Genetics: pedigree following a rare trait autosomal recessive

I have two questions pertaining to this pedigree I believe it to be an autosomal recessive trait.

1. The probability that individuals IV-1 and IV-2 would give rise to an affected individual would be: A. 1/4 b. 1/2 c. 1/8 d. 1/12 e. 1/24

2. Suppose individual IV-2 was unaffected. The probability of his marriage giving rise to an affected child would be: A. 1/4 b. 1/2 c. 1/8 d. 1/12 e. 1/24

I worked on this problem for the last hour or so and figured up question 1 to be 1/2 and question two to be 1/4. I don't feel this is correct. I feel like I am missing something. If you don't mind showing the calculations for how you came about an answer that would be greatly appreciated.

Question 1

Okay, so I'll go through my own process for you here step by step, moving down the tree. Here's an annotated version of your diagram with my own thoughts (it's been a while since I've done this but should hopefully be accurate):

Generation 2: As you realised, the trait is autosomal recessive so the female (II:2) has the genotype aa. The male is considered wild type unless informed otherwise, giving him a genotype of AA. I think you had worked all this out, but they are shown in red on the diagram.

Generation 3: Using the two genotypes of the parents from II (red), we know that all the progeny in generation III are carriers - i.e. Aa genotypes. This is indicated for III:6 in orangey-brown. I think you also worked this out successfully. However in your original chart, III:7 is noted as Aa. III:7 is from outside of the affected family and would therefore again be considered to have be wild type and therefore instead have the genotype AA - shown in purple.

Generation 4: In order to work out the potential genotypes of IV:1 we have to do a cross between the two parents III:6 and III:7 - which works out as follows:

        |          Male          |
|     A            a     |
|------------------------|
F       |                        |
e     A |    AA            Aa    |
m       |                        |
a       |                        |
l     A |    AA            Aa    |
e       |                        |


As you can see, there is a 50% chance of IV:1 being healthy AA and a 50% chance of being a carrier Aa - shown in green. As IV:2 is affected, we know his genotype is aa (pink).

Generation V: Now for answer to your first question. If IV:1 had the AA genotype then the child could never be affected, reducing the odds to 50% immediately. Of this remaining 50%:

        |          Male          |
|     a            a     |
|------------------------|
F       |                        |
e     A |    Aa            Aa    |
m       |                        |
a       |                        |
l     a |    aa            aa    |
e       |                        |


50% of the offspring would be affected.

Therefore the probability of V:I being affected is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

Question 2

The second question seems very odd to me. If IV:2 was unaffected in the literal sense of the word they would be wild-type AA. This would make the probability of them having an affected child as 0.

As this isn't one of your options, I assume they mean that IV:2 is a carrier (Aa). We are this time therefore crossing either AA or Aa (IV:1) with Aa (IV:2). Again, if IV:1 is heterozygous dominant (AA) none of her children will be affected. This means the odds are again reduced by a half. In the case that she is heterozygous Aa then the cross becomes:

        |          Male          |
|     A            a     |
|------------------------|
F       |                        |
e     A |    AA            Aa    |
m       |                        |
a       |                        |
l     a |    Aa            aa    |
e       |                        |


Giving 25% healthy, 50% carrier and 25% affected.

As you're looking for affected children: $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$

• If "affected" means "aa", then "unaffected" should mean "not aa", i.e. "AA or Aa". If neither IV-1 not IV-2 are affected, the offspring could only be affected if both are carriers. The chance of IV-1 being a carrier is 50% whereas the chance of IV-2 being a carrier is 66% (because IV-2 had 1/4 chance to be AA, 1/2 chance Aa and 1/4 chance aa, but we know the genotype isn't aa, thus leaving 1/3 chance AA and 2/3 chance Aa). If they both are Aa, there is a 25% chance of offspring being affected (aa). 1/2 * 2/3 * 1/4 = 1/12 = (D). Edit: Haha excuse my necromancy, was prompted by a flag. – Armatus May 15 '14 at 19:15

Rory gave a good answer to part 1. Your main problem is that you have written Aa for III-7, and that's not the right call. III-7 is not in the family (maybe you missed that detail in the tree), so their odds of carrying a bad allele in the same gene that this family carries a broken gene in is very very small. So III-7 is AA.

So the III-6,7 mating is AA x Aa. That means the odds of their kid also being a carrier is 50/50, the odds of the kid passing on the bad allele to her kid is also 50/50, so the odds of V being affected are .25. (when you have a probability question in the form of "What are the odds of this, then that happening", you multiple the probabilities to get the answer)

In Part 2, if IV-2 is unaffected, then the odds are 2/3 that their genotype is Aa, and 1/3rd that their genotype is AA. (See the Punnett Square, and notice that you disregard the box that leads to an affected phenotype)

So for V to be affected, we need

1) III-6 to have passed on the bad allele to IV 1(50%)

2) IV-1 to have passed on the bad allele (50%)

3) IV-2 to be heterozygous (66.6%)

4) IV-2 to have passed on the bad allele (50%)

The odds of all of that happening is 1/12

From my own point of view, I think the answer to no. 1 is a 50% chance i.e. 2 out of 4. My reason is because, IV-1 could be a carrier of the trait and since the IV-2 is affected that means 2 will be affected. No 2. For the second question, the answer is 1 out of 4, because it's obvious both (IV-1 and IV-2) would be carriers of the trait from their family pedigree, and have 1 in a 4 chance of developing an offspring with the trait.

• This is rather a comment than an answer. Can you elaborate this? – Chris May 15 '14 at 5:38