Consider three independently assorting gene pairs, A/a, B/b, and C/c, where each demonstrates a typical dominance (A-, B-, C-) and recessiveness (aa, bb, cc) relationship. What is the probability of obtaining an offspring that is triply dominant from parents that are AaBbcc and AaBbCc

Assuming complete dominance and following the product probability rule, the probability of obtaining an offspring that is triply dominant AABBCC is 1/32.

What is complete dominance?

We talk about complete dominance when the dominant allele completely masks the recessive allele. It is an inheritance pattern.

This is the case of individuals that are heter0zyg0us for a particular gene and express the dominant phenotype. The dominant allele is hiding the expression of the recessive allele.

Many genes show complete dominance.

In the exposed example, we need to cross AaBbcc x AaBbCc and calculate the probability of getting offspring that is triply dominant. To do it, we will follow the product probability rule.

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)

Since these 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 we are interested in (AABBCC).

Gene A

Parentals) Aa x Aa

Gametes) A a A a

Punnett square) A a

A AA Aa

a Aa aa

F1) 1/4 AA

1/2 Aa

1/4 aa

Gene B

Parentals) Bb x Bb

Gametes) B b B b

Punnett square) B b

B BB Bb

b Bb bb

F1) 1/4 BB

1/2 Bb

1/4 bb

Gene C

Parentals) cc x Cc

Gametes) c c C c

Punnett square) C c

c Cc cc

F1) 1/2 Cc

1/2 cc

The probability of getting and individual with genotype AABBCC is

P(AABBCC) = P(AA) x P(BB) x P(CC)

P(AABBCC) = 1/4 x 1/4 x 1/2

P(AABBCC) = 1/32

The probability of obtaining an offspring that is triply dominant AABBCC is 1/32.

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