genetics - Two mice heterozygote and black are reproduced. Find the probability of getting the filium dark and heterozygote

Well, here's what I did: P: Aa*Aa

F1: 1/4 AA, 1/2 Aa, 1/4 aa, SO the answer is 1/2, but out teacher did this: AA(1) Aa(2) aa(1) and she said we divide the number of filium for the phenotypes number and get 2/3. This is the probability. I don't get it.

Answer

Okay so in your teachers answer there are four offsprings. . You used all offsprings when you made the division so 2/4 = 1/2.Your teachers 2/3 comes from the fact that 3 out of four is black, but only 2 out of that 3 black is heterozygote so 2/3 is good by this logic. I think that your teacher's logic faulty because you have to take into consideration all the offsprings. Basic genetics and Mendel's law tells us that crossing two heterozygotes with independent alleles result in a 3:1 ratio in phenotype separation in the next generation with 1:2:1 genotype separation -- that is what basically both of you have wrote. The fault is in the getting the probability - basic maths tells us that in an independent experiment, the probability variables have to sum up to one. In your teacher's logic this basic rule is not kept, because finding homozygous non-dark offspring would be one (there is only one non-dark filium and that one is homozygous). So by this logic all filium are non dark-homozygous, but then what about the others. I hope you see the problem with that logic.