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### Course: Biology archive > Unit 16

Lesson 1: Mendelian genetics# Probabilities in genetics

The sum rule and product rule. Applying these rules to solve genetics problems involving many genes.

## Introduction

The Punnett square is a valuable tool, but it's not ideal for every genetics problem. For instance, suppose you were asked to calculate the frequency of the recessive class not for an $1024$ boxes. Probably not what you want to draw during an exam, or any other time, if you can help it!

*Aa*x*Aa*cross, not for an*AaBb*x*AaBb*cross, but for an*AaBbCcDdEe*x*AaBbCcDdEe*cross. If you wanted to solve that question using a Punnett square, you could do it – but you'd need to complete a Punnett square withThe five-gene problem above becomes less intimidating once you realize that a Punnett square is just a visual way of representing probability calculations. Although it’s a great tool when you’re working with one or two genes, it can become slow and cumbersome as the number goes up. At some point, it becomes quicker (and less error-prone) to simply do the probability calculations by themselves, without the visual representation of a clunky Punnett square. In all cases, the calculations and the square provide the same information, but by having both tools in your belt, you can be prepared to handle a wider range of problems in a more efficient way.

In this article, we’ll review some probability basics, including how to calculate the probability of two independent events both occurring (event X

*and*event Y) or the probability of either of two mutually exclusive events occurring (event X*or*event Y). We’ll then see how these calculations can be applied to genetics problems, and, in particular, how they can help you solve problems involving relatively large numbers of genes.## Probability basics

Probabilities are mathematical measures of likelihood. In other words, they’re a way of quantifying (giving a specific, numerical value to) how likely something is to happen. A probability of $1$ for an event means that it is guaranteed to happen, while a probability of $0$ for an event means that it is guaranteed not to happen. A simple example of probability is having a $1/2$ chance of getting heads when you flip a coin, as Sal explains in this intro to probability video.

Probabilities can be either empirical, meaning that they are calculated from real-life observations, or theoretical, meaning that they are predicted using a set of rules or assumptions.

- The
**empirical probability**of an event is calculated by counting the number of times that event occurs and dividing it by the total number of times that event could have occurred. For instance, if the event you were looking for was a wrinkled pea seed, and you saw it times out of the$\mathrm{1,850}$ total seeds you examined, the empirical probability of getting a wrinkled seed would be$\mathrm{7,324}$ , or very close to$\mathrm{1,850}/\mathrm{7,324}=0.253$ in$1$ seeds.$4$ - The
**theoretical probability**of an event is calculated based on information about the rules and circumstances that produce the event. It reflects the number of times an event is*expected*to occur relative to the number of times it could possibly occur. For instance, if you had a pea plant heterozygous for a seed shape gene (*Rr*) and let it self-fertilize, you could use the rules of probability and your knowledge of genetics to predict that out of every$1$ offspring would get two recessive alleles ($4$ *rr*) and appear wrinkled, corresponding to a ($0.25$ ) probability. We’ll talk more below about how to apply the rules of probability in this case.$1/4$

In general, the larger the number of data points that are used to calculate an empirical probability, such as shapes of individual pea seeds, the more closely it will approach the theoretical probability.

## The product rule

One probability rule that's very useful in genetics is the $1/6$ chance of getting a six. If you roll two dice at once, your chance of getting two sixes is: (probability of a six on die 1) x (probability of a six on die 2) = $(1/6)\cdot (1/6)=1/36$ .

**product rule**, which states that the probability of two (or more) independent events occurring together can be calculated by multiplying the individual probabilities of the events. For example, if you roll a six-sided die once, you have aIn general, you can think of the product rule as the “and” rule: if both event X

*and*event Y must happen in order for a certain outcome to occur, and if X and Y are independent of each other (don’t affect each other’s likelihood), then you can use the product rule to calculate the probability of the outcome by multiplying the probabilities of X and Y.We can use the product rule to predict frequencies of fertilization events. For instance, consider a cross between two heterozygous ($1/2$ chance of making an $(1/2)\cdot (1/2)=1/4$ .

*Aa*) individuals. What are the odds of getting an*aa*individual in the next generation? The only way to get an*aa*individual is if the mother contributes an*a*gamete and the father contributes an*a*gamete. Each parent has a*a*gamete. Thus, the chance of an*aa*offspring is: (probability of mother contributing*a*) x (probability of father contributing*a*) =This is the same result you’d get with a Punnett square, and actually the same logical process as well—something that took me years to realize! The only difference is that, in the Punnett square, we'd do the calculation visually: we'd represent the $1/2$ probability of an $1$ -square intersect of the column and row (out of the $4$ total squares of the table) represents the $1/4$ chance of getting an

*a*gamete from each parent as one out of two columns (for the father) and one out of two rows (for the mother). The*a*from both parents.## The sum rule of probability

In some genetics problems, you may need to calculate the probability that any one of several events will occur. In this case, you’ll need to apply another rule of probability, the sum rule. According to the

**sum rule**, the probability that any of several mutually exclusive events will occur is equal to the sum of the events’ individual probabilities.For example, if you roll a six-sided die, you have a $1/6$ chance of getting any given number, but you can only get one number per roll. You could never get both a one and a six at the same time; these outcomes are mutually exclusive. Thus, the chances of getting either a one $(1/6)+(1/6)=1/3$ .

