4. So position 7 contributes 3.4 bits of information.
The total information required to open the door is the sum of the
information found at each position or 3.4+5+3+3.4+5+4+3.4 = 27.2 bits. The odds that this
door can be opened by chance are given by1 in 227.2 or 1 in 154 million.
But the above calculation is wrong! The
previous paragraph assigns a probability to a protein evolving based on its information
content today completely ignoring the ability of natural selection to guide evolution.
While the information calculated is correct, 27.2 bits, this information has no
relationship to the probability of any protein evolving. In figure 4.1, only the last door
for each of the three species is shown. There may be many doors leading up to these doors,
and the odds for success may be quite good. Information should never be related to a
probability when natural selection is involved.
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