Bo DengDepartment of Mathematics

UNL

IIT, 14 Feb. 2011http://www.math.unl.edu/~bdeng1

Working Hypothesis

Evolution is driven to maximize biodiversity against constraints in time and energy across all biological scales

Applied to all informational systems: o DNA Replicationo Protein Synthesiso Sexual Reproductiono Speciation to Phylogenetic Treeo Ecological Communityo Animal Braino Consciousnesso Languageo Social, Economical, Political Structures

Claude E. Shannon (1916-2001)

C. E. Shannon, ``A mathematical theory of communication,'' Bell System Technical Journal, vol. 27, pp. 379-423 and 623-656, July and October, 1948.

C. E. Shannon, ``A mathematical theory of communication,'' Bell System Technical Journal, vol. 27, pp. 379-423 and 623-656, July and October, 1948.

Channel

Dial-up 2400 bps ~ 56Kbps

DSL 128 Kbps ~ 8 Mbps

Cable 512 Kbps ~ 20 Mbps

Satellite ~ 6 Mbps

Optic Fiber 45 Mbps ~ 150 Mbps

Transmission Speed Comparison

Internet

All about choices

What is Information? and What Matters the Most?

0 or 1 ……# of sequences of length log

2 n = # of choices n

Bit Unit:

Mathematical Measure of Information: What is in a bit?

One Bit = One Binary Digit

Dead Channel --- Transmit only one kind of symbol all the times e.g. 0000….. 0 bit 0 bit information

Live Channel --- Transmit one of many possible symbols each time, e.g. 011101… in a binary channel Each transmitted symbol is either 0 or 1 Each symbol contains 1 bit informationPop Quiz: How many bits in a quaternary symbol, 1, 2, 3, 4? or in a symbol of n alphabets, 1, 2, 3, …, n?Answer: H4 = 2 bits, and Hn = log2 n bits respectively because 4 = 2 log

2 4, n = 2 log2 n

Ex: { a, b, c, d } = { 00, 01, 10, 11}

Ex: { a, b, c, d } = { 00, 01, 10, 11}

Key Assumption: Each transmitted symbol is just one of n equally probable choices

Let be time needed to transmit symbol k

Then the average transmission time per base is

Tn = (1 + 2 + 3 +…+ n ) / n

And the mean rate is Rn = Hn / Tn = n log2 n / (1 + 2 + 3 +…+ n )

What is in the transmission rate?

The definition implicitly assumes that all symbols occur equally probable. Why, or is it reasonable?

All-purpose Channel

Internet message types: video, audio, pictures, spams, …etc Each has different frequency distribution in the encoding symbols

Example of Possible Non-equiprobability: If we know all video files that have ever transmitted over the internet, then we can make an accurate frequency table: say p1 for Symbol 1, p2 for 2, etc, and pn for symbol n

Recall: Rn = Hn / Tn = n log2n / (1 + 2 + 3 +…+ n )

Equiprobability

Each transmitted Symbol 1 is just one choice out of 1/p1 many possible choices and therefore Symbol 1 contains

log2 1/p1 bits information

since 1/p1 = 2 log2 1/p

1

Similarly, Symbol k contains log2 1/pk bits information

The average bits per symbol for our video only source is H(p) = p1 log2 1/p1 +…+ pn log2 1/pn

Each transmitted Symbol 1 is just one choice out of 1/p1 many possible choices and therefore Symbol 1 contains

log2 1/p1 bits information

since 1/p1 = 2 log2 1/p

1

Similarly, Symbol k contains log2 1/pk bits information

The average bits per symbol for our video only source is H(p) = p1 log2 1/p1 +…+ pn log2 1/pn

Example: Pick a marble from a bag of 2 blue, and 5 read marbles

Probability for picking a blue marble:

pblue

= 2/7

Number of choices for each blue picked

1 / pblue

= 7/2 =3.5

Example: Pick a marble from a bag of 2 blue, and 5 read marbles

Probability for picking a blue marble:

pblue

= 2/7

Number of choices for each blue picked

1 / pblue

= 7/2 =3.5

Important fact: H(p) = p1 log2 1/p1 +…+ pn log2 1/pn <= Hn = log2 n

Important fact: H(p) = p1 log2 1/p1 +…+ pn log2 1/pn <= Hn = log2 n

Conclusion: For an all-purpose channel, the mean rate is calculated not for any particular source entropy but for the maximal source entropy, Hn , which is reached with equaprobability distribution of the transmitting symbols.

