The inheritance pattern of a common trait that shows complete penetrance is shown below:

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What is the kind of inheritance? If III-1 is a carrier, what is the probability of IV-1 (from III-1 and III-2 marriage) showing the trait? a) Y-linked; 0 b) Y-linked; 1/2 c) Autosomal; 1/8 d) Autosomal; 1/6

It cannot be Y-linked because then II-2 could not have transmitted it to her son. It can only be autosomal (recessive). Since III-1 is a carrier (heterozygous, Aa): i)If III-2 is also Aa, the kid will have a 1/4 probability of getting aa ii)If III-2 is AA, the kid will have zero probability of getting aa iii)If III-2 is aa(unlikely because it is not colored in the pedigree), the kid will have a 1/2 probability of getting aa.

I am unsure how 1/8 or 1/6 probability is calculated.


You have correctly identified that the trait is autosomal recessive.

Now for the probabilities. We know that:

  • II-1 is $Aa$ and
  • II-2 is $Aa$ (because they don't show the trait, but their son does),
  • III-1 is $Aa$ (because he is a carrier),
  • III-2 is either $Aa$ or $AA$ (because she doesn't show the trait).

IV-I will show the trait if it is recessive homozygote, $aa$. Because we are not sure about the genotype of III-2, we have two distinct scenarios:

  1. III-1 is $Aa$ and III-2 is $AA$
  2. III-2 is $Aa$ and III-2 is $Aa$

Let's first calculate the probability for each of the scenario. Here, we have to be careful not to fall into the trap by ignoring the conditional probability. The probability of III-2 being $Aa$ is not $1/2$ as we might wrongfully assume from simply drawing a Punnet square of their parents ($Aa \times Aa$). Because we know that III-2 doesn't show the trait, he cannot be $aa$. Therefore, we have to eliminate this possibility from the Punnet square, and we are left with the probabilities $1/3$ for her being $AA$ (scenario 1) and $2/3$ for being $Aa$ (scenario 2).

Scenario 1 (probability 1/3)

III-1 is $Aa$ and III-2 is $AA$.

Their child cannot show the trait.

Scenario 2 (probability 2/3)

III-1 is $Aa$ and III-2 is $Aa$.

Their child will show the trait with the probability $1/4$.


The probability of IV-1 showing the trait is now the product of probabilities for the scenario 2 to be true and the probability for their child to show the trait. Therefore,

$$\text{probability for IV-1 to be } aa = 2/3 \times 1/4 = 1/6.$$

As you see, not being careful about the conditional probability will lead you to the wrong result: $1/2\times1/4 = 1/8$.


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