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## AP®︎/College Biology

### Unit 5: Lesson 2

Mendelian genetics- Introduction to heredity
- Fertilization terminology: gametes, zygotes, haploid, diploid
- Alleles and genes
- Worked example: Punnett squares
- Mendel and his peas
- The law of segregation
- The law of independent assortment
- Probabilities in genetics
- Pedigrees
- Mendelian genetics

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# Probabilities in genetics

AP.BIO:

IST‑1 (EU)

, IST‑1.I (LO)

, IST‑1.I.2 (EK)

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

*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 with 1024 boxes. Probably not what you want to draw during an exam, or any other time, if you can help it!The 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, slash, 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 1, comma, 850 times out of the 7, comma, 324 total seeds you examined, the empirical probability of getting a wrinkled seed would be 1, comma, 850, slash, 7, comma, 324, equals, 0, point, 253, or very close to 1 in 4 seeds. - 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 1 out of every 4 offspring would get two recessive alleles (*rr*) and appear wrinkled, corresponding to a 0, point, 25 (1, slash, 4) probability. We’ll talk more below about how to apply the rules of probability in this case.

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

**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 a 1, slash, 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) = left parenthesis, 1, slash, 6, right parenthesis, dot, left parenthesis, 1, slash, 6, right parenthesis, equals, 1, slash, 36.In 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 (

*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 1, slash, 2 chance of making an*a*gamete. Thus, the chance of an*aa*offspring is: (probability of mother contributing*a*) x (probability of father contributing*a*) = left parenthesis, 1, slash, 2, right parenthesis, dot, left parenthesis, 1, slash, 2, right parenthesis, equals, 1, slash, 4.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, slash, 2 probability of 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 1-square intersect of the column and row (out of the 4 total squares of the table) represents the 1, slash, 4 chance of getting an*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, slash, 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

*or*a six are: (probability of getting a 1) + (probability of getting a 6) = left parenthesis, 1, slash, 6, right parenthesis, plus, left parenthesis, 1, slash, 6, right parenthesis, equals, 1, slash, 3.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, slash, 4. So, the probability of offspring with a dominant phenotype is: (probability of

*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) = left parenthesis, 1, slash, 4, right parenthesis, plus, left parenthesis, 1, slash, 4, right parenthesis, plus, left parenthesis, 1, slash, 4, right parenthesis, equals, 3, slash, 4.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,

*AA*. Two more boxes represent heterozygotes, one with a maternal*A*and a paternal*a*, the other with the opposite combination. Each box is 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, slash, 4, plus, 1, slash, 4, plus, 1, slash, 4, equals, 3, slash, 4) to get the probability of offspring with the dominant phenotype.### The product rule and the sum rule

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

For independent events X and Y, the probability (P) of them both occurring (X and Y) is P, left parenthesis, X, right parenthesis, dot, P, left parenthesis, Y, right parenthesis. | For mutually exclusive events X and Y, the probability (P) that one will occur (X or Y) is P, left parenthesis, X, right parenthesis, plus, P, left parenthesis, Y, right parenthesis. |

## 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

*Bb*genotype, we can draw 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*Bb*genotype is 1, slash, 2. (Alternatively, we could have calculated the probability of*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 1, slash, 2. To get the overall probability of the*BbCc*genotype, we can simply multiply the two probabilities, giving an overall probability of 1, slash, 4.You 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 3, slash, 4. (Draw a Punnett square for*Aa*x*Aa*to confirm for yourself that 3 out of the 4 squares are either*AA*or*Aa*.) - The probability of getting one or more copies of the dominant
*B*allele is 1, slash, 2. (Draw a Punnett square for*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 1. (If one of the parents is homozygous*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 1, slash, 2, as for*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: left parenthesis, 3, slash, 4, right parenthesis, dot, left parenthesis, 1, slash, 2, right parenthesis, dot, left parenthesis, 1, right parenthesis, dot, left parenthesis, 1, slash, 2, right parenthesis, equals, 3, slash, 16.

### Check your understanding

## Want to join the conversation?

- Why do possible outcomes change? Why can you not guarantee the total outcome in a punnett square?(2 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!(6 votes)

- 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)

- 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?(3 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)

- 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?(3 votes)- Not really, but I agree. We need a word to group all these together.(1 vote)

- what is the reason for the 9/16 probability(1 vote)
- Because, 3/4*3/4=9/16. You get multiply these due to the probability rule of multiplication.(3 votes)

- how do solve all recessive traits?(1 vote)
- It 0% since in every possibility of getting CC and Cc, there will be a dominant trait. If you multiply that by anything, the answer will still be 0%.

If you want to know for any case that does not come out to be 0%, then you would make a punnet square for each trait separately and multiply the probabilities. For example, if you have AaBb and aabb, you would make a punnet square for Aa and aa and another punnet square for Bb and bb. Then you calculate the probability of a recessive trait coming up and multiply them. In this case, there is a 2/4 chance for a recessive trait for the first one and a 2/4 chance for a recessive trait for the second one.

2/4 * 2/4 = 4/16 = 1/4

In this case, there is a 1/4 chance for all recessive traits.(3 votes)

- In the probability basics section when explaining empirical probability, why is it 1850/(7324+1850)=0.253 probability and not 1850/7324?(0 votes)
- Yep, totally right, this was a typo! I've corrected it, and the correction should be live on the site soon. Thanks!(2 votes)

- a couple john and mary both have normal colour vision .Elvis their blue eyed baby is colourblind .If john and mary have more children,what genotypic and phenonotypic ratio do you predict for this trait?(2 votes)
- You mean what likelihood of their child being colorblind?

Well, the trait for red-green color-blindness is recessive and sex-linked, meaning it is only on the X chromosome and not on the Y. Remember, girls are XX and boys are XY. So Elvis has the mutant form of this trait on his X. For this to happen, Mary must be a carrier, meaning one of her X's has the trait and one is normal.

John doesn't have color-blindness so he must have the normal trait on his X.

So, since their daughters can't get two copies of the mutant trait since the X they get from dad will be normal, they will never have a color blind daughter.

Then, 1/2 of their sons will be color-blind and 1/2 won't.

Calculating it out, the actual ratio of probability of children will be 1/2 normal girl, 1/4 normal boy, 1/4 color-blind boy.(3 votes)

- Hi , does anyone knows ,

In the offspring of a dihybrid self cross, what percentage of the individuals are themselves

dihybrid? Assume both loci are assorting independently.(1 vote)- From reading this article you already have enough information to answer that question. Since this could easily be a homework question (which are not allowed on Khan Academy) I will only give you hints.

What fraction of the offspring will be "hybrid" (heterozygous) for each locus?

How do we combine probabilities for independent events?

Does that help you work out the answer to your question?(1 vote)

- If mothers genotype data is GG and fathers genotype data is AA and child is GG. What is the probability that the father is not related to the child? I'm confused, in order to get a GG child wont the father have to have a "g" in its genotype?(1 vote)