Pedigree Probability of Autosomal Recessive Trait

Question

Here is a pedigree:

The trait is autosomal recessive.

The question is: What is the probability that the bottom 2 people (4 and 5) have a child with the trait?

I tried doing 2/3 * 2/3 * 2/3 *1/4 and got 2/27 but this is wrong. I thought that the probability of III 1 being a carrier is 2/3. The probability that IV 4 is a carrier is also 2/3. The probability that III 6 is a carrier is 2/3 and IV 5 must be a carrier. So IV 4 and IV 5 must be carriers. The probability that they have a child with the trait is 1/4. But I am not getting the right answer.

The right answer is 1/6.

Answer 1

Starting with the left hand side of the diagram:

• III:2 is definitely a carrier (Tt) as one parent (II:2) is affected (tt).
• III:1 is also definitely a carrier (Tt) as when mating with III:2 they produce an affected (tt) offspring (IV:1)
• This means that we can work out the possibilities for IV:4 as we know the parent genotypes. It follows the standard arrangement for two carrier parents giving the options of:
  • TT (1/4)
  • Tt (2/4 = 1/2)
  • tt (Normally 1/4 but in this case 0 as individual not marked as affected).
• Therefore for this scenario, the probabilities for IV:4 are :
  • TT (1/3)
  • Tt (2/3)

Now if we look at the right hand side of the diagram.

• IV:5 is definitely a carrier (Tt) as one of their parents (III:5) is affected.

This gives two possible Punnett squares to be examined:

|-------------------------------------------------------------------------------|
|                                        ♂ (IV:5)                               |
|                          T                                   t                |
|            -------------------------------------------------------------------|
|           |                                |                                  |
|         T |              TT                |                Tt                |
|           |                                |                                  |
| (IV:4)    |-------------------------------------------------------------------|
|    ♀      |                                |                                  |
|         T |              TT                |                Tt                |
|           |                                |                                  |
|-----------|-------------------------------------------------------------------|

This gives nil affected offspring so we can disregard this option for your question (as we are ONLY looking for scenarios which produce affected individuals).

Therefore the alternative is:

|-------------------------------------------------------------------------------|
|                                        ♂ (IV:5)                               |
|                          T                                   t                |
|            -------------------------------------------------------------------|
|           |                                |                                  |
|         T |              TT                |                Tt                |
|           |                                |                                  |
| (IV:4)    |-------------------------------------------------------------------|
|    ♀      |                                |                                  |
|         t |              TT                |                tt                |
|           |                                |                                  |
|-----------|-------------------------------------------------------------------|

Giving 1/4 affected offspring.

As mentioned above, in order to have affected offspring then IV:4 must be Tt. There is a 2/3 chance of this being the case. If this is the case, then there is a 1/4 chance of the child being tt.

Both conditions need to be true for this to happen so we multiply the fractions:

2/3 * 1/4 = 1/6