It may come as a surprise that our genes and probabilities have some things in common. Due to the random nature of cell meiosis, some aspects to the study of genetics is really applied probability. We will see how to calculate the probabilities associated with dihybrid crosses.

## Definitions and Assumptions

Before we calculate any probabilities, we will define the terms that we use and state the assumptions that we will work with.

- Alleles are genes that come in pairs, one from each parent. The combination of this pair of alleles determines the trait that is exhibited by an offspring.
- The pair of alleles is the genotype of an offspring. The trait exhibited is the offspring's phenotype.
- Alleles will be considered as either dominant or recessive. We will assume that in order for an offspring to display a recessive trait, there must be two copies of the recessive allele. A dominant trait may occur for one or two dominant alleles. Recessive alleles will be denoted by a lower case letter and dominant by an upper case letter.
- An individual with two alleles of the same kind (dominant or recessive) is said to be homozygous. So both DD and dd are homozygous.
- An individual with one dominant and one recessive allele is said to be heterozygous. So Dd is heterozygous.
- In our dihybrid crosses, we will assume that the alleles we are considering are inherited independently of one another.
- In all examples, both parents are heterozygous for all of the genes being considered.

## Monohybrid Cross

Before determining the probabilities for a dihybrid cross, we need to know the probabilities for a monohybrid cross. Suppose that two parents who are heterozygous for a trait produce an offspring. The father has a probability of 50% of passing on either of his two alleles. In the same way, the mother has a probability of 50% of passing on either of her two alleles.

We can use a table called a Punnett square to calculate the probabilities, or we can simply think through the possibilities. Each parent has a genotype Dd, in which each allele is equally likely to be passed down to an offspring. So there is a probability of 50% that a parent contributes the dominant allele D and a 50% probability that the recessive allele d is contributed. The possibilities are summarized:

- There is a 50% x 50% = 25% probability that both of the offspring's alleles are dominant.
- There is a 50% x 50% = 25% probability that both of the offspring's alleles are recessive.
- There is a 50% x 50% + 50% x 50% = 25% + 25% = 50% probability that the offspring is heterozygous.

So for parents who both have genotype Dd, there is a 25% probability that their offspring is DD, a 25% probability that the offspring is dd, and a 50% probability that the offspring is Dd. These probabilities will be important in what follows.

## Dihybrid Crosses and Genotypes

We now consider a dihybrid cross. This time there are two sets of alleles for parents to pass on to their offspring. We will denote these by A and a for the dominant and recessive allele for the first set, and B and b for the dominant and recessive allele of the second set.

Both parents are heterozygous and so they have the genotype of AaBb. Since they both have dominant genes, they will have phenotypes consisting of the dominant traits. As we have said previously, we are only considering pairs of alleles that are not linked to one another, and are inherited independently.

This independence allows us to use the multiplication rule in probability. We can consider each pair of alleles separately from each other. Using the probabilities from the monohybrid cross we see:

- There is a 50% probability that the offspring has Aa in its genotype.
- There is a 25% probability that the offspring has AA in its genotype.
- There is a 25% probability that the offspring has aa in its genotype.
- There is a 50% probability that the offspring has Bb in its genotype.
- There is a 25% probability that the offspring has BB in its genotype.
- There is a 25% probability that the offspring has bb in its genotype.

The first three genotypes are independent of the last three in the above list. So we multiply 3 x 3 = 9 and see that there are these many possible ways to combine the first three with the last three. This is the same ideas as using a tree diagram to calculate the possible ways to combine these items.

For example, since Aa has probability 50% and Bb has a probability of 50%, there is a 50% x 50% = 25% probability that the offspring has a genotype of AaBb. The list below is a complete description of the genotypes that are possible, along with their probabilities.

- The genotype of AaBb has probability 50% x 50% = 25% of occurring.
- The genotype of AaBB has probability 50% x 25% = 12.5% of occurring.
- The genotype of Aabb has probability 50% x 25% = 12.5% of occurring.
- The genotype of AABb has probability 25% x 50% = 12.5% of occurring.
- The genotype of AABB has probability 25% x 25% = 6.25% of occurring.
- The genotype of AAbb has probability 25% x 25% = 6.25% of occurring.
- The genotype of aaBb has probability 25% x 50% = 12.5% of occurring.
- The genotype of aaBB has probability 25% x 25% = 6.25% of occurring.
- The genotype of aabb has probability 25% x 25% = 6.25% of occurring.

## Dihybrid Crosses and Phenotypes

Some of these genotypes will produce the same phenotypes. For example, the genotypes of AaBb, AaBB, AABb, and AABB are all different from each other, yet will all produce the same phenotype. Any individuals with any of these genotypes will exhibit dominant traits for both traits under consideration.

We may then add the probabilities of each of these outcomes together: 25% + 12.5% + 12.5% + 6.25% = 56.25%. This is the probability that both traits are the dominant ones.

In a similar way we could look at the probability that both traits are recessive. The only way for this to occur is to have the genotype aabb. This has a probability of 6.25% of occurring.

We now consider the probability that the offspring exhibits a dominant trait for A and a recessive trait for B. This can occur with genotypes of Aabb and AAbb. We add the probabilities for these genotypes together and have18.75%.

Next, we look at the probability that the offspring has a recessive trait for A and a dominant trait for B. The genotypes are aaBB and aaBb. We add the probabilities for these genotypes together and have a probability of 18.75%. Alternately we could have argued that this scenario is symmetric to the early one with a dominant A trait and a recessive B trait. Hence the probability for this outcomes should be identical.

## Dihybrid Crosses and Ratios

Another way to look at these outcomes is to calculate the ratios that each phenotype occurs. We saw the following probabilities:

- 56.25% of both dominant traits
- 18.75% of exactly one dominant trait
- 6.25% of both recessive traits.

Instead of looking at these probabilities, we can consider their respective ratios. Divide each by 6.25% and we have the ratios 9:3:1. When we consider that there are two different traits under consideration, the actual ratios are 9:3:3:1.

What this means is that if we know that we have two heterozygous parents, if the offspring occur with phenotypes that have ratios deviating from 9:3:3:1, then the two traits we are considering do not work according to classical Mendelian inheritance. Instead, we would need to consider a different model of heredity.