ANALYSIS

It is assumed that backcrossing is practiced recurrently to a completely inbred line. The nonrecurrent and recurrent lines are assumed to be homozygous for different alleles at QTL of interest. A total of M families (lines) are maintained independently. In each generation, in each backcross family, n individuals are recorded for the quantitative trait, and the highest scoring individual is selected and backcrossed again to the recurrent line. Backcrossing and selection are continued for t generations, F₁ being generation 0. Variation that is not explained by the QTL, because of within-family environmental and residual genetic deviations in the trait (likely to be significant only in early backcross generations) is assumed to be normally distributed, N(0,1). Effects, a, of QTL are expressed in units of this within-family (mainly environmental) standard deviation.

Single locus: This simple example serves as a reference for others. It is assumed that a QTL with allele A from the nonrecurrent and A′ from the recurrent parent confers an increase in the heterozygote of a SD on the trait, and that it is unlinked to any other QTL affecting the trait. It is necessary to compute the probability P(n,a) that the offspring of generation t + 1 selected from a heterozygous backcross parent of generation t is itself heterozygous for the locus. This probability can be readily computed using binomial probabilities for the number k of heterozygous offspring among the n recorded and order statistics for the probability that the highest scoring offspring is heterozygous. For example, if one of the k heterozygotes has phenotypic value x for the trait with probability ϕ(x)dx and is the highest in the family, it implies that k − 1 heterozygotes and n − k homozygotes have performance less than x, the probabilities for each individual being Φ(x) and Φ(x + a), respectively, where ϕ(x) and Φ(x) denote the density and cumulative distribution functions, respectively, of the standardized normal distribution. Then, using the method of Hill (1969), the probability an AA′ heterozygote is selected is as follows: P(n,a)=∑k=1nn!k!(n−k)!2nk∫−∞∞[Φ(x)]k−1[Φ(x+a)]n−kϕ(x)dx, which reduces to P(n,a)=(n∕2n)∫−∞∞[Φ(x)+Φ(x+a)]n−1ϕ(x)dx. (1) In the simplified Equation 1, for each of the n individuals in the family, there is a probability of one-half that the individual is an AA′ heterozygote and, if it is AA′, a probability of [Φ(x)/2 + Φ(x + a)/2]ⁿ⁻¹ that all other family members have a lower phenotypic value. For a = 0, the integral in Equation 1 reduces to 2ⁿ⁻¹/n and P(n,a) = ½, as expected for a neutral gene, since the highest scoring individual is equally likely to be AA′ or A′A′; and for a → ∞, the integral reduces to (2ⁿ − 1)/n and P(n,a) = 1 − (½)ⁿ, which is the probability that at least one of the offspring is a heterozygote and available to be selected. Also, the probability that a homozygote A′A′ is selected is given by P(n − a) = 1 − P(n,a). Equation 1 can readily be evaluated using Simpson's rule numerically. Results are given in Table 1, and these and later results obtained using order statistics were checked by Monte Carlo simulation.

These values can be approximated for small values of a, by P(n,a)∼0.5+ia∕4, (2) where i is the standardized selection intensity for the normal distribution with finite numbers recorded. For n = 2, 3, 4, 6, 8, 12, and 20, i = 0.56, 0.85, 1.03, 1.27, 1.42, 1.63, and 1.87, respectively (Falconer and Mackay 1996). Comparing these values with those in Table 1, it is seen that the approximation is adequate for a ≤ 0.5.

Repeated backcrossing: In the first few generations of backcrossing, the variance is likely to be inflated by segregation at other loci in a way that the accuracy of selection and probabilities of retention will be lower than those shown in Table 1, where the effect of the gene is expressed in terms of the within-family environmental SD. This problem is visited again later, but first, let us assume for simplicity that the (unit) within-family variance remains constant. The probability P(n,a)_t that the QTL remains segregating for t generations in a line maintained with one parent and the same selection intensity each generation is therefore P(n,a)=[P(n,a)]t (3) If the numbers available for selection, n_t, vary among generations, Equation 3 has to be replaced by the product of the appropriate values or, equivalently, the harmonic mean of P(n_t,a) used. Unless n_t varies greatly, simply using an approximate mean of n in Equation 3 suffices. For example, if n_t takes values of 2, 3, 4, 5, and 6 in successive generations and a = 1, the probability the QTL is retained for five generations is 0.19, whereas inserting n = 4 in each generation in Equation 3 gives 0.21.

