Genetics: Dihybrid Cross and the Law of Independent Assortment
While performing monohybrid crosses and studying one trait at a time, Mendel observed that the expression of one trait seemed independent of other traits. For example, a plant could have inflated or constricted pod regardless of whether or not it had yellow or green seeds. The shape of pod did not seem to affect the expression or inheritance of the color or seed, or vise versa. These observations made him think that “factors” (alleles) from each pair separate independently of members of another pair during the formation of gametes (meiosis). To test if his hypothesis is correct, Mendel performed dihybrid crosses.
A dihybrid cross is so called because it is a cross between two parents that differ in two traits. For example, he crossed a purebred plant having inflated pod and yellow seeds with a purebred plant having constricted pod and green seeds. Because inflated pod and yellow seeds are dominant over constricted pod and green seeds respectively, all the F1 offspring had inflated pods and yellow seeds. Mendel allowed the F1 plants to self-pollinate. He observed that there were four different phenotypes in the F2 generation: inflated pods with yellow seeds, inflated pods with green seeds, constricted pods with yellow seeds, and constricted pods with green seeds. This made him conclude that the four factors (IiYy) in both parents segregated independently into the F1 gametes. This is known as The Law of Independent Assortment – during the formation of gametes (meiosis), factors (genes) for different traits segregate/assort independently from one another. Therefore, all possible combinations of factors can occur in the gametes.
Solving a Dihybrid Cross
There are two ways by which one can figure out the probable results of a given dihybrid cross. We can use a Punnet square, shown previously, or employ the multiplicative and additive rules of probability. Consider the sample problem below:
In horses, black coat is dominant to brown coat, while trotter is dominant to pacer. A horse, which is heterozygous for both traits, is mated with a brown pacer. Predict the genotypic and phenotypic ratios of the offspring.
Steps in Solving the Problem:
Step 1 : Prepare a key to the traits.
Step 2 : Write the genotype of each parent.
Step 3 : Determine the possible gametes from each parent.
Step 4 : Construct a Punnet square.
Hint: To determine the number of squares, multiply the number of types of gametes from each parent. Thus, 4 x 1 = 4. Draw 4 squares.
Step 5 : Determine the genotypic ratio and the phenotypic ratio.
Key: B = black coat
b = brown coat
T = trotter
t = pacer
B. Using the Multiplicative and Additive Laws of Probability.
1. Consider the given in the problem: P1 : BbTt x BbTt. Solve the two monohybrid crosses separately:
2. How do we solve for the GR and PR of dihybrid cross using the laws of probability? To get the genotypic ratio, the following steps are used:
a. In the separate monohybrid crosses, determine the chance of each gamete receiving an allele.
P1: Bb x bb P1: Tt x tt
G: ½ B ½ b 2/2 b G: ½ T ½ t 2/2 t
b. Using the multiplicative law, multiply the gametes in each monohybrid cross.
½ B ½ b ½ T ½ t
x 2/2 b x 2/2 t
2/4 Bb 2/4 bb 2/4 Tt 2/4 tt
c. Multiply the product of each monohybrid cross.
2/4 Bb 2/4 bb
x 2/4 Tt 2/4 tt
4/16 BbTt 4/16 bbTt 4/16 Bbtt 4/16 bbtt
d. Reduce the ratio into its lowest term. Thus, using the additive law, we can conclude that the genotypic ratio is 1 BbTt : 1 bbTt : 1 Bbtt : 1 bbtt
To solve for the phenotypic ratio, we know from the two separate monohybrid crosses that:
· The chance for black = ½
The chance for brown = ½
· The chance for trotter = ½
The chance for pacer = ½
Using the multiplicative law,
· The chance of black trotter = ½ x ½ = ¼
· The chance of black pacer = ½ x ½ = ¼
· The chance of brown trotter = ½ x ½ = ¼
· The chance of brown pacer = ½ x ½ = ¼
Using the additive law, we conclude that the phenotypic ratio is 1 : 1 : 1 : 1.