*or*a six are: (probability of getting a 1) + (probability of getting a 6) =You can think of the sum rule as the “or” rule: if an outcome requires that either event X

*or*event Y occur, and if X and Y are mutually exclusive (if only one or the other can occur in a given case), then the probability of the outcome can be calculated by adding the probabilities of X and Y.As an example, let's use the sum rule to predict the fraction of offspring from an

*Aa*x*Aa*cross that will have the dominant phenotype (*AA*or*Aa*genotype). In this cross, there are three events that can lead to a dominant phenotype:- Two
*A*gametes meet (giving*AA*genotype),*or* *A*gamete from Mom meets*a*gamete from Dad (giving*Aa*genotype),*or**a*gamete from Mom meets*A*gamete from Dad (giving*Aa*genotype)

In any one fertilization event, only one of these three possibilities can occur (they are mutually exclusive).

Since this is an “or” situation where the events are mutually exclusive, we can apply the sum rule. Using the product rule as we did above, we can find that each individual event has a probability of $1/4$ . So, the probability of offspring with a dominant phenotype is: (probability of $(1/4)+(1/4)+(1/4)=3/4$ .

*A*from Mom and*A*from Dad) + (probability of*A*from Mom and*a*from Dad) + (probability of*a*from Mom and*A*from Dad) =Once again, this is the same result we’d get with a Punnett square. One out of the four boxes of the Punnett square holds the dominant homozygote, $1$ out of the $4$ boxes in the whole Punnett square, and since the boxes don't overlap (they’re mutually exclusive), we can add them up ($1/4+1/4+1/4=3/4$ ) to get the probability of offspring with the dominant phenotype.

*AA*. Two more boxes represent heterozygotes, one with a maternal*A*and a paternal*a*, the other with the opposite combination. Each box is### The product rule and the sum rule

Product rule | Sum rule |
---|---|

For independent events X and Y, the probability (and Y) is | For mutually exclusive events X and Y, the probability (or Y) is |

## Applying probability rules to dihybrid crosses

Direct calculation of probabilities doesn’t have much advantage over Punnett squares for single-gene inheritance scenarios. (In fact, if you prefer to learn visually, you may find direct calculation trickier rather than easier.) Where probabilities shine, though, is when you’re looking at the behavior of two, or even more, genes.

For instance, let’s imagine that we breed two dogs with the genotype

*BbCc*, where dominant allele*B*specifies black coat color (versus*b*, yellow coat color) and dominant allele*C*specifies straight fur (versus*c*, curly fur). Assuming that the two genes assort independently and are not sex-linked, how can we predict the number of*BbCc*puppies among the offspring?One approach is to draw a $16$ -square Punnett square. For a cross involving two genes, a Punnett square is still a good strategy. Alternatively, we can use a shortcut technique involving four-square Punnett squares and a little application of the product rule. In this technique, we break the overall question down into two smaller questions, each relating to a different genetic event:

- What’s the probability of getting a
*Bb*genotype? - What’s the probability of getting an
*Cc*genotype?

In order for a puppy to have a

*BbCc*genotype, both of these events must take place: the puppy must receive*Bb*alleles,*and*it must receive*Cc*alleles. The two events are independent because the genes assort independently (don't affect one another's inheritance). So, once we calculate the probability of each genetic event, we can multiply these probabilities using the product rule to get the probability of the genotype of interest (*BbCc*).To calculate the probability of getting a $4$ -square Punnett square using the parents' alleles for the coat color gene only, as shown above. Using the Punnett square, you can see that the probability of the $1/2$ . (Alternatively, we could have calculated the probability of $1/2$ . To get the overall probability of the $1/4$ .

*Bb*genotype, we can draw a*Bb*genotype is*Bb*using the product rule for gamete contributions from the two parents and the sum rule for the two gamete combinations that give*Bb*.) Using a similar Punnett square for the parents' fur texture alleles, the probability of getting an*Cc*genotype is also*BbCc*genotype, we can simply multiply the two probabilities, giving an overall probability ofYou can also use this technique to predict phenotype frequencies. Give it a try in the practice question below!

