Accepted 5 January 2015

Published online 29 January 2015 in Wiley Online Library

(wileyonlinelibrary.com) DOI: 10.1002/sim.6429

Random recombination and evolution of drug resistance Alan Kleinman*† The effects of random recombination on the random mutation and natural selection phenomenon can be understood by considering the mathematical behavior of this phenomenon. This phenomenon operates in a mathematically predicable behavior, which when understood, explains the empirical observations of this phenomenon. The mathematical behavior of random recombination is derived using the principles given by probability theory. The derivation of the equations describing the random recombination phenomenon is done in the context of an empirical example. Copyright © 2015 John Wiley & Sons, Ltd. Keywords:

random recombination; evolution; probability theory; drug resistance; mathematics

1. The mathematical model of random recombination Random recombination differs from non-random recombination. Random recombination occurs when the parent chromosomes meet in a random manner such as what occurs with the recombination associated with HIV and the pollination of weeds. On the other hand, non-random recombination occurs in breeding programs and other selective mating situations. This research article addresses the mathematics of random recombination because this case is of importance in the evolution of HIV drug resistance, herbicide resistance, pesticide resistance, and other replicators that do recombination. The mathematical equations of random recombination are obtained from the text, Advanced Engineering Mathematics [1] by Erwin Kreyszig. The equations used are based on the empirical example of the random recombination of HIV. The mathematical model is based on this empirical example, but the principles are generally applicable to other randomly recombining replicators. 1.1. The empirical example of random recombination The empirical example that we will use to frame the equations that describe random recombination is the random recombination of HIV [2, 3]. Consider the following: Assume that we have a population of HIV viral particles in a population that is doing random recombination. Assume that the population size is size ‘n’. In this population, assume that some members of the population have a beneficial allele for a protease inhibitor, call this allele ‘A’. Other members of the population have a beneficial allele for a reverse transcriptase inhibitor, call this allele ‘B’. The remaining members of the population have neither allele A nor allele B, call these non-A and non-B alleles ‘C’. If a member of the population with beneficial allele A at one genetic locus meets another member of the population with beneficial allele B at a different genetic locus, we assume that these members will recombine giving an offspring with both alleles A and B and thus, a more fit member of the population. The probability that a member of the population with allele A meeting and recombining randomly with a member with allele B can then be computed using the principles of probability theory. The probability distribution for this empirical example is identical to the probability distribution given for random card drawing when there are only three different cards in the deck, A, B, and C.

1977

PO Box 1240 Coarsegold, CA 93614, U.S.A. PO Box 1240 Coarsegold, CA 93614, U.S.A.

*Correspondence to: Alan Kleinman, † E-mail: [email protected]

Copyright © 2015 John Wiley & Sons, Ltd.

Statist. Med. 2015, 34 1977–1980

A. KLEINMAN

1.2. The mathematics of random recombination To derive the mathematical behavior of the empirical example of the random recombination, we first define some terms. n nA nB nC

– is the total population size. – is the number of members in the population with beneficial allele A. – is the number of members in the population with beneficial allele B. – is the number of members in the population that have neither beneficial allele A nor beneficial allele B.

Assume that in this population n that members in the population can randomly meet and recombine. What are the probabilities for these members to randomly meet and recombine? First, one needs to recognize that there are multiple different permutations of selecting the individual members in the population. In this case, we are selecting two members of the population from three classes (members with beneficial allele A, members with beneficial allele B, and members that have neither beneficial allele A nor B, those are members C). The appropriate equation for computing the number of permutations for this case is given by Theorem 2 on page 718 of Reference [1]. Then it must be noted that we are dealing with the probabilities of multiple random variables. Therefore, the probability function that describes this selection process is given as follows: f (x, y) =

( n )x ( n )y ( n )2−x−y 2! C A B x!y!(2 − x − y)! n n n

(x + y ⩽ 2)

(1)

and (2)! . (x)!(y)!(2 − x − y)!

