12 Sept. 1964
Dear Emile:
Let the chance of a mutation effecting one amino-acid site in time t be αt.
For N sites the total number of mutations is about Nαt.
Let us consider the time t, at which Nαt mutations have occurred. We assume equal
probability for all sites.
The probability that the first site is unchanged after the first mutation is
P1 = N - 1/N = 1 – 1/N
and after Nαt mutations is
PNαt = PiNαt = (1-1/N)Nαt
lin
N→∞ (1 – 1/N)N = 1/e
Hence PNαt = (1/e) αt = At, A = (1/e) α
Let us use the expression P(t) as average probability of unchanged site at time
t.
We see that P(nτ) = Ptn
t P
τ 0.90
2τ .81
3τ .729
4τ .656
6τ .532
8τ .430
10τ .348
12τ .282
14τ .228
16τ .185
18t .150
20τ .122
The prob. that the first site is unchanged after N & t mutations is (1/e) αt
= e –αt
The prob. that it is changed is 1-e-αt
The number changed is N-λ = N(1-e-αt)
The number of mutations is Nαt = r
N-λ = N(1-e-αt/N) λ= no. not changed
N-λ/N = 1-e-αt/N
λ/N = e-r/N
-r/N = ln (λ/N)
r/N = ln N/λ
N-λ = 30, N = 70[?]
r/N = 2.303 log 100/70 = 2.303 x 0.152 = 0.35
r = 35 OK
I ignore back mutations, because of uncertainty of their probability and because
accuracy of experimental points does not justify a refinement of the theory.
Consider a young protein (Hb, Mb) with all sites changeable.
Differences amid man : (horse, cow, etc.) separation epoch are
α man – α horse 17
β “ - β “ 26 average 21.5
β “ - β cattle 27
21.5/148.5= .15 We assume this epoch to be at -80M years.
The average α – β and α – t change is 55%. On curve this gives α – β separation
at – 375 M years.
The average man – (horse, etc) cytochrome c change is 10% of total residues,
rather than 15% (Hb).
Of the two obvious explanations, 1, lower mutation rate for all residues, and
2, very low mutation rate for certain functionally unimportant residues in this old
protein and normal for others, I accept the latter.
The 10% difference becomes 15% if 1/3rd of the chain is immutable.
The difference 39 residues for man—yeast is [39 / (2/3 x 104)] x 100 = 56%
On this basis, epoch – 350M years. This is not acceptable.
[page 5, graph]
R = n ln n/ λ
αt = N ln N/ λ = N
N/ λ = e(αt/N) λ = Ne- (αt/N)
No. of changes / N = ln (N / no. of apparent changes)
N = 60 λ=30
60/30 = N2.303 log 2 = 2.303 x 0.301 = 0.69.60 = 41
Cytochrome c 10 out of 50 in 200 M years = prob. Per year is 10-9
20 “ ” 100
30 “ “ 150
30 “ “ 150 Agrees roughly with Hb- 22 out of 144 in 80M
100% in 520 M years, vs 1000M for cyt. C
Cytochrome c seems to have some immutable loci. For the others, rate is about
half that of the “young” α + β of Hb.
Back mutations
Coincidences.
