# 4.5 Calculating Probabilities Using Pedigree Charts

Once the mode of inheritance of a disease or trait is identified, some inferences about the genotype of individuals in a pedigree can be made, based on their phenotypes and where they appear in the family tree. Given these genotypes, it is possible to calculate the probability of a particular genotype being inherited in subsequent generations. This can be useful in genetic counselling, for example when prospective parents wish to know the likelihood of their offspring inheriting a disease for which they have a family history.

Probabilities in pedigrees are calculated using knowledge of Mendelian inheritance and the same basic methods as are used in other fields.

The first formula is the **product rule**: the joint probability of two independent events is the product of their individual probabilities; this is the probability of one event AND another event occurring.

For example:

The probability of a rolling a “five” with a single throw of a single six-sided die is 1/6, and the probability of rolling “five” in each of three successive rolls is 1/6 x 1/6 x 1/6 = 1/216.

The second useful formula is the **sum rule**, which states that the combined probability of two independent events is the sum of their individual probabilities. This is the probability of one event OR another event occurring.

For example:

The probability of rolling a five or six in a single throw of a dice is 1/6 + 1/6 = 1/3.

With these rules in mind, we can calculate the probability that two carriers (i.e. heterozygotes) of an AR disease will have a child affected with the disease as ½ x ½ = ¼, since for each parent, the probability of any gametes carrying the disease allele is ½. This is consistent with what we already know from calculating probabilities using a Punnett square (e.g. in a monohybrid cross *Aa* x *Aa*, ¼ of the offspring are *aa*).

We can likewise calculate probabilities in the more complex pedigree shown in **Figure 4.5.1**.

Assuming the disease has an AR pattern of inheritance, what is the probability that individual #14 will be affected? We can assume that individuals #1, #2, #3 and #4 are heterozygotes (*Aa*), because they each had at least one affected (*aa*) child, but they are not affected themselves. This means that there is a 2/3 chance that individual #6 is also *Aa*. This is because according to Mendelian inheritance, when two heterozygotes mate, there is a 1:2:1 distribution of genotypes *AA*:*Aa*:*aa*. However, because #6 is unaffected, he can’t be *aa*, so he is either *Aa *or *AA*, but the probability of him being *Aa *is twice as likely as *AA*. By the same reasoning, there is likewise a 2/3 chance that #9 is a heterozygous carrier of the disease allele.

If individual #6 is a heterozygous for the disease allele, then there is a ½ chance that #12 will also be a heterozygote (i.e., if the mating of #6 and #7 is *Aa *× *AA*, half of the progeny will be *Aa*; we are also assuming that #7, who is unrelated, does not carry any disease alleles). Therefore, the combined probability that #12 is also a heterozygote is 2/3 x 1/2 = 1/3. This reasoning also applies to individual #13, i.e., there is a 1/3 probability that he is a heterozygote for the disease. Thus, the overall probability that both individual #12 and #13 are heterozygous, and that a particular offspring of theirs will be homozygous for the disease alleles is 1/3 x 1/3 x 1/4 = 1/36.

**Media Attribution**

**Figure 4.5.1**Original attributed to Unknown (2017), CC BY-NC 3.0, Open Genetics Lectures

**Reference**

Unknown. (2017). Figure 15. Individuals in this pedigree are labeled with…[digital image]. In Locke, J., Harrington, M., Canham, L. and Min Ku Kang (Eds.),* Open Genetics Lectures, Fall 2017* (Chapter 23, p. 7). Dataverse/ BCcampus. http://solr.bccampus.ca:8001/bcc/file/7a7b00f9-fb56-4c49-81a9-cfa3ad80e6d8/1/OpenGeneticsLectures_Fall2017.pdf