(2)

In this probability function, we assume that n ≫ 1 so that selection of a single individual has little effect on the value of n and the frequency of individual variants in the population. The first term Equation (2) in this probability function Equation (1) takes into account the different possible permutations of ( ) ( ) ( ) selecting A’s, B’s, and C’s particular members from the population. The nA ∕n , nB ∕n , and nC ∕n terms are the frequencies of each of the variants in the population. We also assume that when each of the members are selected, the correct crossing over and recombination will always occur. So if a member with allele A is selected with a member with allele B, the resulting offspring will have recombined alleles A and B. And likewise, if a member with allele B is selected with a member with allele C, the offspring will have recombined alleles B and C, and so on for the other possible selections. In addition, we have the following condition: nA + nB + nC = n.

(3)

If we set x = y = 1, we can compute the probability of selecting an A and a B variant from the population. Equation (1) then reduces to f (x = 1, y = 1) = P(A, B) = 2

(n ) (n ) A B . n n

(4)

1978

The factor ‘2’ is included because of the two possible permutations of selections of an A and a B variant or a B(and an ) A variant. ( ) Immediately, we know that the maximum probability for Equation (4) occurs when nA ∕n = nB ∕n = ½. This occurs when half the population has allele A and the other half of the population has allele B, and there are no members that have neither allele A nor B. In this case, the probability of a member with allele A meeting and recombining with a member with allele B is 0.5. The other two possible selections would be an A variant meeting and recombining with another A variant and a B variant meeting and recombining with another B variant. Neither of those last two recombination possibilities would give the improved fitness of an A and a B variant meeting and recombining. Following is a graph for Equation (4) for various values of nA ∕n and nB ∕n. To examine the detail of P(A,B) when the frequencies of alleles nA and nB are less than 0.1, consider the data in Table I. Copyright © 2015 John Wiley & Sons, Ltd.

Statist. Med. 2015, 34 1977–1980

A. KLEINMAN

Graph 1. P(A,B) as a function of nA ∕n for various values of nB ∕n.

Table I. P(A,B) for various values of nA ∕n and nB ∕n ⩽ 1E − 01. na ∕n|nb ∕n 1E − 01 1E − 02 1E − 03 1E − 04 1E − 05 1E − 06 1E − 07 1E − 08 1E − 09 1E − 10 1E − 01 1E − 02 1E − 03 1E − 04 1E − 05 1E − 06 1E − 07 1E − 08 1E − 09 1E − 10

2E − 02 2E − 03 2E − 04 2E − 05 2E − 06 2E − 07 2E − 08 2E − 09 2E − 10 2E − 11

2E − 03 2E − 04 2E − 05 2E − 06 2E − 07 2E − 08 2E − 09 2E − 10 2E − 11 2E − 12

2E − 04 2E − 05 2E − 06 2E − 07 2E − 08 2E − 09 2E − 10 2E − 11 2E − 12 2E − 13

2E − 05 2E − 06 2E − 07 2E − 08 2E − 09 2E − 10 2E − 11 2E − 12 2E − 13 2E − 14

2E − 06 2E − 07 2E − 08 2E − 09 2E − 10 2E − 11 2E − 12 2E − 13 2E − 14 2E − 15

2E − 07 2E − 08 2E − 09 2E − 10 2E − 11 2E − 12 2E − 13 2E − 14 2E − 15 2E − 16

2E − 08 2E − 09 2E − 10 2E − 11 2E − 12 2E − 13 2E − 14 2E − 15 2E − 16 2E − 17

2E − 09 2E − 10 2E − 11 2E − 12 2E − 13 2E − 14 2E − 15 2E − 16 2E − 17 2E − 18

2E − 10 2E − 11 2E − 12 2E − 13 2E − 14 2E − 15 2E − 16 2E − 17 2E − 18 2E − 19

2E − 11 2E − 12 2E − 13 2E − 14 2E − 15 2E − 16 2E − 17 2E − 18 2E − 19 2E − 20

1.3. An empirical example of random recombination

Copyright © 2015 John Wiley & Sons, Ltd.

Statist. Med. 2015, 34 1977–1980

1979

Consider now the empirical example of a well-treated patient suffering with HIV. Assume that the A allele gives improved fitness of replication against the reverse transcriptase inhibitor and the B allele give improved fitness of replication against the protease inhibitor. What is the estimate of the probability for an A variant virus meeting and recombining with a B variant virus giving an offspring with both the A and B alleles giving improve fitness of reproduction against both the reverse transcriptase inhibitor and the protease inhibitor in a well-treated patient with HIV? The viral load in a well-treated HIV patient is less than 50 viral particles per mL [4]. The average plasma volume in an adult human is about 3 L of plasma [5]. This gives 50 viral particles per mL × 1000 mL∕L × 3 L plasma = 150 000 viral particles in a well-treated HIV patient. Thus, we have n = 150 000. This number does not include the extra-vascular viral load, but the computation can be modified for this condition. Table II contains values for P(A,B) as a function of nA for various values of nB generated using Equation (4).

