Calculating Heterozygous alleles


After assuming the black mice are dominant (BB) and white mice are recessive (bb),
I keep getting 9, which is not in the option.

The answer is H. Thank you!

hi since we are looking fo the scenario with the maximum amount, every black mouse could potentially be heterozygous
bb x Bb gives four Bb children
Bb x Bb gives four Bb children
Bb x bb give two Bb children for a total of 12


Umm… I’m not sure that’s quite right.
bb x Bb does not give 4 Bb offspring, it actually gives 2 Bb mice and 2 bb mice.

Also, Bb x Bb does not give 4 Bb children. It gives 1 BB, 2 Bb, and 1 bb mice.

your third cross is correct. Notice how it’s technically the same as the first one, but you got two different outcomes.

In addition, how does 4 and 4 and 2 equal to 12?

Anyways, I think winter solved the question correctly but somehow miscalculated the number of mice.

In the first cross it is obvious that the white mouse is recessive, since its offspring with the black mouse are all black. And that indicates that the black mouse has to be pure dominant (BB). Here’s why:

And since we’ve established that the second mouse from left to right is pure dominant, we won’t be able to change it in the next cross.

In the next cross, the other black mouse (3rd in order from left to right), is heterozygous-dominant since its offspring with the white mouse (the coat we’ve established to be recessive) include recessive mice (white) which is only possible if the black mouse was heterozygous, (Bb).

Therefore, the second cross, BB x Bb should look like this:

Lastly, the third and final cross, is between Bb and bb. We have already discovered their alleles in previous crosses. So Bb x bb is going to go like this:

To reiterate, we get 4 heterozygous mice in the first cross, 2 in the second, and 2 in the last.

So the answer is 8.

Hi i understand your reasoning, but the pheno ratio is only a probability

Bb x Bb doesn’t necessarily give 2 Bb mice and 2 bb mice in real life, it gives a probability of 1/2 Bb and 1/2 bb
Two Bb parents could give birth to 4 Bb mice, it’s less statistically likely, but still possible

Here is an easier example:
XX x XY give 2 XX and 2 XY in the punnet square, so there is 1/2 chance of having a girl and 1/2 chance of having a boy
However, if they have 4 children, they could all be girls or all be boys, it doesn’t have to be the same as the punnet square predictions

1 Like

But if you don’t follow the most likely statistics, how else are you gonna know?

The key word in the question is “what is the maximum possible”
This means that we need to find the hypothetical scenario that will give us the max number of heterozygous mice, even if that scenario is very very unlikely

Ari calls these tricky questions “is it real questions”, he goes more into depth in this video:


You are right. Thanks for the explanation, you’ve convinced me.

1 Like