Conclusion: For an all-purpose channel, the mean rate is calculated not for any particular source entropy but for the maximal source entropy, Hn , which is reached with equaprobability distribution of the transmitting symbols.

0 or 1 …… 1/p

1 = # of sequences of length log

21/p

1 Bit Unit:

Design Criterion

To choose n so that Rn = Hn / Tn

is the largest!

Example

Encoding states:

Symbols: 1 2 3 …. n Trans. Times: 1 2 3 … n

Assume: 1 = 1 sec, 2 = 2 sec, 3 = 3 sec, … , n = n sec

Then Rn = Hn / Tn = n log2n / (1 + 2 + 3 +…+ n ) = 2 log2 n / (n+1)

....

http://www.mun.ca/biology/scarr/An11_01_DNA_replication.mov

James D. Watson (1928 -), Francis Crick (1916 - 2004), Molecular structure of nucleic acids, Nature, 171(1953), pp.737--738.

James D. Watson (1928 -), Francis Crick (1916 - 2004), Molecular structure of nucleic acids, Nature, 171(1953), pp.737--738.

DNA Replication

Fact: DNA replication is the same for all genomes Replication is a sequential process – one base a time

Observation: Each species genome is an information source Genome upon replication is a transmitted message

Conceptual Model:

DNA replication is an all-purpose channel

Questions: Why 4 bases: A, T , C , G?

Communication Model for DNA Replication

Replication Mean Rate: Rn = Hn / Tn , (per-base diversity rate)

Assumption: Weaker chemical bonds take longer to replicate (Heisenberg’s Uncertainty Principle: t E ~ constant )

Paring times of high energy bonds are ignored (as a first attempt/order approximation for the pairing time)A = T = pairing time of one H…O bond = t0

G = C = pairing time of two H…O bond = 2 t0

5 = 6 = pairing time of three H…O bond = 3 t0, etc.

(by Watson and Crick’s base paring principle)

Time scale of a single Hydrogen bond pairing: 4X10-15 sec.

Time scale of a single Hydrogen bond pairing: 4X10-15 sec.

Let k = # of base pairs, and n = # of basesThen

n = 2 k Since 2m-1 =2m = m t0 for m = 1,2, …, k

Rn = Hn / Tn = log2 n / [2(1 + 3 + …+ 2k-1) /n]

= log2 n / [(n/2+1) t0/2]

The Result

A further refined model predicts

1.65 < C,G / A,T < 3 R4 = the optimal rate

1.8267

2 Sexes Problem

Sexual Reproduction is a process of information exchange

Reproduction Mean Ratio: Sn = Hn / En ,

Assumption: Information payoff per-crossover base for n sexes: Hn = log2 n

1:1 sex ratio with M members for each sex Cost to sexual reproduction in energy and time is inversely proportional to the probability of having a reproductive group of n members having exactly one sex each Reproductive group is formed by random encounter

Reproductive Probability:

Reproductive Group in k Tries:

Expected Tries for One Reproductive Group :

Expected Tries for One Reproductive Group for Large Population :

The Result: Entropy-to-Cost Ratio: Sn = Hn / En ,

M = 10m

Genetic Entropy Exchange without Sexual but Existential Cost :

Multiparous Strategy

Multiparous Entropy:

Multiparous Cost :

Multiparous Entropy to Cost Ratio :

With Mixed (Random & Wedlock) Cost :

Discussions

> 80 million yrs> 2%< 0.98n = 6

> 1 billion yrs> 25%< 0.75n = 2

Evolutionary Set-back by

Slower byn = 4Rn / R4

= 2

Life on Earth could have not evolved faster and have had a richer diversity at the same time

Evolutionary Clock Set-back with 3 Sexes:

Consistent with Darwinian Theory of Survival-of- the-Fittest theory but at the molecular levelQuestion: Was the origin of life driven by informational selection?

