If a father and mother are planning a child, but the father has an autosomal dominant disease, and the mother does not have it, then what is the chance the child does? We do not know the fathers genotype; we know the mothers (aa).

So I broke it down into:

Father could have AA or Aa: $50\%$ chance of each. If its Aa, child has $50\%$ chance of getting it. But if its AA, child has $100\%$ chance. Considering this, what is the chance the child has the disease?

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    $\begingroup$ What all have you tried? $\endgroup$ – Aaron Zolotor Oct 23 '18 at 3:25
  • $\begingroup$ I made punnet squares to find the probability of the father's different genotypes and how they would affect the child's probability of having the disease $\endgroup$ – IntelliJ_Please Oct 23 '18 at 3:26
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    $\begingroup$ @Ryhthmink not a PSQ. $\endgroup$ – The Long Night Oct 23 '18 at 3:28
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    $\begingroup$ PSQ? Punnet square question? $\endgroup$ – IntelliJ_Please Oct 23 '18 at 3:28
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    $\begingroup$ Why do you suppose that the father is equally likely to be double-dominant as single-dominant? $\endgroup$ – Graham Kemp Oct 23 '18 at 3:48

As you have already figured out, since the father has a disease expressed by a dominant gene, there are 2 possible genotypes AA and Aa (A=Dominant Diseased, a=recessive normal).

Now, since the mother is healthy (aa genotype), the transmission probability is -

$$P(Transmission/AA) = 1$$ (Diseased gene shall be passed by father)


$$ P(Transmission/Aa) = 0.5$$ (Father passes diseased and healthy genes with equal probability)

At this stage, we use Bayes' Law to get- (See update 2 for alternative method)

$$P( Transmission ) = P(AA).P(Transmission/AA) + P(Aa).P(Transmission/Aa)$$

Now, in the absence of any evidence to the contrary, it is reasonable to assume that the father can have Aa and AA genotypes with equal probability (See update 1), so we have-

$P(AA) = P(Aa) = 1/2$

$P(Transmission) = 1/2.(1)+1/2.(0.5) = 0.75$.

Thus, the transmission probability is 75%.

Updates -

  1. Regarding the father's genotype-

In light of @Gerry Myerson's comments, there is need for some clarification, as the assumption of equal probability for AA and Aa seems shaky. There are two aspects to this issue-

  1. Mathematical -

    From the mathematical point of view this is simply a question of the relative probabilities of the father being of AA and Aa genotypes ( or equivalently, the statistical weights of these two genotypes while taking the weighted average of the transmission probabilities to get the effective probability, (see update 2 ). This one seems to boil down to a question of context - suffice to say, if this was actually a mathematical problem posed without extra data, the above solution giving 75% probability would stand. If on the other hand side data is given, it would be used to calculate the probability.

  2. Context or The biological problem -

    As Gerry said, this trait would be selected against (perhaps deleterious), and as such, we can assume that $P(AA) \lt P(Aa)$. To me the relation, $P(AA) = ( P(Aa) )^2$, which gives $P(Aa) = \frac {\sqrt{5}-1}{2}$, $P(AA)=\frac {3-\sqrt{5}}{2}$ and $P(Transmission) = \frac {5-\sqrt{5}}{4}$ seems particularly promising.

    Of course, in reality, we would expect $P(AA)$ to be some function of $P(Aa)$ and of the chances of reproduction between and survival of people having various genotypes. Even in this case, empirical data would be essential for getting the required probabilities.

    In fact, empirical data seems to be the principal missing requirement to getting a definitive answer here. I would say that just in case this an actual practical problem, you should look for some data giving the genotypical distribution for this phenotype. Of course, any analysis that is more detailed than this belongs to the field of biology and not mathematics.

    1. Alternative method-

An alternative (though equivalent) approach can be to interpret this as a statistical problem and thus use the formula -

$$E(Transmission) =\Sigma w_i \cdot E_i$$

where, $E(Transmission)$ isbge statistical average expectation of transmission and $w_i$ is the statistical weight corresponding to each expectation (expected chance if transmission for each genotype). Ckrarky, each $w_i$ shall equal the corresponding probability from the first approach, so that both approaches yield the same answer, as expected.


The answer is either $0.75$ or $\frac {5-\sqrt{5}}{4}=0.691$(approx) or the given data is insufficient.

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    $\begingroup$ Disease-causing genes tend to be rare in the population, so I'd think that the father would be more likely to have $Aa$ than $AA$. $\endgroup$ – Gerry Myerson Oct 23 '18 at 5:59
  • $\begingroup$ $\LaTeX \text{ Tip}:$ Use \text{ bleh bleh bleh} to obtain $\text { bleh bleh bleh}$. Not necessary but it does make posts neater. $\endgroup$ – Mohammad Zuhair Khan Oct 27 '18 at 3:38

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