Question

pThe probability of having an offspring with both the dominant phenotypes for both genes is equal to the product of the individual probabilities. This means, the probability of having a dominant phenotype for both traits would be

Answer 1

The product probability rule states P(A∩B) = P(A) x P(B). In the exposed example, the probability of having a dominant phenotype is 9/16.

What is the product probability rule?

The product probability rule allows us to calculate the probability of occurrence of event A and event B at the same time.

It is about a joint probability of two or more events that might happen simultaneously, not excluding each other.

This rule is based on the dependence or independence of the events. If two events A and B are independent of each other, the occurrence of one of them does not affect the occurrence of the other one.

For independent events, the rule establishes that, given two different events A and B, the probability of occurrence of both together is

P(A∩B) = P(A) x P(B)

The probability of inheriting one of the alleles from each parent is 0.5 (beacause each parent has two alleles).

According to this, the probability of having a dominant phenotype for both traits depends on the paretal genotypes.

For instance, let us analyze the cross between two dihybrid individuals AaBb  x  AaBb.

Assuming the involved genes code for different traits and both are independent genes, for easier calculations, we can perform crosses that involve each of the genes separately -events-, and then multiply de results according to the genotype and phenotype we are interested in (A-B-)

Gene A

Parentals) Aa   x   Aa

Gametes) A   a     A   a

Punnett square)       A(1/2)      a(1/2)

                    A(1/2)    AA(1/4)     Aa(1/4)

                    a(1/2)     Aa(1/4)      aa(1/4)

F1) Genotype

      1/4 AA ⇒ 1/2A x 1/2A

      2/4 = 1/2 Aa ⇒ (1/2A x 1/2a) + (1/2A x 1/2a)

      1/4 aa ⇒ 1/2a x 1/2a

      Phenotype

      3/4 A- ⇒ 1/4AA + 2/4Aa

      1/2 aa

     

Gene B

Parentals) Bb   x   Bb

Gametes) B   b     B    b

Punnett square)   B      b

                   B       BB    Bb

                    b       Bb   bb

F1 Genotype

      1/4 BB ⇒ 1/2B x 1/2B

      2/4 = 1/2 Bb ⇒ (1/2B x 1/2b) + (1/2B x 1/2b)

      1/4 bb ⇒ 1/2b x 1/2b

      Phenotype

      3/4 B- ⇒ 1/4 BB + 2/4 Bb

      1/2 bb

The probability of having a dominant genotype and phenotype for both traits is,

• Genotpe AABB ⇒ 1/4 AA x 1/4 BB = 1/16 AABB

• Phenotype A-B- ⇒ 3/4 A-  x  3/4 B- = 9/16 A-B-

You can learn more about the product probability rule in crosses at

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