Why is the per-base diversity measure by Hn = log2 n or H ( p ) = pk log2 1/pk

log2 1/(p1 p2) = log2 1/p1 + log2 1/p2 Information is additive

The Role of Mathematics

Mathematics is driven by open problems Science is driven by existing solutions Mathematical modeling is to discover the mathematics to which Nature fits as a solution Exception to the rule is the rule in biology

Acknowledgements

Dr. Reg Garrett, Department of Biology, University of Virginia, regarding the GC transcription elongation problem

Dr. David Ussery, Center for Biological Sequence Analysis, Technical University of Denmark, on most base frequency data

Dr. Daniel Smith, Department of Biology, Oregon State University, regarding the base frequencies of P. ubique

Dr. Tony Joern, Department of Biology, UNL, Kansas State University

Dr. Etsuko Moriyama, the Beadle Center for Genetics Research, University of Nebraska-Lincoln

Dr. Hideaki Moriyama, Dr. Xiao-Cheng Zhen, Department of Chemistry, University of Nebraska-Lincoln

Irakli Loladze, David Logan, Department of Mathematics, UNL

The show of life is on your DNA channel

We are consumers of reproductive entropy

Genome

Base Frequency

A T G C

S. coelicolor 13.9 14.0 36.1 36.0 0.1% -44.2%

E. coli K-12 24.6 24.6 25.4 25.4 0.0% -1.6%

E. coli O15:H7 24.8 24.7 25.2 25.2 0.1% -1.0%

Human* 29.4 29.7 20.5 20.4 0.3% 18.2%

P.ubique 35.3 35.0 14.9 14.8 0.3% 40.6%

W. glossinidia 38.8 38.7 11.2 11.3 0.1% 55.0%

d

)()( CGTA pppp

|}| |,|max{ CGTA ppppd

* Base frequency for the chromosome 14 which has the largest d.

Genome

Base Frequency

A T G C

phage P1 26.1 26.6 23.5 23.8 0.5% 5.4% 1.9978

phage T4 31.8 32.9 16.5 18.8 2.3% 29.5% 1.9355

phage VT2-Sa 25.6 24.5 26.9 23.0 3.9% 0.2% 1.9976

phage 933W 27.6 22.8 27.4 22.2 5.2% 0.8% 1.9927

phage phiX174 24.0 31.3 23.3 21.5 7.3% 10.6% 1.9846

max. 2.0000

d )( pH

Viruses are taking advantage of the replication system by having the near maximal per-base diversity entropy and having their hosts do the replication for them.

To Maximize Stationary Entropy: H(p) = p1 log2 1/p1 +…+ pn log2 1/pn

Genome

Base Frequency

A T G C

S. coelicolor 13.9 14.0 36.1 36.0 0.1% -44.2% 1.8538 1.1623

E. coli K-12 24.6 24.6 25.4 25.4 0.0% -1.6% 1.9998 1.4093

E. coli O15:H7 24.8 24.7 25.2 25.2 0.1% -1.0% 1.9999 1.4122

Human* 29.4 29.7 20.5 20.4 0.3% 18.2% 1.9834 1.4005

P.ubique 35.3 35.0 14.9 14.8 0.3% 40.6% 1.8774 1.5081

W. glossinidia 38.8 38.7 11.2 11.3 0.1% 55.0% 1.7688 1.4921

d

8267.1 ** * Base frequency for the chromosome 14 which has the largest d.

)( pH**

, )( pRTA

1.82671.8267

Others have to scramble with individual and absolute Channel Capacities, i.e., Objective: Max. R(p) = H (p) / T (p) Subject to: p1 + p2 + …+ pn = 1, pk > 0

Optimization Result:

pA pTpG pC

pG pA, G,C /A,T

K = max R(p) = (log2 1/pA) /A,T

Optimization Result:

pA pTpG pC

pG pA, G,C /A,T

K = max R(p) = (log2 1/pA) /A,T