I am confused about how to compute probabilities for linked genes.
The problem that I am solving is this:
- P generation: AAbbddEEff X aaBBDDeeFF. If A and B are 40 centimorgan apart, what is the probabilty to get aabbddeeff in F2?
Here is what I have:
- F1 is: AaBbDdEeFf.
- The next cross it goes through is this: AaBbDdEeFf X AaBbDdEeFf.
- To compute this cross, I drew the 2 chromosomes for AaBb (in coupling), and then the probability for a is 0.5 (A or a) and the possibility for no recombination is 0.6. So 0.5 x 0.6 = 0.3. The same goes for b, so aabb is: 0.3 x 0.3 = 0.09.
- For ddeeff: 0.25x0.25x0.25 = 0.015625.
- Multiplying these gives 0.09 x 0.015625 = 0.00140625.
- Extending that and for 2 gametes gives this: (0.00140625)^2 = 0.00197x10^-3.
But the correct answer is not what I have computed, but instead is 0.000625.
I'm really confused with this question. If we know that only 2 of the 5 genes are linked, is it still correct to multiply: 0.09 x 0.015625 = 0.00140625? Maybe I should use a Punnett square somehow? Because in that case, the probability for a recessive homozygous is 0.0625 but I'm not sure it fits to this question.