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Figure 1.

Probability that a QTL of effect a remains segregating for t generations of backcrossing when the best individual of n is selected each generation: P(n,a)_t plotted against t for a = 0, 0.5, 1, 1.5, and 2, and n = 4 and 12.

Some examples using Equation 3 are given in Figure 1. The main point is that the differences between the probability of continued segregation of a QTL of large effect and a QTL of small effect becomes wider as more generations of backcrossing are undertaken and more intense selection is practiced. Of course, if backcrosssing is continued too long, even those of very large effect are lost. In principle, there is some intermediate optimum time for discriminating among QTL of specified effects, if such can be defined.

Since a number of independent replicate lines can be kept, it is useful to reconsider these results in terms of the distribution of the number of lines in which a QTL of specified effect would be segregating. If M lines are maintained, then the expected number in which there is segregation at generation t is MP(n,a)_t. The actual number segregating, m, has a binomial distribution, but the Poisson distribution gives an adequate approximation and results are more readily generalized. Hence, we assume that m has a Poisson distribution with parameter MP(n,a)_t. Some examples are given in Table 2, for experiments in which 10 or 20 lines are maintained. For example, if 10 lines are maintained with selection of one from n = 4 for t = 4 generations, a QTL with effect of 0.25 SD has a probability of <2% of remaining segregating in four or more lines, whereas a QTL with effect of 1.5 SD or more, has a <1% chance of being lost from all 10 lines and a 66% chance of remaining segregating in four or more lines.

Correction for background genetic variation: With segregation at other loci than that being analyzed, there will be additional variation within families, particularly in early generations. With additive genes and genetic variation V_A caused by background genetic variation in the F₂, and assuming for simplicity that the background variation is caused by very many unlinked loci, each of very small effect, there will be V_A/2 in the first backcross and V_A/2^t in the tth backcross. There will also be variance a²V_E/4 caused by the QTL under consideration in the first backcross, which with additive gene action implies a²V_E/2 in the F₂, where V_E is the within-family environmental variance. Consider a simple case, where V_A = V_E and a = 1, so the total genetic variance in the F₂ would be 3V_E/2, and the within-family environmental plus background genetic variance in the backcross would be [1 + (½)^t]V_E. Hence, the effect of the QTL in environmental SD units would be 0.816, 0.894, 0.942, … in backcross generations t = 1, 2, 3,… For example, P(4, 0.816) = 0.698, P(4, 0.894) = 0.714, and P(4, 0.942) = 0.724, whereas P(4,1) = 0.735. Hence, the probability that the QTL remains segregating to generations 1, 2, 3, and 4 is 0.697, 0.498, 0.361, and 0.263, respectively, whereas the equivalent values assuming that P(4,1) is appropriate each generation are 0.735, 0.541, 0.397, and 0.292, respectively. It seems that the quantitative differences are rather small, and that the simple calculations are adequate unless the segregation variance, V_A, is much larger than V_E (i.e., heritability in the F₂ considerably in excess of one-half) and if, because of the selection, it declines much more slowly than one-half per generation, i.e., there are other QTL of large effect or linked in coupling phase.

Two loci: The usefulness of the method depends not only on keeping QTL of large effect segregating, but also on losing those of small or negative effect that may be maintained by linkage to a QTL of large effect. Let us consider a model where there are two additive (i.e., nonepistatic) QTL, A₁ and A₂, on the same chromosome, with effects a₁ and a₂ within family standard deviations, respectively. The recombination fraction is r between the loci. In any generation where the backcross parent is a double heterozygote, A₁A₂/A′₁A′₂, there are four possible offspring genotypes selected in the next generation: the double heterozygote with both A₁ and A₂ present, i.e., A₁A₂/A′₁A′₂ with probability P¹²(n,a₁,a₂), or the single heterozygote with only A₁, i.e., A₁A′₂/A′₁A′₂ with probability P^12′(n,a₁,a₂), or only A₂, or both lost. An extension of Equation 1 can be used. For example, the probability that the double heterozygote is selected is given by P12(n,a1,a2)=[n(1−r)∕2n]∫−∞∞[(1−r)Φ(x+a1+a2)+rΦ(x+a2)+rΦ(x+a1)+(1−r)Φ(x)]n−1ϕ(x)dx. (4) Similar equations apply for sampling the other genotypes, the term in (1 − r) before the integral being replaced by r for recombinant types. If only one QTL remains segregating, in subsequent generations its behavior in that line is described as would be for single loci (1).

