Some rabbits have a genetic condition. The dominant allele codes for this condition. A homozygous dominant rabbit mated with a rabbit that did not have the condition. They had three offspring. One of the offspring then mated with a rabbit that did not have the condition and they also produced three offspring. Two of the offspring had the condition and one did not. One body cell that is in early interphase is taken from each of the rabbits in these three generations. What is the total number of copies of the allele for the condition in this collection of cells.

Answer:

9) B

Please remember: the DOMINANT allele codes for the condition.
Let dominant allele = C
Let recessive allele = c
If a rabbit does not have a condition, they must be homozygous recessive = cc
If a rabbit is heterozygous, they will have the condition since the dominant allele causes the condition.
1st Generation:
Homozygous dominant rabbit = CC
Rabbit without condition = cc
After drawing a punnett square, we realise that all offspring would be heterozygous = Cc
2nd Generation:
First rabbit = Cc (from above offspring)
Second rabbit without condition = cc
After drawing a Punnett square, we realise 50% would be Cc and 50% would be cc Two of the offspring had the condition (Cc)
First homozygous dominant rabbit has 2 copies of allele.
In the 2nd generation, 3 heterozygotes, each with 1 copy of the allele are born. In the 3rd generation, 2 heterozygotes, each with 1 copy of the allele are born.
Total copies = 2+3+3=7