## Laws of Probability in Genetics

There are **two laws involving calculating probabilities and proportions in genetics**. The first is **the sum law,** and the second is **the product law**. Let's define them one by one before we use them to examine some problems and cases.

### Key Definitions of Probability in Genetics

The s**um law** (or sum rule) - To find the probability of two (or more) events occurring, as long as all the events are **mutually exclusive** (meaning that either one event can happen, or the other can happen, but not both), you must add the probabilities of each individual event occurring together.

The sum law is also known as the **"OR" rule**. If you see phrases containing the word "or", you should use the sum law. For example, "what is the probability of this OR that occurring?" or "what is the probability of expressing this OR that trait?".

The** product law** (or product rule) - To find the probability of two (or more) events occurring, as long as all the events are** independent of one another **(they can all happen at the same time), multiply the probabilities of all individual events occurring.

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).

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!

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**.

Please refer to the "Non-Mendelian Genetics" article for a refresher on these concepts.

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.

The two pure-breeding parents are called P generation, and when crossed, their descendants are F1. If the F1 x F1 cross is performed, their descendants are F2. If you cross F2 x F2, their descendants are F3, and so on.

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) = $\frac{3}{4}\times \frac{3}{4}=\frac{9}{16}$

- Each offspring has a 3/4 chance to be tall.

- 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**.

Dihybrids are organisms that are heterozygotes at two different alleles.

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.

What would happen if two people who are dihybrids for widow's peaks (Ww) and freckles(Ff) traits married and had children? What kind of alleles could they make? What is the possible genetic outcome for their offspring? Let's examine and answer such questions.

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.

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.

A dihybrid cross arises when you cross two organisms with genotypes like this: AaBb x AaBb, and gives the phenotypic ratio 9:3:3:1. A monohybrid cross arises when you cross two organisms with genotypes like this: Aa x Aa, and gives the genotypic

ratio 1:2:1.

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.

- This is Pr (FF or Ff) and Pr (WW or Ww)
- 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!

- First, we need to know the probability of having a single child with freckles and a widow's peak.

## Probabilities in Genetics - Key Takeaways

- There are
**two key laws of probability in genetics**:**the sum law**and**the product law**. - The sum law is also called the
**OR law**, meaning you use it to add two probabilities together when determining probabilities of one OR the other possibility. - The product law is also called the
**AND law**, meaning you use it to multiply probabilities with each other when determining the probabilities of one AND another possibility. **Punnett squares**are best used for single-gene analysis of traits that obey Mendelian genetics,- If more complex genetic assessments are required, probabilities and simple mathematics are necessary to help analyze offspring potential genotypes and phenotypes.

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##### Frequently Asked Questions about Probabilities in Genetics

What is probability in genetics?

Probability in genetics are simple mathematical analysis of proportions to help understand and postulate future inheritance patterns.

How is probability used in the study of genetics?

Probability is used in genetics to help analyze possible offspring genetic outcomes.

How can the principles of probability be used in genetics?

The principles of probability, namely the sum law and the product law, can help to calculate the likelihood of a genotype or phenotype in offspring of a cross.

How is probability important in genetics?

Probability is important in genetics because it helps us determine genotypic and phenotypic outcomes for future offspring.

How is probability used to predict traits?

Probability helps predict traits by making a mathematical calculation of the likelihood of a traits existence in a given offspring, via the sum law and the product law.

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