Examples are given in Table 3 of the probabilities for a series of examples in which the sum of the effects of the two loci are the same (a₁ + a₂ = 1), but their relative sizes differ (a₁ = 0.5, 0.75 and 1.5). Results for a₁ = 1 and a₂ = 0 can be obtained from Table 1 by noting that P ¹²(n,1,0) = (1 − r)P(n,1), where P(n,1) is given by Equation 1. If linkage is loose, it is seen that the extreme cases of equal effects (a₁ = a₂) and a₂ < 0 give quite different outcomes, but as linkage becomes tight, the survival probability of the double heterozygote depends little on the relative size of effects of the two loci.

These probabilities can be approximated, providing values of ia₁ and ia₂ are not too large, for example: P12(n,a1,a2)∼(1−r)[1+i(a1+a2)∕2]∕2,P12′(n,a1,a2)∼r[1+i(a1−a2)∕2]∕2, (5) and similarly for P^1′2(n,a₁,a₂). For example, with r = 0.2, these approximations give values of P¹²(4, 0.75, 0.25) = 0.606, P^12′(4, 0.75, 0.25) = 0.126, and P^1′2(4, 0.75, 0.25) = 0.074, compared with the exact values of 0.594, 0.120, and 0.072, respectively, in Table 3.

The examples in Table 3 are for nonepistatic loci, i.e., with additive effects in heterozygotes over loci. Equation 4 changes in a straightforward way if this is not the case, and probabilities that one or both of the loci continue to segregate change correspondingly. For example, consider the case where the double heterozygote is 1 SD superior to each single heterozygote and the double homozygote. For complete linkage (r = 0), results are therefore the same as in each example in Table 3, whereas for r = 0.2, P¹² = 0.6376, P^12′= P^1′2 = 0.0604, and P^1′2′ = 0.2416, and for unlinked loci (r = 0.5), P¹² = 0.4502, and P^12′ = P^1′2 = P^1′2′ = 0.1832, i.e., single heterozygotes are less likely to be selected than in the additive case.

To consider the passage over several generations of each of the genotypic classes, it is necessary to include the probabilities of the single locus segregants. We construct the 3 × 3 transition matrix B, for which the rows and columns identify the following states: (1) A₁ and A₂, (2) A₁ but not A₂, and (3) A₂ but not A₁; the elements b_i,j specify the transition probability from state i at generation t to state j at generation t + 1. (Alternatively, a 4 × 4 matrix can be used, with the fourth row and column denoting the case where neither A₁ nor A₂ are segregating; because this is an absorbing state, the computation of segration probabilities are not affected.) Hence, using the single locus formulas from the preceding section, B=(P12(n,a1,a2)P12′(n,a1,a2)P1′2(n,a1,a2)0P(n,a1)000P(n,a2)) (6) Assuming the population starts in the state where both A₁ and A₂ are segregating, from Equation 6, it can be shown that at generation t, P12(n,a1,a2)=b11t, (7a) P1(n,a1,a2)t=b12(b11t−b22t)∕(b11−b22),and (7b) P2(n,a1,a2)t=b13(b11t−b33t)∕(b11−b33). (7c)

For the example given in Table 3, results for segration probabilities for four and eight generations are given in Table 4. Although the probability that both loci remain segregating is not greatly affected by the relative magnitude of the gene effects (and no probabilities of retention are high in this example because the total effect is only a₁ + a₂ = 1 and n = 4), the QTL of smaller or negative effect has a low probability of remaining segregating alone for many generations, so the method does have some discriminating power.

