# Question Video: Calculating the Probability of a Child Inheriting a Disease Caused by a Dominant Allele Biology

Huntington’s disease is an inherited disease caused by a dominant allele (H). The Punnett square provided shows the genotypes of a male and a female and the predicted genotypes for their offspring. What is the probability, in percent, that a child born to these parents would inherit Huntington’s disease?

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### Video Transcript

Huntington’s disease is an inherited disease caused by a dominant allele, uppercase H. The Punnett square provided shows the genotypes of a male and a female and the predicted genotypes for their offspring. What is the probability, in percent, that a child born to these parents would inherit Huntington’s disease?

The question tells us that Huntington’s disease is caused by the presence of a dominant allele, indicated by an uppercase H. This means that if a person has at least one copy of this allele in their genotype, they will inherit the disease. Let’s have a look at the different genotypes shown in our Punnett square. We can see that the mother has one copy of this dominant allele, shown by an uppercase H, but the father only has recessive alleles, indicated by lowercase h. An offspring born to these parents will inherit one allele from each of their parents at random. We can see that this results in the offspring having one of four possible genotypes.

It’s important to note that the Punnett square shows the probability of each individual child inheriting these genotypes, not the outcome if this couple were to have four children. Since the allele from each parent is inherited completely at random, this means that each entry in the Punnett square is equally likely to occur. Of the four possible genotypes, two have the dominant uppercase H allele present. This means that two of the four genotypes would result in the inheritance of Huntington’s disease. The other two genotypes only have recessive lowercase h alleles present and so would not result in Huntington’s disease.

So, two out of the four genotypes result in Huntington’s disease, and we need to express this as a percent. We recognize that two is half of four, so this is 50 percent. If the numbers hadn’t been quite so easy to work with, we could’ve done a slightly longer calculation. That is, we could’ve divided two by four and then multiplied by 100 to get the value as a percent. So, our final answer is that the probability of a child born to these parents inheriting Huntington’s disease is 50 percent.