A. KLEINMAN

Table II. Probability of an A member meeting and recombining with a B member as a function of number of members with A allele and number of members with B allele for various frequencies of members. Total population size is n = 150 000. nA

nB

frequency nA

frequency nB

75000 7500 750 75 8 75000 7500 750 75 8 8 8 8 100000 50000 125000 25000 140000 10000

75000 7500 750 75 8 8 8 8 8 75000 7500 750 75 50000 100000 25000 125000 10000 140000

0.500000000 0.050000000 0.005000000 0.000500000 0.000053333 0.500000000 0.050000000 0.005000000 0.000500000 0.000053333 0.000053333 0.000053333 0.000053333 0.666666667 0.333333333 0.833333333 0.166666667 0.933333333 0.066666667

0.500000000 0.050000000 0.005000000 0.000500000 0.000053333 0.000053333 0.000053333 0.000053333 0.000053333 0.500000000 0.050000000 0.005000000 0.000500000 0.333333333 0.666666667 0.166666667 0.833333333 0.066666667 0.933333333

P(A,B) 0.5000000000 0.0050000000 0.0000500000 0.0000005000 0.0000000057 0.0000533333 0.0000053333 0.0000005333 0.0000000533 0.0000533333 0.0000533333 0.0000053333 0.0000005333 0.4444444444 0.4444444444 0.2777777778 0.2777777778 0.1244444444 0.1244444444

1.4. Discussion Does recombination have a significant effect on the evolution of drug resistance in the treatment of HIV? The mathematics of probability theory, and the empirical evidence appears to be saying no. Combination therapy for the treatment of HIV has now shown success for more than two decades. The mathematics of probability theory is telling us why. Unless members of the population with beneficial allele A and other members of the population with beneficial allele B can amplify (increase in frequency), the probability of a member with allele A meeting and recombining with a member with allele B remains small. In the empirical example presented, even if the population of n = 150 000 where nA = nB = 750, the probability of an A member meeting and recombining with a B member is still only P(A, B) = 0.00005. The multiplication rule of probabilities rapidly reduces the probabilities of an A member randomly meeting a B member (as frequencies decrease) to recombine to give a descendent with both A and B alleles. The importance of this is significant when designing combination selection pressures for suppressing the evolution of selection pressure resistant variants. As shown in Reference [6], the multiplication rule of probabilities has significant effect on any evolutionary process by the random mutation and natural selection phenomenon, but the multiplication rule of probabilities also has significant impact on random recombination. Random recombination will only have a significant effect on the evolution of drug resistance against multiple selection pressures when populations have high frequencies of both resistance alleles.

References

1980

1. Kreyszig E. Advanced Engineering Mathematics, Third Edition. John Wiley and Sons Inc.: New York·London·Sydney·Toronto, 1972. 2. Burke DS. Recombination in HIV: an important viral evolutionary strategy. Emerging Infectious Diseases 1997; 3(3): 253–259. http://wwwnc.cdc.gov/eid/article/3/3/pdfs/97-0301.pdf [Accessed on 23 January 2015]. 3. Jetzt AE, Yu H, Klarmann GJ, Ron Y, Preston BD, Dougherty JP. High rate of recombination throughout the human immunodeficiency virus type 1 genome. Journal of Virology 2000; 74(3):1234–1240. http://jvi.asm.org/content/74/3/1234. abstract [Accessed on 23 January 2015]. 4. NAM aidsmap. http://www.aidsmap.com/Viral-load/page/1044622/ [Accessed on 23 January 2015]. 5. Wikipedia. Blood. http://en.wikipedia.org/wiki/Blood [Accessed on 23 January 2015]. 6. Kleinman A. The basic science and mathematics of random mutation and natural selection. Statistics in Medicine 2014; 33(29):5074–5080.

Copyright © 2015 John Wiley & Sons, Ltd.

Statist. Med. 2015, 34 1977–1980