Marker segregation: The previous analyses have been restricted to the fate of the QTL, but their segregation has to be detected by means of molecular markers. The calculations for individual markers are straightforward and are simply a special case of the two-locus analysis given above. Let us assume that A₁ is the QTL and A₂ is the marker, i.e., a₁ = a and a₂ = 0, with the recombination fraction between the loci equal to r. Then the elements of B are given by b11=(1−r)P(n,a),b12=rP(n,a),b13=r[1−P(n,a)],b22=P(n,a),b33=1∕2 where P(n,a) is given by Equation 1. An example for a marker locus is given as part of Table 3 (a₁ = 1, a₂ = 0). The probability, from Equations 7a and 7c, respectively, that the marker remains segregating in coupling with the QTL is P12(n,a,0)t=[(1−r)P(n,a)]t the probability it remains segregating without the QTL is P1′2(n,a,0)t=[r(1−P(n,a))]{(1∕2)t−[(1−r)P(n,a)]t}∕{1∕2−[(1−r)P(n,a)}, and their sum, P¹²(n,a,0)_t + P^1′2(n,a,0)_t, is the overall probability the marker is retained. Because the probabilities that QTL are retained for many generations are already small unless the QTL has an effect as large as 2 SD or so (Figure 1), it is clear that only very tightly linked markers are likely to be of value in QTL detection. An illustration is given in Table 4 (a₁ = 1, a₂ = 0). Hence, the marker analysis after backcrossing needs to be done with very closely spaced markers in a genomewide scan.

If there are two QTL, the fate of alleles at a marker locus, say A₃, depends on whether it is between or outside this pair. If A₃ is outside the interval A₁–A₂, then its probability of segregation is given by expanding the formulas given in Equation 3. If A₃ lies outside the interval, but nearer to A₂, for example, the probability that all three loci remain segregating is given by (1 − r₂₃)P¹²(n,a₁,a₂). If A₃ lies between A₁ and A₂, then, for example, the probability that all three loci remain segregating is [(1 − r₁₃)(1 − r₂₃)/(1 − r₁₂)]P¹²(n,a₁,a₂). The matrix has to be formally extended to consider seven possible classes (all three loci, three pairs, and three singles), but the calculations are straightforward.

Multiple linked QTL: The preceding analysis is concerned with the outcome of backcrossing when there are only one or two QTL on the chromosome affecting the trait under selection. An alternative model is that the difference in performance between the recurrent and nonrecurrent parent lines are caused by many QTL of (mainly) small effect on each chromosome. Some examples have been considered using the Monte Carlo simulation, with a model of a chromosome of map length LcM and typically 21 loci simulated at equal spacing, with the most distant at the ends of the chromosome. Ten of these loci, numbers 1 (i.e., at position 0.05L), 3,…, 19, were assumed to be QTL of equal effect, and the remaining 11 loci, numbers 0, 2,…, 20, were assumed to be markers with no effect on the trait. The probabilities that individual QTL remain segregating do not differ greatly whether or not they are near the middle or the end of the chromosome, unless the chromosome is of length 200 cM or more. Similarly, the probability that individual markers remain segregating is little different from that of the QTL between which they lie. Hence, only summary figures are given for the examples in Table 5. If the chromosome is short (L < 100), the probability that loci on it remain segregating is quite substantial, even if the individual loci have effects of 0.4 SD or less. If it is long (L > 200), the probability of continued segregation is a little higher than for neutral genes (6% in the example of t = 4 generations), even if the individual effects are as large as 0.4 SD and the total difference between the ancestral chromosomes is 4 SD. In general, there will be greater discrimination if selection is more intense and the numbers of generations are longer than the examples in Table 5 (n = 4, t = 4). A model with very many QTL of very small effect in coupling would give similar results to those in Table 5, as seen by the similar segregation probabilities for QTL and markers.

Unless the chromosome is very long (L > 200) and individual QTL are all of large effect (say a > 1), an alternative extreme model in which there is complete repulsion of QTL, i.e., alternating positive and negative effects on the trait along the chromosome with no net effect of the QTL together on the chromosome, would behave in a very similar way to the case in which all loci are neutral. In general, there will be greater discrimination if selection is more intense and the numbers of generations are longer than the examples in Table 5 (n = 4, t = 4).

