If two organisms, both with the genotype AaBb, are mated,what is the probability of obtaining the genotypes AABB andAaBb in the F2 generation?

To solve this problem, we need to use the principles of Mendelian genetics.

Step 1: Determine the possible gametes for each parent.

For a parent with the genotype AaBb, the possible gametes are AB, Ab, aB, and ab.

Step 2: Create a Punnett square.

A Punnett square is a diagram that is used to predict the genotypes of a particular cross or breeding experiment. In this case, we will create a 4x4 Punnett square because each parent can produce four different types of gametes.

Step 3: Fill in the Punnett square.

Each cell in the Punnett square represents a possible genotype for the offspring. Fill in each cell by combining the alleles from the corresponding row and column.

Step 4: Count the number of each genotype.

Once the Punnett square is filled in, count the number of each genotype.

Step 5: Calculate the probability.

The probability of a particular genotype is the number of cells with that genotype divided by the total number of cells.

For the genotype AABB, there is only one cell with this genotype out of 16 total cells, so the probability is 1/16 or 0.0625.

For the genotype AaBb, there are four cells with this genotype out of 16 total cells, so the probability is 4/16 or 0.25.

So, the probability of obtaining the genotypes AABB and AaBb in the F2 generation is 0.0625 and 0.25, respectively.