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Gametic Probabilities for Three Linked Loci
The gametic products resulting from self-fertilization of an AaBbCc F1 plant when there is coupling-phase linkage between all three loci and the gene order is A-B-C are provided. For purposes of the numerical example let rab = rbc = 0.2. Double crossovers exist, but there is an assumption of no interference.
Table 1. Gametic products, probabilities and numerical example
for three linked loci.
| |
Observed |
Expected Frequency |
Gametic genotype |
Count |
Probability |
Numerical Example |
ABC |
f1 |
½ (1-rab)(1-rbc) |
0.32 |
ABc |
f2 |
½(1-rab)rbc |
0.08 |
AbC |
f3 |
1/2(rabrbc) |
0.02 |
Abc |
f4 |
1/2rab(1-rbc) |
0.08 |
aBC |
f5 |
1/2rab(1-rbc) |
0.08 |
aBc |
f6 |
1/2(rabrbc) |
0.02 |
abC |
f7 |
½(1-rab)rbc |
0.08 |
abc |
f8 |
½ (1-rab)(1-rbc) |
0.32 |
The P(ABC) = (1 - rac)/2 because the P(ABC and abc) = (1 -rac). We know that
(1 - 2rac) = (1 - 2rab)(1 - 2rbc) which simplifies to rac = rab + rbc - 2rabrbc = 0.2 + 0.2 - 0.8 = 0.32.
The probability of recombination between the A and C loci = rac = 0.32
= P(ABc) + P(Abc) + P(aBC) + (abC)
= 0.08 + 0.08 + 0.08 + 0.08.
The reason that P(ABC) does not equal
is that the probability that a gamete contains the A-B combination is not independent of the probability that the same gamete will contain the B-C combination. If a gamete has the
A-B combination, the probability that it will have the B-C combination is increased.
Table 2. Numerical example of the proportion of each gametic type for three loci considered jointly.
|
BC |
bc |
Bc |
bC |
AB |
0.32 |
---- |
0.08 |
---- |
ab |
---- |
0.32 |
---- |
0.08 |
Ab |
---- |
0.08 |
---- |
0.02 |
aB |
0.08 |
---- |
0.02 |
---- |
|
0.4 |
0.4 |
0.1 |
0.1 |
Table 3. Probability of each gametic type, assuming pair-wise independence.
| |
BC (0.4) |
bc (0.4) |
Bc (0.2) |
bC (0.2) |
| AB (0.4) |
0.16 |
0.16 |
0.08 |
0.08 |
| ab (0.4) |
0.16 |
0.16 |
0.08 |
0.08 |
| Ab (0.2) |
0.08 |
0.08 |
0.04 |
0.04 |
| aB (0.2) |
0.08 |
0.08 |
0.04 |
0.04 |
The results of Table 3 would suggest that the probability that a gamete has the AB and bc genotype is 0.16, assuming independence. However, the probability that a gamete has both the B and b alleles in the same gamete is zero.
The probability of an ABC gamete is zero if there has been recombination between the A and B loci. In other words, the probability of the ABC gamete is zero when the gamete has the Ab or aB genotype. Therefore, the probability of an AB gamete is not independent of the probability of a BC gamete. The probability of an ABC gamete is the P(AB gamete) X P(BC gamete/AB). An ABC gamete can only occur when there has been no recombination between the A and B loci. The probability of an ABC gamete depends on whether the AB combination is present.
Table 4. Gametic products, probabilities and numerical example for the A-B loci.
Gametic genotype |
Expected frequency |
Numerical example |
AB |
1/2(1-rab) |
0.4 |
ab |
1/2(1-rab) |
0.4 |
Ab |
1/2rab |
0.2 |
aB |
1/2rab |
0.2 |
If we only consider the B and C loci, we can determine the expected proportions of each gamete and these are provided in Table 5.
Table 5. Gametic products, probabilities and numerical example for the B-C loci.
Gametic genotype |
Expected frequency |
Numerical example |
BC |
1/2(1-rbc) |
0.4 |
bc |
1/2(1-rbc) |
0.4 |
Bc |
1/2rbc |
0.2 |
bC |
1/2rbc |
0.2 |
An ABC gamete cannot occur unless there is no recombination between the A and B loci and no recombination between the B and C loci. However, the probability of an ABC gamete is not equal to
The probability of an AB gamete is not independent of the probability of a BC gamete, this is why we cannot multiply the separate probabilities to get the probability of an ABC gamete.
The combination of ab and BC loci cannot occur. The combination of AB and BC loci can occur. The probability of an ABC gamete depends on whether the AB combination is present.
P(ABC gamete) = P(AB gamete) P(BC gamete/AB).
P(AB) = ½(1-rab) = 0.5(1 - 0.2) = 0.4.
P(BC/AB) = P(AB and BC)/P(AB)
= [½ (1 - rab)(1 - rbc)]/ [1/2(1 - rab)]
= (1 - rbc) = 0.32/0.4 =0.8
P(AB and BC) = P(ABC) = P(AB gamete) P(BC gamete/AB)
= 0.4(0.8) = 0.32.
However, the probability of an ABC gamete is not equal to
(1-rab) |
(1-rbc) |
= 0.4(0.4) = 0.16 |
2 |
2 |
The probability of an ABC gamete is 0.32, not 0.16. The probability of an ABC gamete would be 0.16 only if the probability that a gamete has the A-B combination was independent of the probability that a gamete had the B-C combination.
P(ABC gamete) = P(AB gamete) P(BC gamete/AB)
| |
= [½(1-rab)] [½ (1 - rab)(1 - rbc)]/ [1/2(1 - rab)] |
| |
= [½ (1 - rab)(1 - rbc)]. |
If the probability of a BC gamete was independent of the probability of an AB gamete, then
P(BC) = P(BC/AB). P(BC) = 1/2(1- rbc).
| P(BC/AB) = P(AB and BC)/P(AB) |
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= [½ (1 - rab)(1 - rbc)]/ [1/2(1 - rab)] = (1 - rbc). |
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We can see that
P(BC) = P(BC/AB) because 1/2(1- rbc) = (1 - rbc).
If P(BC) and P(AB) were independent, then it follows that P(BC) and P(Ab) should be independent. If P(BC) and P(Ab) were independent, then
| P(BC and Ab) = P(BC) P(Ab) = 1/2(1- rbc)½ rab |
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| |
= (½)(0.8)(½)(0.2) = 0.04. |
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However, the P(BC and Ab) = 0 because a gamete cannot have both the B and b alleles.
If a gamete contains the B allele it cannot also have the b allele.
Copyright
2000©, Ted Helms
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