Interspersed inter se matings: QTL of small effect, particularly if they are partially recessive to that in the recurrent parent, have a low probability of retention. It is possible to increase the probability of QTL segregation by increasing the strength of the selection in increasing frequency relative to that of backcrossing in reducing frequency. One possible method, which fits within the independent family (line) structure discussed here, is to intersperse a generation of inter se mating between each generation of backcrossing, i.e., to allow two generations of selection per generation of backcrossing. For simplicity, it is assumed that the same family size is used in each case: in the backcrossing generations, the best male (or female) from n of that sex is selected; in the inter se generation, the best male from n males recorded and the best female from n females recorded are selected and mated. The important difference from the previous analysis is that now matings can be made between two heterozygotes so that homozygotes for the QTL of interest may then be selected for the next backcross generation.

In the previous analysis of backcrosses alone, the quantity a was used for simplicity to define the homozygote–heterozygote difference; a fuller definition is now required, and the notation of Falconer and Mackay (1996) is adopted, adding an asterisk to distinguish values from those above: the genotypic values (in within family SD units) are AAa*, AA′ d*, and A′A′ −a*. Hence, in the previous analysis, a = a* + d*. If the parents of the inter se generation are a homozygote and a heterozygote, the calculations given in Equation 1 still apply. If the mating is between two heterozygotes, the probabilities that the individual selected for the next backcross generation is AA, AA′, or A′A′ are readily computed by extending the formula, as exemplified by Equation 4. Let these probabilities be, for example, P^AA(n,a*,d*). The full two-generation process can be described by a transition matrix C, in which the rows and columns denote the genotype of the backcross parent, AA for the first and AA′ for the second (a third row and column could be added for the absorbing state when the backcross parent is A′A′): C=(PAA(n,a∗,d∗)[P(n,a∗+d∗)]2PAA(n,a∗,d∗)PAA′(n,a∗,d∗)[P(n,a∗+d∗)]2{PAA′(n,a∗,d∗)+2[1−P(n,a∗+d∗)]}). In the first row of C, because the backcross parent is AA, the mating for the inter se generation is always AA′ × AA′, so the probabilities c₁₁ and c₁₂ refer to the selection of AA and AA′, respectively, from among their offspring. In the second row, where the backcross parent is the AA′ heterozygote, the probability is [P(n,a* + d*)]² that both the selected male and female for the inter se mating are heterozygote, which leads immediately to c₂₁ and the first term in c₂₂. The second term in c₂₂ is the product of the probabilities, 2P(n,a* + d*)[1 − P(n,a* + d*)], that the inter se mating is between a homozygote and heterozygote, and P(n,a* + d*), that a heterozygous offspring is selected from the mating.

In Table 6, examples are given assuming that n = 4 individuals are recorded per family in both the backcross and inter se mating generations, for different degrees of dominance. The value of t refers to the number of backcross generations completed, and the probabilities given are that the QTL is still segregating, i.e., the individual selected from the inter se generation is either AA or AA′. For reference, to define the rate of loss after a few generations, values of 1 − λ, where λ is the larger eigen value of C, are also given and can be compared directly with the probabilities 1 − P(n,a), which equal 1 − eigenvalue for a 1 × 1 matrix), from Table 1 when only backcrossing with selection is practiced.

The greatest benefits from the inter se mating arise, of course, when the QTL of high value is recessive and therefore neutral during the selection among backcrossed individuals. In the additive case, the interspersed generation does not half the rate of loss (i.e., 1 − λ) unless the effect of the QTL is quite large. Because the time (i.e., total number of generations) required to reduce the probability of segregation of neutral background genes is doubled, it is moot whether there is benefit in inserting the inter se generation. Maintaining more replicate lines or increasing the number of individuals from each family recorded for the quantitative trait, each generation, family size may make more efficient use of resources. Where reproductive rate limits selection intensity, for example in mice, it might be increased by keeping two litters from each female or by selecting only among males that have been given multiple matings.