Quote:

Original post by TimkinQuote:

Original post by Jotaf

You're assuming that a # means there's a 50% chance of occuring a 1 (P(1)=50%) and 50% of a 0 (P(0)=50%).

No, I'm not assuming that. I'm simply saying that '#' is a wildcard, meaning that any gene depicted with that allele in the schema can take on any of its allele values in the chromosome. As to the probability of each allele value, that obviously depends on the frequency of specific allele values in the (sub)population that the schema represents (so I'm not assuming 2 or even 3 members... I'm not assuming any number at all).

Ok. Let me rephrase.

Choose 2 parents in the population, P1 and P2.

For any given allele, if it has the same value in both barents, crossing-over will always chose that value. If the values are different (eg. P1=0 and P2=1), and assuming your random number generator is good, there's a 50% chance that one will be chosen over the other. The schema for P1=1101 and P2=1000 is 1#0#, and each # has a 50% chance of being replaced by a 0 or a 1.

Now choose 3 parents. 1/3 probability of choosing a value from either one. Sure, there's a lot more # in any genotype: P1=1101, P2=1000, P3=0101 -> ##0#.

It seems, as you suggested, that this is a worse representation since there are many more random elements present. However, hidden in there are the probabilities I mentioned:

1st allele: P1=1, P2=1, P3=0 -> P(0)=33%, P(1)=67%

2nd allele: P1=1, P2=0, P3=1 -> P(0)=33%, P(1)=67%

4th allele: P1=1, P2=0, P3=1 -> P(0)=33%, P(1)=67%

As you can see, the result is biased towards having more Ones, because they appear more often in those positions in the parents. Actually, as you take into account more and more parents, the result is more likely to be the "weighted average" of the whole population. Interesting :)

It depends on the specific application whether this is good or bad. But more parents isn't generally "bad" in all cases like you're sugesting.