Probabilities in Genetics

The product law is also known as the "AND" or "BOTH" rule. When you see phrases containing "and" or "both", you should use the product law. For example, "What is the probability of this AND that occurring?" or "What is the probability of expressing BOTH traits?".

Application of Probability in Genetics

One of the reasons these mathematical formulae are useful is the Punnett square. It is best used for examining crosses of two alleles at one or a maximum of two genes. If you use a Punnett square to examine this sort of cross: Aa x aa, you will get four boxes to examine the genotypes of offspring. In other words, you will have four squares for a cross at one gene with two possible alleles (Fig. 1).

Figure 1: Simple Punnett Square. Chisom, StudySmarter original

If you use a Punnett square to examine a dihybrid cross: AaBb x AaBb, for example, you'd require 16 boxes to examine the genotypes of offspring (Fig. 2). That is sixteen squares for a cross at two genes with two possible alleles each. Can you see the exponential increase in the number of squares, hence the meticulous work a Punnett square of that size would require? Let's not even imagine how many squares a three-gene cross would need!

Figure 2: Dihybrid cross Punnett square. Socratic.

Ultimately, beyond a certain point, Punnett squares are not a feasible option. Therefore, probability and simple mathematics are required. Even for simple, single-gene crosses, probability can be used to double-check our Punnett squares.

Examples of Probability in Genetics

Let's start with a simple cross of two homozygotes, or pure-breeding, plants. These plants follow the principles of Mendelian inheritance (this is an assumption to make in probability calculations unless stated otherwise), so there is complete dominance, independent assortment, and segregation of alleles.

The tall plants (TT) are dominant over the short plants (ss). Without doing a Punnett square, we know that each parent must have one of their two alleles in their gametes. Because both parents are homozygotes, they only have one type of allele to give. Thus we know that the tall plants must give T alleles and short plants must give s alleles, and the first filial generation has offspring with only one genotype: Ts (Fig. 3).

Now let us continue by crossing two F1 plants together. This kind of cross (F1 x F1) is a monohybrid cross because both plants are hybrid (heterozygous) for the same gene.

Without doing a Punnett square, we know that each parent can give either T or s alleles, and those two alleles can combine with the other parent's allele (Fig. 4).

Ts and sT are the same; thus, there are three possible genotypes: TT, Ts, ss. We know the proportions they occur in as well because, for this cross, we get the TT genotype once, the ss genotype once, and the Ts genotype twice.

So the probability of getting a homozygous tall plant: Pr (TT) = 1/4. The probability of getting a heterozygous tall plant: Pr (Tt) = 2/4 = 1/2. The probability of getting a homozygous short plant: Pr (tt) = 1/4.

We discovered these probabilities through reasoning (Fig. 5). Now, let's use the sum and product rule to answer further questions. It's important to analyze each part of each question before combining them for your final answer.

• What is the probability of having a tall plant in the F2 generation?
  • Tall plants can have TT or Ts genotypes. So we are looking for Pr (TT or Ts).
  • We remember that OR in probability signifies addition.
  • Therefore, according to the sum rule:
    • Pr (TT or Ts) = Pr (TT) + Pr (Ts) = $\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$

• What is the probability of having two tall offspring in the F2 generation?
  • Each offspring has a 3/4 chance to be tall.
    • We discovered this in the previous question.
  • But, we want two tall offspring this time.
  • Wwe need offspring #1 AND offspring #2 to be tall. This is Pr (TT or Ts) AND Pr (TT or Ts)
  • Therefore, according to the product rule:
    • Pr (tall) x Pr (tall) = $\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$

• What is the probability of having one short and one tall offspring in the F2 generation?
  • We know each offspring has Pr (tall) = 3/4.
  • Because there are only two possible phenotypes for this trait, to find the probability of short offspring, we simply do:
    • Pr (short) = 1 - Pr (tall) = $1-\frac{3}{4}=\frac{1}{4}$
  • We need to find Pr (short) AND Pr (tall), which means multiplication.
  • Therefore according to the product rule:
    • Pr (short) x Pr (tall) = $\frac{1}{4}\times \frac{3}{4}=\frac{3}{16}$

Now, let's look at a more challenging example—the dihybrid cross.

Let's use human genetics and traits for this example to demonstrate that this kind of probability analysis is possible in higher organisms. Many genes in human beings do not follow the principles of Mendelian genetics, but here are two that do. The alleles for freckles and for a widow's peak (that unique V-shaped hairline) are dominant at their respective gene loci. Therefore, the alleles for no freckles and no widow's peak (a more rounded or straight hairline) are recessive.

Q: What do dihybrids for widow's peaks and freckles look like?

A: They would both have freckles and widow's peaks because they are heterozygotes for those traits, which are dominant.

Q: What are the genotypes for these parents?

A: FfWw for both parents.

Q: Since both parents have the same genotype (for these two traits), they can make the same possible alleles in their gametes. What are these possibilities?

A: Each gamete must contain one allele of each gene; therefore, FW, Fw, fW, and fw are the possible gametes.

Q: What are the possible genetic outcomes for their offspring?

A: Let's use a Punnett square to examine this. For a dihybrid cross, Punnett squares are quite large and unwieldy (16 squares), but we will simplify this with probabilities later.

Figure 6: Human Dihybrid cross.

We can see the possible genotypes and thus the possible phenotypes (Fig. 6). As usual, with a dihybrid cross, the phenotypic ratio is 9:3:3:1, where 9/16 have dominant/dominant phenotypes, 3/16 have dominant/recessive phenotypes, 3/16 have recessive/dominant phenotypes, and 1/16 have recessive/recessive phenotypes.

But we know that, even without a Punnett square, we can determine genetic outcomes for offspring using our probability laws. Let's try this.

• What is the probability these dihybrids have two children with freckles and widow's peaks?
  • First, we need to know the probability of having a single child with freckles and a widow's peak.
    • This is Pr (FF or Ff) and Pr (WW or Ww)
      • From the monohybrid genotypic ratio, we know Pr (FF) = 1/4. Pr (Ff) = 1/2. Pr (WW) = 1/4. Pr (Ww) = 1/2.
    • According to the sum and product law, we get: Pr (FF + Ff) x Pr (WW + Ww)
    • This is $(\frac{1}{4}+\frac{1}{2})\times (\frac{1}{4}+\frac{1}{2})=\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$
      • Remember, this is the probability of having freckles and a widow's peak in a SINGLE child. Let's check the accuracy with the above Punnett square.
  • Now, let's find the probability of having two children both with these traits.
    • This Pr (freckles and widow's peak) AND Pr (freckles and widow's peak)
    • According to the product law, we get: Pr (9/16) x Pr (9/16)
      • This is $\frac{9}{16}\times \frac{9}{16}=\frac{81}{256}$
  • Even though these are two dominant traits, the odds of both occurring in two children of these same parents is less than 1/3!