We will discuss a solution to this problem here.
Let’s try to understand the problem by illustrating the example case
- AA-AA: 1
- AA-Aa: 0
- AA-aa: 0
- Aa-Aa: 1
- Aa-aa: 0
- aa-aa: 1
Solution: 3.5, which is expected number of offspring containing dominant allele “A”
Let’s see how.
It is guaranteed that the pair AA-AA will produce only AA. Since every couple has exactly two offspring, so expected number is 2.
In the case of Aa-Aa, 0.25 possibilities of getting AA, 0.5 possibilities of getting Aa, and 0.25 possibilities of getting aa. Again, since every couple has exactly two offspring, the expected number of AA, Aa, and aa are 0.5, 1, and 0.5 respectively. So, the expected number containing “A” is 0.5 + 1 = 1.5.
For aa-aa, we don’t even need to consider this case since it will produce only aa, and will have no effect on the expected number of offspring containing “A”.
So, total expected number: 2 + 1.5 = 3.5
Any questions — I’m just a comment away!