## Beyond dihybrid crosses

The probability method is most powerful (and helpful) in cases involving a large number of genes.

For instance, imagine a cross between two individuals with various alleles of four unlinked genes:

*AaBbCCdd*x*AabbCcDd*. Suppose you wanted to figure out the probability of getting offspring with the dominant phenotype for all four traits. Fortunately, you can apply the exact same logic as in the case of the dihybrid crosses above. To have the dominant phenotype for all four traits, and organism must have: one or more copies of the dominant allele*A**and*one or more copies of dominant allele*B**and*one or more copies of the dominant allele*C**and*one or more copies of the dominant allele*D*.Since the genes are unlinked, these are four independent events, so we can calculate a probability for each and then multiply the probabilities to get the probability of the overall outcome.

- The probability of getting one or more copies of the dominant
*A*allele is . (Draw a Punnett square for$3/4$ *Aa*x*Aa*to confirm for yourself that out of the$3$ squares are either$4$ *AA*or*Aa*.) - The probability of getting one or more copies of the dominant
*B*allele is . (Draw a Punnett square for$1/2$ *Bb*x*bb*: you’ll find that half the offspring are*Bb*, and the other half*bb*.) - The probability of getting one or more copies of the dominant
*C*allele is . (If one of the parents is homozygous$1$ *CC*, there’s no way to get offspring without a*C*allele!) - The probability of getting one or more copies of the dominant
*D*allele is , as for$1/2$ *B*. (Half the offspring will be*Dd*, and the other half will be*dd*.)

To get the overall probability of offspring with the dominant phenotype for all four genes, we can multiply the probabilities of the four independent events: $(3/4)\cdot (1/2)\cdot (1)\cdot (1/2)=3/16$ .

### Check your understanding

## Want to join the conversation?

- What is the probability of having 5 boys in a roll(11 votes)
- Probability for sex ratio is always 1/2, but since we are speaking of having 5 boys in a row, probability changes.

1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32(33 votes)

- Why do possible outcomes change? Why can you not guarantee the total outcome in a punnett square?(4 votes)
- There's still chance involved. It's like if you tossed two coins. Each has a 50% chance of landing heads side up, but you can't say that there will definitely be one that lands heads up and one that lands tails up.

The same goes for genetic probabilities. In the article's example with the dogs' fur color, there might be a litter of eight puppies in which five puppies have the lighter color. Even though each puppy only has a 1/4 probability that it would have light fur, it's not like the genes are sorting themselves out and saying "Well we already have two puppies with light fur, so we have to stop allowing those genotypes to happen now." Each is an independent event.

Hope this helps!(9 votes)

- If monohybrid crossing is something like this Aa x Aa and dihybrid crossing is something like this AaBb x AaBb, then is this AaBbCc x AaBbCc called trihybrid crossing and this AaBbCcDd x AaBbCcDd called tetrahybrid crossing?

Or is there a specific word for crossings that involve more than one pair of alleles?

Perhaps polihybrid crossing?(4 votes)- Not really, but I agree. We need a word to group all these together.(3 votes)

- Why isn't there a review article for this important lesson?(5 votes)
- I'm still a little confused about the whole probability thing could somebody explain it in a more understandable way please?(3 votes)
- basically, you use the punnets squares to determine all the different possible combinations u have. u line them up and cross them to tell the combos. and then u look and see: how many times does, lets say, AA occur? u look and it occurs 2/4 times in the box, so u know there is a 50% chance of the cross producing an AA set. do u understand the whole punnett square thing, or...?(5 votes)

- If I am given 50% of AD and 30% of BC and 30% of AC and 10% of AB, (not precise percentages). How can I figure out the alleles of the "parents" that have ABCD combinations? Or work the math backwards?(4 votes)
- If both parents had ABCD alleles, then it would be equal percent for all combinations, which is not given by this task.

Also, I see no CD combination.

Since there is no CD, my projection is it is 0% that both parents have ABCD alleles.(3 votes)

- What? I feel like I need a bit more learning about this.(4 votes)
- I always had confusion with recessive traits and dominate traits. If you can simplify of for me. I am still confused(3 votes)
- recessive traits are traits that can only be seen when an organism has two recessive alleles for that trait. For example, Mendel's plants were only short when their alleles were tt. Dominant traits are when a dominant allele is one of the allels in a trait. Dominant traits "overrule" the recessive traits.(1 vote)

- What's an example of an "or" situation (either Event X or event Y) that is NOT mutually exclusive?(3 votes)
- If two events are not mutually exclusive, it would be the same as "and" situation.(2 votes)

- what is the reason for the 9/16 probability(2 votes)
- Because, 3/4*3/4=9/16. You get multiply these due to the probability rule of multiplication.(3 votes)