To determine the probability that the baby rabbits will be brown when two brown rabbits are crossed, we need to consider their possible genotypes.
Assuming one brown rabbit has the genotype BB and the other has the genotype Bb, we can set up a Punnett square to find the outcomes:
- BB x Bb:
- Possible gametes from the first parent (BB): B, B
- Possible gametes from the second parent (Bb): B, b
The Punnett square would look like this:
B | b
----------------
B | BB | Bb
----------------
B | BB | Bb
From this Punnett square:
- BB (homozygous brown) is produced in 2 out of 4 cases.
- Bb (heterozygous brown) is also produced in 2 out of 4 cases.
- Total brown rabbits (BB + Bb) = 4.
Thus, the probability that the baby rabbits will be brown is \( \frac{4}{4} = 100% \).
So, the answer is D: 100%.
If both parents were Bb (heterozygous), the Punnett square would give a different outcome:
- Bb x Bb:
- Possible gametes from both parents (Bb): B, b
The Punnett square would look like this:
B | b
----------------
B | BB | Bb
----------------
b | Bb | bb
From this Punnett square:
- BB = 1 (homozygous brown),
- Bb = 2 (heterozygous brown),
- bb = 1 (white rabbit).
So in this case, brown rabbits would be produced in 3 out of 4 cases.
Thus, if both parents were Bb, the probability that the baby rabbits will be brown is \( \frac{3}{4} = 75% \).
To summarize:
- BB x Bb = 100% brown offspring
- Bb x Bb = 75% brown offspring
Without knowing the exact genotypes of the two brown rabbits in your question, we can only conclude based on the scenarios. If one of them is homozygous BB, the answer is D. If both are heterozygous Bb, the answer would be C.
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